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Cell biology/Membrane Transport: Nucleocytoplasmic Exchange

Here is the link to the ITunes U Lecture from Berkeley. | Membrane Transport: Nucleocytoplasmic Exchange This is a challenging lecture because of the lack of class notes. Hopefully, I have simplified some of the concepts below.

Symport
In the last lecture, symport or using an existing concentration gradient to concentrate a needed substance against its concentration gradient. Remember from last lecture:
 * The sodium-potassium pump expelled 3 sodium ions and imported 2 potassium ions creating a charge on the membrane and establishing a gradient for sodium to pass into the cell.
 * The sodium-glucose symport used the existing sodium gradient (it wants inside the cell) to get glucose in the cell. Sodium could only enter the cell with a glucose escort. This increased the glucose concentration on the cytoplasmic side of the membrane.

Antiport
Antiport is very similar to symport but focuses on ridding the cell of ions.
 * Calcium-sodium antiporter-Three sodium ions are taken into the cell for each calcium ion expelled. This lowers the intracellular calcium concentration.
 * Sodium-Proton antiport-One sodium ion is allowed into the cell for each proton expelled. This is used to maintain pH.

Application

 * Digitalis (from the Foxglove plant) is a cardiac glycoside like quabain (although slightly less toxic). It works in the same mechanism, as quabain, by binding to the external surface of the alpha subunit of the sodium-potassium ATPase and blocks the hydrololysis of aspartyl-phosphate intermediate on the other side of the membrane. This stops ATPase from functioning. If you remember, the sodium-potassium pump is responsible for creating the sodium gradient that allowed the symporter and antiporters, just discussed, to function.
 * Digitalis helps with heart problems after a heart attack. After a heart attack there is a great deal of weakened tissue and digitalis strengthens the contraction of remaining heart muscle. At a low dose, digitalis diminishes the magnitude of the sodium gradient, which reduces the function of the calcium-sodium antiporter and the calcium level rises inside the cell. Calcium is important in muscle contraction (calcium binds to troponin-tropomyosin to be discussed in a later lecture). Since there is more calcium within the cell, the cell will contract more easily and more strongly.
 * Now there are drugs (called calcium channel blockers) that target the calcium-sodium antiporter which lowers the risk of the drug because it is more specific. In other words, it blocks the activity of this one type of channel rather than blocking a channel that allows many other cellular functions and transport mechanisms to work.

Bacterial Channels
These are much better understood because bacteria are easier to culture. E. coli has been extensively studied.

Cold Osmotic Shock Susceptible Transport Proteins

 * E. coli has a two part system to collect histidine from its environment. Where the histidine binds first to a J protein, which then transports the histidine to a membrane bound complex that transports the histidine into the cell.
 * First of all remember that E. coli has two membranes, an inner and outer membrane. In between these membranes is the periplasmic space (see diagram). The cell produces a great deal of J proteins and secretes them into the periplasmic space.
 * When histidine enters the periplasmic space, the J proteins bind to the histidine.
 * The J protein then delivers the histidine by binding to the MPQ complex. The P protein is the active part of this channel. It is an ABC protein (ATP-binding cassette transporter, like most active transporters).
 * This is a complex and unique type of channel because it requires a transporter to bind the desired molecule before attaching to the channel.
 * How did they figure out that they needed a soluble periplasmic protein, the J protein?
 * They ruptured the outer membrane with a technique like cold osmotic shock.
 * This process requires the bacteria to be cultured, centrifuged to create a pellet, extraction of the pellet, and resuspension in low osmotic strength solution (like water with small amounts of salt) over ice.
 * This creates large cracks in the membrane and the periplasm then flows out. When this solution is centrifuged, the J protein is found in the supernatant.
 * When the cells are returned to normal growth medium and regular temperature the membrane anneals back together (heals) and for a while the cell is unable to take up histidine.
 * After time passes, new J protein is produced and allows histidine to be picked up by the cell. Such transport systems are titled as sensitive to cold osmotic shock, which means they require a soluble periplasmic component.

Lactose Permease

 * This transport protein is not sensitive to cold osmotic shock.
 * The lac Y protein is a small hydrophobic protein that snakes through membrane many times and it is the integral protein for lactose permease.
 * If you treat intact cells with lysozyme (which breaks through the peptidoglycan barrier found in the periplasm, effectively removing the outer membrane), the cell looses its shape and become a spheroplast. (Which needs to be maintained with a high osmotic level buffer).
 * If you add C14 (radioactive) labeled lactose, and then you can measure the C14 lactose that is consumed by the cell. When you sediment the cells (centrifuge them to the bottom of a test tube), the C14 lactose will be a part of the cells. If they are not active, lactose will not be taken up. Studies have show that if the cells are metabolically active protons will be taken up with the lactose.
 * From this experiment we know that this permease functions as a symport carrier that takes up both a proton and lactose.
 * Another experiment can be performed with these spheroplasts, where you pop the spheroplasts with a low osmotic pressure solution (low salt) allowing the cytoplasm to spew out. Under specific conditions, the cell membrane can be resealed to form large cell ghosts (or large vesicles). If you add a low pH solution (high in protons), the cell membrane will take up lactose again.

"Group Translocation" Pathway

 * Another E. coli transport pathway that is used for glucose and other sugars.
 * Phophoenolpyruvate (PEP) is used as a phosphate (P) donor which is given to histidine on Enzyme 1.
 * Enzyme 1 transfers the phosphate to HPR. (HPR is a barometer of the cellular metabolism-if HPR is phosphorylated then the cell needs sugar).
 * HPR+P transfers the phosphate to Enzyme 3 ( peripheral membrane protein connected to a glucose channel).
 * Enzyme 3 adds the phosphate group to glucose so it cannot escape and glucose accumulates in the cell.
 * For other sugars like manose, the HPR is still used but Enzyme 3 is not used.

Nucleocytoplasmic Exchange
The nuclear envelope continues to form the endoplasmic reticulum. The nuclear envelope is distinct because it possesses Nuclear pores.
 * Nuclear pores are large complexes and span both membranes that create the nuclear envelope. 125,000,000 kilodaltons, 30 times the size of a ribosome, and contain 50-100 different proteins.
 * This is a permissive channel and small molecules and small proteins can flow through these pores.
 * Larger proteins must be actively transported in or out of the nucleus.
 * Ribosomes (created with RNA and proteins) require the RNA and proteins to be moved in and out of the nuclear pore multiple times.
 * Messenger RNA must be transported out of the nucleus, via the nuclear pore, to be translated into proteins.
 * The protein subunits travel to the nucleus, via the nuclear pore, for further assembly.
 * In the nucleolus, the ribosomal RNA and protein subunits are assembled to create a nearly complete ribosome which then travels back into the cytoplasm through the nuclear pore.

How do you study the nuclear pore and its proteins?

 * Nuclear proteins can be tagged with a florescent dye and microinjected into the cytoplasm. Over time it is possible to visualize the concentration of the flourescent dye within the nucleus.
 * Xenopus, a type of African toad, oocytes (eggs) are ideal for this because they are very large and their nucleus can be visualized with the naked eye.

''Figure of nucleoplasmin protein uptake through the nuclear pore. In the above diagram, nucleoplasmin is a large nuclear protein with distinct head and tail domains that can be separated by proteolysis. 1. The whole nucleoplasmin molecules rapidly accumulate in the nucleus when they are injected into the cytoplasm despite the fact that they are too large to passively diffuse into the nucleus.2. There is no nuclear uptake when only the heads are injected. 3. When the tails are injected they are taken up into the nucleus, because the tails have the nuclear import signal. 4. The final part of the diagram shows colloidal gold particles that can be imported to the nucleus with the attached tail particles.''

How big does a protein have to be before it can enter the nuclear pore?

 * Colloidal gold particles can be created in different sizes, from 5nm to 100nm.
 * These gold particles can be attached to nuclear proteins and then injected into the cytoplasm.
 * Smaller particles can find their way into the nucleus. 100nm particles cannot make their way in, while 50nm particles are lined up along the membrane half-way through the pore (visualized by an electron microscope).
 * So 50nm appears to be the limit for what can be transported through the pore.

What is special about nuclear proteins that allows them access into the nuclear pore?
Entry into the nucleus is provided by DNA region called a signal (the peptide it produces is actually called the signal but the DNA region might have the same designation). In fact, it is possible to take a cytoplasmic protein and add in the sequence of amino acids that the signal codes for and produce a protein that can travel to the nucleus through the nuclear pore.
 * The first study to prove this used SV40, a monkey tumor virus, which produces a T antigen (a protein necessary for replication of the viral nucleic acid). This T antigen must travel through the nuclear pore so it can replicate within in the nucleus (the only place it can replicate).
 * Sidebar: The T antigen protein is the only thing required to get access to the cell. How do we know? The T antigen protein was tagged and injected into the cytoplasm and it traveled to the nucleus (like the previous experiment).
 * Experiment
 * For this experiment you take the gene for T antigen and induce deletions along the gene and fuse the resulting genes to GFP (from jellyfish codes for a fluorescent protein).
 * Then each one of the genes are injected into the cytoplasm. In the cytoplasm, the genes and GFP are translated into proteins.
 * The proteins are now florescent (thanks to the GFP portion) and the proteins are monitored to find which ones enter the nucleus.
 * The proteins (created from the broken T antigen gene) that cannot gain access to the nucleus are missing an essential signal.
 * Finally you are able to match up the proteins that could not access the nucleus with the deletion in the gene and determine the location of the signal in the gene.
 * This experiment shows that the signal is necessary for transport but not sufficient.
 * If you fuse the signal and the GFP gene alone, the protein created will be capable of gaining access to the nucleus. This shows that the signal is necessary and sufficient for entry into the nucleus.

Nucleocytoplasmic Exchange is continued in the next lecture.

Please feel free to add details or make changes where necessary. @undefined

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Cell biology/Membrane Assembly: Signal Hypothesis

Here is the link to the ITunes U Lecture from Berkeley. | Membrane Assembly: Signal Hypothesis

Review:

 * SV40 T antigen
 * mutate sequences to find one (or more) regions of genes required for import.
 * "isolate" signal protein-fuse it to a reporter protein (like GFP) to see if the signal is sufficient.
 * FYI: What happens when there are 2 different regions of the primary sequence are required for import?
 * Sometimes the secondary or tertiary structure has these two regions located close together. So when the protein is folded, the 2 regions are side by side and provide the signal for import.
 * The T antigen sequence required for entry is: proline-lysine-lysine-lysine-arginine-lysine-valine.
 * This sequence has 5 basic charges in a row, which is common for a signal protein.
 * Ribosomes:
 * Ribosomes require protein subunits to travel to the nucleus so they must have a nuclear localization signal (NLS).
 * In the nucleus, the protein subunits are assembled in complex with the ribosomal RNA.
 * Finally, the ribosomes must have another signal, nuclear export signal (NES) to get out of the nucleus

Nucleocytoplasmic Exchange
Or Mechanism of Import and Export of Proteins Into/From the Nucleus.

 * Involves a several carrier proteins and a GTP binding (regulatory) protein. The GTP protein must be bound to the cargo that is imported or exported out of the nucleus.
 * The central feature of this regulatory pathway can be reduced to the conversion of protein, called Ran, from a guanosine diphosphate (GDP) bound state to a guanosine triphosphate (GTP) bound state (GDP or GDP are bound to Ran) or the reverse reaction.
 * There are many small GDP/GTP binding proteins that regulate the traffic of proteins and cytoskeletal events throughout the cell.
 * This small regulatory protein, Ran, is important because it controls the entry and exit of all proteins that must enter/exit the nucleus.
 * This system is maintained by creating a gradient, of GDP/GTP, within the cell. Ran-GDP is the preferred type of the protein in the cytoplasm, while Ran-GTP is the preferred type of the protein in the nucleus.
 * How does this happen?
 * In the nucleus, Ran-GDP → Ran-GTP, by Ran-GEF (guanine nucleotide exchange factor).
 * FYI: Ran slowly hydrolyzes GTP but at an insignificant rate.
 * Ran-GEF (always in the nucleus) binds in the nucleus to Ran-GDP and displaces the GDP allowing GTP to replaces it.
 * In the cytoplasm, Ran-GTP → Ran-GDP, by Ran-GAP (GTP hydrolysis activating protein).
 * Ran-GAP (always in the cytoplasm) binds Ran-GTP and stimulates GTP hydrolysis, so a phosphate group drops off, creating GDP.

 * FYI there are other small GDP/GTP binding proteins with their own specific GEF and GAP enzymes.

What happens when a protein (cargo), in the cytoplasm wants to be imported into the nucleus?
See Diagram above.
 * 1) The cargo protein has an NLS (nuclear localization signal) and alpha importin binds to the NLS.
 * 2) Beta importin binds to the alpha importin.
 * 3) Beta importin is responsible for driving the protein complex just created through the pore. It makes contact with the inside of the pore complex (mildly hydrophobic), and binds and dissociates with protein sequences of the pore.
 * 4) Ran-GDP has already diffused in.
 * 5) Ran-GEF acts on (activates) Ran-GDP converting it to Ran-GTP.
 * 6) Ran-GTP binds with the alpha beta cargo complex and causes it to dissociate.
 * 7) The cargo molecule does whatever it was supposed to to (i.e. transcription or ribosome production).
 * 8) Ran-GTP binds to the beta importin (alpha importin follows) and the complex travels through the nuclear pore.
 * 9) On the cytoplasmic side the Ran-GAP pulls off a phosphate group, from Ran-GTP creating Ran-GDP
 * Cells have lots of importin alphas which are devoted to importing a dozen (or two) cargo molecules.

What happens when a protein (cargo)wants to be exported from the nucleus?
*Cells have lots of exportin alphas which are devoted to importing a dozen (or two) cargo molecules.
 * 1) Exportin binds to NES (nuclear export signal) on the cargo protein.
 * 2) Exportin has an affinity for Ran-GTP and the complex travels out into the cytoplasm.
 * 3) *Exportin is responsible for interacting with the nuclear pore.
 * 4) When the complex exits the pore, Ran-GAP removes a phosphate group from Ran-GTP, creating Ran-GDP.
 * 5) Ran-GDP then detaches from the complex leaving the cargo protein in the cytoplasm along with exportin.
 * 6) Exportin must be transported back into the nucleus with Ran-GDP.
 * How can you regulate import/export of molecules from the nucleus?
 * Regulation of import/export of certain molecules is managed by regulating importin and exportin proteins.
 * What keeps cargo molecules from being exported/imported early?
 * The NES or NLS is exposed late in the assembly of the desired protein/complex (ribosome) and so their is no attachment point for the transportation enzymes until the cargo molecules are ready for import/export.

Secretory or Membrane Protein Transport
Proteins that are transported to the into or out of cell membrane are very different from nucleocytoplasm transport. The pores that are on the cell membrane are quite small (diameter = 8 angstroms) and only allow proteins to pass as they are being synthesized (compared to the nuclear membrane that allows all small proteins to pass and larger proteins to pass with Ran-GDP or Ran-GTP).

Enter the Endoplasmic Reticulum
The endoplasmic reticulum is a conveyor belt wrapped within the cell on which ribosomes are deposited. In pancreatic cells, 90% of the ribosomes are touching the cytoplasmic surface of the ER. Protected in the interior of the ER is a clear space called the lumen. The transfer of secretory proteins (as they are being made by the ribosomes) from the membrane to the lumen determines their future as a secretory protein. When the proteins complete production, the ER fuses with the cell membrane and releases the proteins to the outside of the cell. George Palade confirmed the function of the endoplasmic reticulum and Golgi Apparatus for producing secretory proteins. He also found that ribosomes are responsible for protein manufacturing. He studied the movement of proteins within the pancreas (90% of their energy is devoted to making secretory proteins and their endoplasmic reticulum (ER) has 25X the surface area of the plasma membrane).
 * In the 1940s it first became possible to prepare cells and tissues to study with the electron microscope.
 * Palade performed an experiment where he homogenized pancreatic tissue and created vesicles that have ribosomes still bound (with their messenger RNA) and show that the ribosomes are caught in the act of translating a nascent polypeptide chain.
 * He then took these vesicles and added translation factors, ATP, and GTP, transfer RNA and other materials necessary for protein synthesis. The vesicles would then create full proteins from their nascent state inside the vesicle.
 * How do you prove that the proteins are discharged into the interior of the vesicle via vectorial translocation?
 * Vectorial translocation is the name of transfer in a unidirectional manner.
 * To prove this you can use an earlier study (protease protection). For this study, you would take the vesicles, ATP, GTP, transfer RNA and radiolabeled amino acids. Then treat one group with trypsin (protease) and another with trypsin and a detergent. This would show that trypsin can only act upon the amino acids when the detergent is used.

 * From this we know that we can create a protein in a test tube.

How do these frequently polar molecules cross the lipid bilayer membrane?
To be continued in the next lecture.
 * Secretory proteins have an N-terminal extension that is clipped after entry into the endoplasmic reticulum.
 * Cesar Milstein won the Nobel prize for monoclonal antibodies. But we are more interested in his work with the antibody light chain because he discovered the N-terminal extension on an immature protein, antibody light chain.
 * Earlier work with SDS gels gave the mature light chain's structure and weight of 23.5kD.
 * Milstein took light chain mRNA and used it to create a protein with an in vitro synthesis reaction. He used translation materials from reticulocytes and radioactive amino acids.
 * It is possible to get ribosomes and translation necessities, without a membrane, by breaking open reticulocytes.
 * He found that the in vitro created protein, a precursor protein, had a weight of 25.5kD.
 * So there are an extra 20 amino acids at the N-terminus of precursor secretory proteins that is later cleaved before the protein becomes fully developed.
 * Finally, Günter Blobel was able to isolate the biochemical components necessary for the production of the light chain from beginning to end.
 * Günter Blobel (Palade's protege) also won the Nobel Prize.
 * His study involved stripping ER vesicles of their ribosomes and adding mRNA and ribosomes to create fully formed proteins (without their N-terminus extension) inside of the ER vesicle.

Please feel free to add details or make changes where necessary. @undefined

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Cell biology/Membrane Assembly: Mechanism

Here is the link to the ITunes U Lecture from Berkeley. | Membrane Assembly: Mechanism

Review
Blobel developed a way of taking ER membrane vesicles from pancreatic tissue. He then stripped the ribosomes off leaving the integral membrane proteins in tact. He then found that if he added ribosomes and translation material he could recreate the protein translation process.

How did he remove the ribosomes and the mRNA?
Blobel then added mRNA (for a secretory protein) and necessary translation factors (ATP, GTP, ect.) and fresh ribosomes. When all of this was mixed together, the mature protein was produced inside the vesicle.
 * It is harder than you might think to remove the ribosomes and mRNA. The ribosomes are attached electrostatically to the channel (so they are sensitive to ionic changes).
 * The mRNA that is being translated acts like an anchor and it is hard to remove.
 * Puromycin (antibiotic that is not that useful for human consumption) looks like the 3' end of the tRNA molecule. Puromycin tricks the ribosome into attaching it to the C-terminus of the nascent chain. Additionally, Puromycin does not hydrogen bond to the ribosome, which causes the release of the nascent chain.
 * Now that the anchoring nascent chain is gone, the ribosome can be removed with a high salt buffer. This is how Blobel created virgin or smooth ER vesicles.

What is the sequence for the creation of a mature secretory protein?

 * You need the membrane to get the mature protein. In a solution with out the membrane, the precursor form is produced.
 * His first experiment looked for a co-translational translocation, that is he looked to see if the mature protein could be produced within the vesicle if all of the materials were added at the beginning.
 * His next experiment looked for post-translational translocation, that is he allowed the protein to be transcribed and then added the ER membrane vesicles. Both of these experiments were tested with an SDS page. (see graph of results)
 * Looking at the results, you can see that experiment 1 produced mature proteins that were protected from protease because they were inside the vesicle. In experiment 2, you can see that no mature proteins were produced and the precursor protein was found only on the outside of the vesicle because it was not protease protected.

So we have learned that post-translational translocation does not occur, Why is that?

 * The secretory proteins have 6 or more hydrophobic amino acids at the N-terminal domain (the part that would be synthesized first). *This hydrophobic core constitutes the information necessary to guide the nascent chain (from the cytoplasm) to the ER membrane so it can be translocated into the lumen of the ER while the protein is still being created.

What happens if the protein doesn't get to the ribosome early in the translation?
The hydrophobic core will hide itself in the protein or attach to other hydrophobic places in the cell and not be available to initiate translocation. So translocation occurs in the beginning or not at all.

Signal Hypothesis

States that there is special information (hydrophobic core) at the N-terminus of the secretory proteins to guide the nascent protein ribosome complex to a site on the ER membrane that allows it to be translocated to the ER lumen (and ultimately allows secretion from the cell).

What molecules are responsible for this reaction?

 * Salt wash strips away anything peripherally bound to the ER membrane, and the previous experiment has used a crude mix of translation factors. If you utilized all pure translation factors, only the precursor molecule will be produced outside the vesicle.
 * So what was present in the crude mix that is not found in the purified mix?

 * If you just perform a salt wash of an ER membrane, to remove peripheral proteins but not ribosomes, and gather the supernatant you can add it to the purified factors and create mature proteins inside the vesicle. So obviously a peripheral protein is necessary for the translocation process.

 * The important protein is called signal recognition particle (SRP) This complex recognizes the hydrophobic core and locks onto the nascent chain guiding it to the ER membrane. The signal recognition particle is the major ribonucleoprotein particle found within the cytoplasm. It is largely found on the cytoplasmic surface of the ER.
 * SRP binds to the hydrophobic N-terminal of the nascent protein and then hydrogen bonds with the large subunit of the ribosome. This arrests translation.[[image:PDB 1wgw EBI.jpg|thumb|200px|right|54KD SRP subunit]]

Why arrest peptide development?

 * Arresting nascent chain development allows the process of translation and translocation to be synchronized.

 * SRP's function: bind the nascent peptide and guide the nascent protein to the correct location.
 * The 54KD subunit is the most well known has a GTP binding/hydrolysis site that opens a spot for the hydrophobic core and that location binds tightly and arrests development.

Are there other proteins necessary for this process?

 * We have identified soluble proteins involved in the process but now we need to find the integral proteins necessary for this process. There are two. How were they identified?
 * Again the Blobel lab used a technique called affinity chromatography.
 * Make an affinity column packed with beads (solid support) to which a ligand is covalently attached (SRP). A buffer with ER membranes (that had been salt washed-no peripheral proteins left) and a detergent, that releases proteins from ER membrane, is ran through the affinity column. Proteins that are not attracted to SRP will flow through the affinity column but some proteins will be bound and can only be removed with harsh methods. So at the end you add harsh conditions such as pH changes or very ionic solutions which will break through protein protein attachments. This releases the necessary proteins that have bound to SRP.
 * From this technique it was learned that SRP binds to an SRP receptor, which has two subunits that are integral membrane proteins. Both subunits are GTP binding proteins. When the nascent chain-SRP-ribosome complex attaches to the SRP receptor, it causes all the GTP to be hydrolyzed, which in turn, pushes the nascent chain out of the opening in SRP. The nascent chain is poised at the surface of the membrane and can now travel inside the ER lumen and once the nascent chain is freed protein synthesis can begin again.

Sadly, The SRP receptor is necessary but not sufficient for translocation.
Schekman's lab figured out that there is also a channel necessary. He used genetic research. Yeast needs to secrete in order to grow. His lab found a temperature sensitive lethal secretion mutant at sec61. The normal gene is SEC61. This will be explained in the next lesson. Look at [|NCBI Bookshelf, Alberts Et. Al Textbook] for a description of experimental techniques including Schekman's laboratory technique with yeast cells.

Please feel free to add details or make changes where necessary. @undefined

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Cell biology/Membrane Assembly: Topography

Here is the link to the ITunes U Lecture from Berkeley. | Membrane Assembly: Topography

ER Translocation Membrane Channel

 * Schekman's lab figured out that there is also a channel necessary using genetic research. He used genetic methods to identify the part of genetic sequence that creates this channel.
 * Background:Cells must be able to secrete in order to grow, or to rephrase they must be able to perform protein translocation.
 * His lab found a temperature sensitive lethal secretion mutant at sec61 (the normal gene is SEC61). This mutation allows the cell to function at room temperature but the cell dies at higher temperatures. Additionally the mutation causes the cell to produce secretory precursor protein in the cytoplasm (rather than fully functional secretory protein within the ER lumen).
 * His laboratory introduced bits of genes to this mutant until it was restored to wild-type function. In other words they took the broken (mutant) yeast cell and added genes until it was fixed.
 * Experimental Steps
 * Use a plasmid vector with a selectable marker (complements a mutation in the host enzyme (i.e. Leu-creates leucine)-restore function that the host does not have without integrating this plasmid).
 * Create a library of DNA from the yeast cell (by cutting up the genome with restriction enzymes).
 * Insert pieces of the wild-type yeast DNA (from the library) into the plasmid.
 * Use a double mutant yeast cell (sec61ts and leu mutation-cell cannot create leucine and dies at a high temperature) and infect with the plasmids created.
 * Grow infected yeast cells on leucine deficient petri plate at room temperature. If the yeast cell did not take up a plasmid, they will die. If they took up any plasmid they will grow colonies on this plate.
 * Take the plate and make a copy (by using filter paper or velvet to touch the plate transferring colonies onto the paper/velvet and then placing the paper/velvet onto another petri dish in the same orientation).
 * Incubate the second plate at body temperature, 37 degrees.
 * Only a few colonies will grow on the second plate but they are likely to have the wild-type SEC61 gene and the LEU gene that the yeast cell was missing.
 * Now to get the DNA sequence. Clone the surving colony, extract plasmid, use original restriction enzyme, and perform DNA sequence analysis.
 * Result: this gene produces a 53KD multi-spanning hydrophobic membrane protein, which is completely separate from SRP and SRP receptor and similar to SecY (Sec Y is required for secretion of proteins in bacteria).
 * An antibody can be created to this protein and this antibody binds to and precipitates with 3 proteins called Sec61α,β,γ in a trimeric complex.

 * This channel has been crystallized (from a thermophile) and the structure reveals an 8 angstrom pore aligned with secretory proteins.

Can charged ions (glutamate) penetrate a polar channel responsible for protein translocation?
Comprehension Question
 * Use a two part chamber that measures conductance through a membrane channel. This chamber has 2 halves filled with buffer and the divider has a small aperture connecting the two sides.
 * This aperture is covered with an artificial lipid bilayer.
 * Experiment 1: Add ER membrane vesicles (with ribosomes and nascent chains) and a chemical that will fuse vesicles to the artificial bilayer. Add glutamate and if there are ionic channels in the ER vesicles that are fused to the lipid bilayer, they will flow into the second chamber and create a change in ionic concentration and conductivity.
 * When nascent chains are present the ions will not transfer to the next chamber.
 * Experiment 2: Same procedure except the ER vesicles are treated with puromycin (so ribosomes are attached but nascent chains are removed).
 * Under these conditions, the glutamate ions will transfer to the next chamber.
 * Experiment 3: Same as above except this time remove the ribosomes and nascent chains with puromycin and high salt solution.
 * There will be no transfer of glutamate to the second chamber.
 * What if you did this reaction with the vesicles of a sec61 temperature mutant (ts) and fused them with the bilayer at room temperature? In the puromycin treated trial, what would you expect to happen?
 * At room temperature, the experiment should progress but at higher temperatures it should not work.

Mechanism of signal hypothesis

 * Secretory protein is being produced in the cytoplasm by a ribosome and forms a hairpin loop through the membrane the N-terminus remains on the cytoplasmic side and the apolar amino acids slip through a hole in the channel and enter the lipid bilayer (escaping the polar environment of the channel).
 * The rest of the chain continues to grow into the ER lumen.
 * When a certain sequence is presented to an enzyme, signal peptidase, resting on the lumen side of the channel signal peptidase clips the polypeptide chain at the signal polypeptide, leaving the secretory protein growing into the ER lumen. The signal peptide then floats into the lipid bilayer and is degraded into amino acids.

How do sugars get attached

 * Some proteins acquire N-glycosylatedly linked oligosaccharides.
 * The oligosaccharide is ready to be linked to the polypeptide waiting for the consensus sequence for N-glycosylation (Asn-X-Ser or Asn-X-Thr).
 * The N-terminus of asparagine is acted upon by the dolichol polysaccharide creating an N-linkage.
 * Many cytoplasmic proteins have this sequence but they are not glycosylated because they are translocated into the endoplasmic reticulum.

Types of membrane proteins

 * Type I proteins have their N-terminus facing the outside of the cell.
 * Type II proteins have their C-terminus facing the outside of the cell.
 * Polytopic proteins loop through the cytoplasm multiple times.

Please feel free to add details or make changes where necessary. @undefined

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Cell biology/Cholesterol Regulation

Here is the link to the ITunes U Lecture from Berkeley. | Cholesterol Regulation

Download the article for the this lecture at the | Nobel Prize website. Michael Brown and Joseph Bolstein studied Cholesterol regulation and won the Nobel prize.

Cholesterol

 * Can be derived from dietary sources.
 * Can be synthesized by our cells.
 * Cholesterol is important in cellular membranes, component of bile acids, and as a hormone precursor.
 * Cholesterol is a hydrophobic molecule with a small polar group (hydroxyl).

The cholesterol biosynthetic pathway

 * HMG-CoA reductase is the rate limiting enzyme and is tightly regulated.
 * It is an integral protein in the ER with catalytic domain on the cytoplasmic side. It acts upon a water soluble substrate and produces a water soluble product.
 * It consists of 8 transmembrane alpha helices.
 * 5 of these have a sterol-sensing domain, which binds cholesterol. When cholesterol is bound to the sterol sensing domain, the ubiquitination of HMG-CoA reductase results and the enzyme is degraded by the attachment of a small molecule called ubiquitin (negative feedback mechanism).

How do you transport the hydrophobic cholesterol in the blood?

 * It is carried around by lipoproteins These are a group of lipid-protein complexes filled with cholesterol esters (very hydrophobic in the center) on the surface there are phospholipids and other molecules with hydrophilic regions. The cholesterol molecules can also be found on the surface but not in an ester form.
 * The lipoprotein responsible for transporting the majority of cholesterol is a low-density lipoprotein (LDL). LDLs have apolipoprotein B which wraps around the sphere and has a role in bringing the cholesterol into the cell.

Brown and Goldstein studied Familial Hypercholesterolemia (FH)

 * Described, in 1930, by Carl Muller, in Oslo, as a disorder that caused people to have heart attacks at a young age (30-40). 1/500 people are heterozygotes for this disorder.
 * As a heterozygote people have 2X the number of LDL particles in their plasma.
 * Of the people who have heart attacks before the age of 60, less than 5% are FH heterozygotes. So diet/lifestyle must be a contributing factor to heart attacks.
 * FH homozygotes are 1 in a million.
 * They have a 6-10x increase in plasma LDL levels from birth and have heart attacks in their childhood.

Brown and Goldstein used fibroblasts to study FH.
They developed sensitive assays for HMG-CoA reductase and studied its response to lipoprotein levels. The activity of the enzyme changes in normal cells when cholesterol {from lipoproteins) is available. The cell decreases its production of cholesterol because they can get it from the environment. However, in FH cells, HMG-CoA reductase is unaffected by lipoprotein (cholesterol) levels in the medium.

 * Possibilities
 * Can the cholesterol get in the cell?
 * Is the cholesterol unable to influence the regulation of the HMG-CoA reductase enzyme?
 * To test the second possibility they mixed up cholesterol in an emulsion so the cells would be able to take up the cholesterol through a different pathway. When the cholesterol entered the cell it affected the regulation of HMG-CoA reductase enzyme. So the HMG-CoA reductase enzyme is sensitive to internal cholesterol levels, and there must be a problem getting cholesterol into the cell.

Are the FH cells defective in their ability to bind LDL?

 * They labeled LDL and compared cells with the FH disorder to normal cells.
 * They found that most cells with the FH disorder did not bind LDL at all.
 * In the patient, JD, who had FH, the cells were able to bind LDL normally but his cells could not take up the LDL.
 * They found this out by comparing JD's cells and normal cells first at low temperatures and then increasing temperatures. In normal cells the LDL bound decreases, at higher temperatures, because it is internalized. In JD's cells there is little internalization of LDL.

FH causes a defect in the LDL receptor.
LDL receptor structure:
 * They cloned and sequenced the DNA that codes for this receptor protein.
 * It is a Type 1 transmembrane protein.
 * 1) The ligand binding domain which consists of 292 amino acids with 40 amino acids repeating 7 times (with little variation). This domain has many cysteine residues, which cluster with negatively charged amino acids, to form binding sites for an alpha helix in apolipoprotein.
 * EGF precursor homology region (containing approximately 400 amino acids) is like the EGF precursor and shares genetic material that is used to create this separate protein. This is explained by the theory of mosaic genes where the introns(noncoding DNA) can rearrange so different exons (coding DNA) can be placed side by side in the resulting protein. In this way, segments of genes can be rearranged to create different proteins
 * O-linked sugar chains bond to the 58 amino acids found in the third domain.
 * 1) The hydrophobic domain consists of 22 hydrophobic amino acids that span the cell membrane.
 * 2) The cytoplasmic domain or the cytoplasmic tail (containing 50 amino acids) is important in attaching the catharin molecules and creating the endocytosis vesicles.
 * there is also a precursor region that is cleaved when the protein enters the ER (as discussed in the previous lecture).

They then made antibodies to the LDL receptor and studied it's activity within the cell.

 * They found that the LDL receptor was responsible for receptor mediated endocytosis.
 * The receptor binds LDL-apoprotein complex, then clatharin molecules bind causing a vesicle to form. Which in turn pinch off from the membrane.
 * The clatharin coat is then shed off the vesicle.
 * The vesicle moves towards the center of the cell and ATPases act upon the vesicle lowering the pH inside the vesicle.
 * The LDL particle is released when the beta propeller domain (of the LDL receptor) is protanated which causes the ligand binding domain of the LDL receptor to bind to itself. Once the receptor releases the LDL ligand particle (by binding to itself), it recycles back to the membrane.
 * Even in the absence of bound LDL this internalization and cycling still happens.
 * The LDL receptor molecule does this every 5 minutes for its 20 hour lifespan.
 * The patient, JD was unable to import LDL into the cell due to a single point mutation in they cytoplasmic domain. N-P-X-Y, asparagine-proline-any-tyrosine, is an essential motif to import LDL and in JD the tyrosine was replaced by cysteine making the receptor inoperable.

How Intracellular Cholesterol Regulates Gene Expression

 * There are many genes regulated by the cholesterol levels including the LDL receptor and HMG CoA reductase.
 * When cholesterol levels are low transcription of the genes for LDL receptor and HMG CoA reductase are increased, so they in turn can import/create more cholesterol (feedback loop).
 * Upstream of these genes is SRE (Sterol regulatory element) where a regulatory protein can bind.
 * This regulatory protein is called SREBP (Sterol Regulatory Element Binding Protein) and increases transcription when it binds.
 * When cholesterol is high, SREBP remains complexed (with insig and SCAP) in the ER.
 * In this complex SREBP cannot be trafficked to the Golgi Apparatus.
 * When cholesterol levels drop, insig is cleaved and SCAP and SREBP travel to the Golgi Apparatus.
 * SREBP is then cleaved from SCAP.
 * SREBP then travels to the nucleus.
 * This process is called intramembrane proteolysis.
 * SREBP has a nuclear localization signal and goes to the nucleus when it is released.
 * In summation, low cholesterol levels cause SREBP to activate transcription of HMG-CoA reductase and LDL receptor.

Statins

 * Inhibit HMG-CoA reductase and reduce intracellular cholesterol levels,
 * Increase SREBP-mediated transcription (including HMG-CoA reductase which offsets the inhibition is has already performed),
 * Increase expression of LDL receptors in the liver and
 * LDL particles are cleared from the blood rather than clotting blood vessels.

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title: Cell biology/Cholesterol Regulation


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Cell biology/Lysosomal Protein Transport

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Cell biology/Action Potential

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Cell biology/Synaptic Transmission

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Channel Receptor Types (Overview)
There are several types of known synaptic channel types.
 * Ionotropic channels involved in (or inhibiting) an action potential or train of action potentials, are: M-currents (or M-channels), T-currents, L-currents, C-currents, AHP currents, N-currents, GIRK channels, GABAA, GABAB. These can exist presynaptically or postsynaptically.
 * G-Protein Coupled Receptors (GPCRs) are metabotropic second messenger channels (usually postsynaptic). Ordinarily, a ligand will bind and the alpha subunit will have one effect, and the beta-gamma subunit may have another.  The GPCR type is generally named after its alpha unit's effect and are referred to as Gs (or Gαs), Gq, Gi, and Go (the latter 2 are frequently lumped together as Gi/o.  In many sensory organs, the GPCR is a 7-transmembrane protien unit comprised of alpha helices.  There are certain exceptions.
 * Noteworthy example: Interneurons may innervate the presynaptic sensory terminals (as in Aplysia), to use a GPCR to alter motor neuron output (short-term facilitation) via change in open channel probability. Subsequent stimulation of transcription further affects long-term facilitation.

''Author's note: this is a broad overview of a few types of synaptic transmission with a few examples that do not divulge much detail. I included key words to encourage pages to be created as branches, explaining in more detail, each part of synaptic transmission.''

title: Cell biology/Synaptic Transmission


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Cell biology/Cell Communication: Ligand and Receptors

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Cell biology/G-Protein Coupled Receptor Signaling I

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Cell biology/G-Protein Coupled Receptor Signaling II

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User:Egm6322.s09.bit.sahin/r7

We know that $$\lambda^2-\lambda(a+c)+(ac-b^2)=0$$

and

$$\left (\lambda-\lambda_1  \right )\left (\lambda-\lambda_2   \right )=\lambda^{2}-\left ( \lambda_{1}+ \lambda_{2}\right )\lambda+\lambda_{1}\lambda_{2}=0$$

therefore

$$\lambda_{1}\lambda_{2}=ac-b^{2}$$

So,

Case#1: $$\lambda_1 \lambda_2 < 0$$ (hyperbola) or $$\lambda_1 \lambda_2 > 0$$ (ellipses)

Case#2: $$\lambda_1 \lambda_2 = 0$$(parabolas)

Section 2
$$\mathbf{J}=\begin{bmatrix} \phi _{x} &\phi _{y} \\ \psi _{x}&\psi_{y} \end{bmatrix}$$, $$\mathbf{J^{T}}=\begin{bmatrix} \phi _{x} &\psi _{x} \\ \phi _{y}&\psi_{y} \end{bmatrix}$$ and $$\mathbf{A}=\begin{bmatrix} a &b \\ b &c \end{bmatrix}$$

$$\mathbf{\bar{A}}=\mathbf{JAJ^T}=\begin{bmatrix} \bar{a} &\bar{b} \\ \bar{b} &\bar{c} \end{bmatrix} $$

Substituting these matrices into second equation

$$\mathbf{\bar{A}}=\begin{bmatrix} \phi _{x} &\phi _{y} \\ \psi _{x}&\psi_{y} \end{bmatrix} \begin{bmatrix} a &b \\ b &c \end{bmatrix} \begin{bmatrix} \phi _{x} &\psi _{x} \\ \phi _{y}&\psi_{y} \end{bmatrix}$$

$$\mathbf{\bar{A}}=\begin{bmatrix} a\left (\phi _{x} \right )^{2}+2b\phi _{x}\phi _{y}+c\left (\phi _{y}  \right )^{2} & a\phi _{x}\psi _{x}+b\phi _{x}\psi _{y}+b\psi _{x}\phi _{y}+c\phi _{y}\psi _{y} \\ a\phi _{x}\psi _{x}+b\phi _{x}\psi _{y}+b\psi _{x}\phi _{y}+c\phi _{y}\psi _{y} & a\left ( \psi _{x} \right )^{2}+2b\psi _{x}\psi _{y}+c\left ( \psi _{y} \right )^{2} \end{bmatrix}$$

Therefore,

$$\bar{a}=a\left (\phi _{x} \right )^{2}+2b\phi _{x}\phi _{y}+c\left (\phi _{y}  \right )^{2}$$

$$\bar{b}=a\phi _{x}\psi _{x}+b\phi _{x}\psi _{y}+b\psi _{x}\phi _{y}+c\phi _{y}\psi _{y}$$

$$\bar{c}=a\left ( \psi _{x} \right )^{2}+2b\psi _{x}\psi _{y}+c\left ( \psi _{y} \right )^{2}$$

Section 3
 Hyperbolic PDEs

Method of characteristics applied in two different places

1) Finding canonical forms (Form 1 by hyperbolic PDEs) by setting

$$\bar{a}=\bar{c}=0$$

2) Solving Form 1 of hyperbolic PDEs

Setting $$\bar{a}=\bar{c}=0$$

Remembering that

$$\bar{a}=a\left (\phi _{x} \right )^{2}+b\phi _{x}\phi _{y}+c\left (\phi _{y}  \right )^{2}$$

and assuming that $$\phi _{x}$$ and $$\phi _{y}$$ are proportional

$$\phi _{x}=k\phi _{y}$$

substituting second equation to the first one, we have

$$\bar{a}=\left ( \phi _{y} \right )^{2}\left ( ak^{2}+bk+c \right )=0$$

the solution of $$\left ( ak^{2}+bk+c \right )=0$$ is

$$k=\frac{-b\pm \sqrt{\Delta }}{2a}$$, $$\Delta =b^{2}-4ac$$

so

$$k_{+}:=\frac{-b+ \sqrt{\Delta }}{2a}$$ and $$k_{-}:=\frac{-b- \sqrt{\Delta }}{2a}$$

it can be written that

$$d\phi =\phi _{x}dx+\phi _{y}dy$$ (2)

from the proportionality $$\phi _{x}=k\phi _{y}$$

$$0=\phi _{x}-k\phi _{y}$$ (3)

comparing Eq(2) and Eq(3) gives

$$\frac{d\phi }{0}=\frac{dx}{1}=\frac{dy}{-k}$$

title: User:Egm6322.s09.bit.sahin/r7


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User:Rijaz

Hello

title: User:Rijaz


text:
Survey research and design in psychology/Tutorials/Multiple linear regression/General steps

General steps
The general recommended steps for conducting a multiple linear regression analysis are:

 * 1) Conceptualise the model (e.g., draw a path diagram or Venn diagram to indicate the IVs and the DV) and establish research questions and/or hypotheses.
 * 2) Check assumptions:
 * 3) Levels of measurement
 * 4) Sample size
 * 5) Normality
 * 6) Linearity
 * 7) Homoscedasticity
 * 8) Multicollinearity
 * 9) Multivariate outliers
 * 10) Normality of residuals
 * 11) Choose the type of MLR:
 * 12) Standard
 * 13) Hierarchical
 * 14) Stepwise / Forward / Backward
 * 15) Interpret statistical output and psychological meaning of results. Consider:
 * 16) Overall model
 * 17) R, R2, Adjusted R2, sig. of R
 * 18) Change in R2 and the sig. of this change (if a hierarchical MLR is conducted)
 * 19) Regression coefficients
 * 20) Y-intercept (labelled "Constant" in the SPSS MLR Coefficients table output)
 * 21) Unstandardised (B)
 * 22) Standardised (β (beta))
 * 23) t and significance for each predictor
 * 24) Zero-order correlations (r) and semi-partial correlations squared (sr2) for each IV in each model
 * 25) Depict the relationships in a path diagram or Venn diagram (if useful/relevant)
 * 26) Regression equation: Present a prediction equation for Y (if useful/relevant)


text: Category:Multiple linear regression/Data analysis tutorial


text:
Multiple linear regression/Multivariate outliers

==Multivariate outliers== Check whether there are influential multivariate outlying cases using Mahalanobis distance & Cook’s D. Using PASW:
 * Linear Regression - Save - Mahalanobis and Cook's D - OK
 * SPSS will create new variables called mah_1 and coo_1.
 * Check the Residuals Statistics table in the output for the maximum Mahalanobis and Cook’s distances.
 * The maximum Mahalanobis distance should not be greater than the critical chi-square value with degrees of freedom equal to number of predictors, with critical alpha (α) =.001.
 * Cook’s D should not be greater than 1.
 * If outliers are detected, check each case, and consider removing the case from the analysis.
 * See also Francis 5.1.4.2 Screening for influential case


text:
Index:Novial/Novial Lexike/a

Back to the Novial front page

Back to the Novial Lexike front page

NOVIAL LEXIKE (1930) INDEX:

Parts of speech: adj. = adjektive; adv. = adverbie; sb. = substantive; prp. = prepositione; vb. = verbe.

NOVIAL LEXIKE INDEX:

Back to the Novial Lexike front page

Back to the Novial front page

title: Index:Novial/Novial Lexike/a


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Template:Protection templates

 * See also, Protection policy.

title: Template:Protection templates


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Template:Pp-template/doc

Usage

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This template is not to be used as a bluff. Only use it on FULLY PROTECTED pages. Please note that only custodians and curators can protect pages.

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Digital Electronics/Lecture High Level Language Logic Modeling and Synthesis

 * 1) Introduction to Hardware Description languages (HDLs)
 * 2) Modeling Digital Circuits using VHDL ( V ery High Speed Integrated Circuits H ardware D escription L anguage)
 * 3) Modeling Digital Circuits using Verilog


text:
User:Calebrw

Please see w:User:Calebrw or n:User:Calebrw for more about me. Thanks. Calebrw 05:48, 24 April 2009 (UTC)

Projects

 * Adobe InDesign
 * Helping out a bit with template stuff.

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User:Eurominuteman~enwikiversity

Welcome

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File:Adobe-InDesign-Default-Workspace.jpg

Summary
Default workspace of Adobe InDesign CS4, running on Windows Vista. Windscreen monitor.


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User:Murakumon/Adobe InDesign/Workspace

The Adobe InDesign workspace features five main areas:
 * Menu bar
 * Control bar
 * Tool bar
 * Pallets bar
 * Document work area

Tool bar
the tool bar is usually located at the left side of te screen. The little tabs that allow the user to create, edit, manipulate, etc are tools that one can use in indesign

title: User:Murakumon/Adobe InDesign/Workspace


text:
User:Stevenarntson/humanitiesii/perception

Hegelian Dialectic

thesis/antithesis

 * 1) the book
 * 2) the rabbit/duck
 * 3) annie dillard's american childhood
 * "Once upon a time, I, Chuang Chou, dreamt I was a butterfly..."
 * 1) excerpts from Emerson's self-reliance

synthesis

 * 1) 1907: protocubism. les demoiselles d'avignon
 * 2) 1910: analytic cubism. ma jolie
 * 3) 1921: synthetic cubism. three musicians


text:
User:Stevenarntson/humanitiesii/perception/self reliance

The other terror that scares us from self-trust is our consistency; a reverence for our past act or word because the eyes of others have no other data for computing our orbit than our past acts, and we are loath to disappoint them.

But why should you keep your head over your shoulder? Why drag about this monstrous corpse of your memory, lest you contradict somewhat you have stated in this or that public place? Suppose you should contradict yourself; what then? It seems to be a rule of wisdom never to rely on your memory alone, scarcely even in acts of pure memory, but to bring the past for judgment into the thousand-eyed present, and live ever in a new day. Trust your emotion. In your metaphysics you have denied personality to the Deity, yet when the devout motions of the soul come, yield to them heart and life, though they should clothe God with shape and color. Leave your theory, as Joseph his coat in the hands of the harlot, and flee.

A foolish consistency is the hobgoblin of little minds, adored by little statesmen and philosophers and divines. With consistency a great soul has simply nothing to do. He may as well concern himself with his shadow on the wall. Out upon your guarded lips! Sew them up with pockthread, do. Else if you would be a man speak what you think to-day in words as hard as cannon balls, and tomorrow speak what to-morrow thinks in hard words again, though it contradict every thing you said to-day. Ah, then, exclaim the aged ladies, you shall be sure to be misunderstood! Misunderstood! It is a right fool's word. Is it so bad then to be misunderstood? Pythagoras was misunderstood, and Socrates and Jesus, and Luther, and Copernicus, and Galileo, and Newton, and every pure and wise spirit that ever took flesh. To be great is to be misunderstood.

I suppose no man can violate his nature. All the sallies of his will are rounded in by the law of his being, as the inequalities of the Andes and Himmalch are insignificant in the curve of the sphere. Nor does it matter how you gauge and try him. A character is like an acrostic or Alexandrian stanza;-read it forward, backward, or across, it still spells the same thing. In this pleasing contrite wood-life which God allows me, let me record day by day my honest thought without prospect or retrospect, and, I cannot doubt, it will be found to be symmetrical, though I mean it not and see it not. My book should smell of pines and resound with the hum of insects. The swallow over my window should interweave that thread or straw he carries in his bill into my web also. We pass for what we are. Character teaches above our wills. Men imagine that they communicate their virtue or vice only by overt actions, and do not see that virtue or vice emit a breath every moment.

Fear never but you shall be consistent in whatever variety of actions, so they each be honest and natural in their hour. For of one will, the actions will be harmonious, however unlike they seem. These varieties are lost sight of when seen at a little distance, at a little height of thought. One tendency unites them all. The voyage of the best ship is a zigzag line of a hundred tacks. This is only microscopic criticism. See the line from a sufficient distance, and it straightens itself to the average tendency. Your genuine action will explain itself and will explain your other genuine actions. Your conformity explains nothing.


text:
Diophantine equations

Abstract
Our purpose in this article which has been published is to show how much diophantine equations are rich in analytic applications. Effectively, those equations allow to build amazing sequences, series and numbers. The question of the proof of some theorems remains of course, we will see it in this communication. We will make also an allusion to the very known Fermat numbers ($$2^{2^n}$$). We will see how this problem of the proof is actual and how it can be solved using amazing sequences and series.

The Ghanouchi's sequences
Let us begin by Fermat equation (E), it is $$U^n=X^n+Y^n$$

with GCD(X,Y)=1

We pose $$u=U^{2n}$$

$$x=U^nX^n$$

$$y=U^nY^n$$

$$z=X^nY^n$$

then

$$u=U^{2n}=U^n(X^n+Y^n)=x+y\quad\quad{(1)}$$

and

$$\frac{1}{z}=\frac{1}{X^nY^n}=\frac{U^{2n}}{U^nX^nU^nY^n}=\frac{u}{xy}=\frac{x+y}{xy}=\frac{1}{x}+\frac{1}{y}\quad\quad{(2)}$$

If U, X, Y are integers verifying equation (E), then u, x, y, z as defined verify

LEMMA 1
$$u=x+y\quad\quad{(1)}$$

$$\frac{1}{z}=\frac{1}{x}+\frac{1}{y}\quad\quad{(2)}$$

Firstly

$$z=X^nY^n$$

$$u=(X^n+Y^n)^2$$

$$x=X^n(X^n+Y^n)$$

$$y=Y^n(X^n+Y^n)$$

We pose

$$x_1=x$$

and

$$y_1=y$$

but

$$\forall{x_1,y_1}$$

$$\exists{z_1}$$

verifying

$$\frac{1}{z_1}=\frac{1}{x_1}+\frac{1}{y_1}$$

and

$$z_1=\frac{xy}{x+y}=z$$

then

$$(x_1+y_1)z_1=x_1y_1$$

or

$$x_1(y_1-z_1)=z_1y_1$$

we pose

$$y_2=y_1-z_1=\frac{z_1y_1}{x_1}$$

and

$$y_1(x_1-z_1)=z_1x_1$$ also

$$x_2=x_1-z_1=\frac{z_1x_1}{y_1}$$ and

$$x_2y_2=z_1^{2}$$

which means

$$x_1=x_2+z_1=x_2+\sqrt{x_2y_2}$$

and $$y_1=y_2+z_1=y_2+\sqrt{x_2y_2}$$

and

$$u_1=u=(x_1+y_1)=(\sqrt{x_2}+\sqrt{y_2})^2>x_2+y_2>0$$

$$(x_1+y_1)$$ is an integer

$$x_1=\sqrt{x_2}(\sqrt{x_2}+\sqrt{y_2})>x_2>0$$ $$x_1$$ is an integer

$$y_1=\sqrt{y_2}(\sqrt{x_2}+\sqrt{y_2})>y_2>0$$

$$y_1$$ is an integer

$$z_1=\frac{x_1y_1}{x_1+y_1}=X^aY^b=\sqrt{x_2y_2}>z_2=\frac{x_2y_2}{x_2+y_2}>0$$

$$z_2$$ is rational, because

$$\forall{x_2,y_2}$$ rationals

$$\exists{z_2}$$ rational verifying

$$\frac{1}{z_2}=\frac{1}{x_2}+\frac{1}{y_2}$$

until infinity. For i

$$(x_i+y_i)=(\sqrt{x_{i+1}}+\sqrt{y_{i+1}})^2>x_{i+1}+y_{i+1}>0$$

$$x_i+y_i$$ is rational for i>1

$$x_i=\sqrt{x_{i+1}}(\sqrt{x_{i+1}}+\sqrt{y_{i+1}})>x_{i+1}>0$$

$$x_i$$ is rational for i>1

$$y_i=\sqrt{y_{i+1}}(\sqrt{x_{i+1}}+\sqrt{y_{i+1}})>y_{i+1}>0$$

$$y_i$$ is rational for i>1

$$z_i=\frac{x_iy_i}{x_i+y_i}=\sqrt{x_{i+1}y_{i+1}}>z_{i+1}=\frac{x_{i+1}y_{i+1}}{x_{i+1}+y_{i+1}}>0$$

$$z_i$$ is rational for i>1, and

$$\frac{1}{z_{i+1}}=\frac{1}{x_{i+1}}+\frac{1}{y_{i+1}}$$

$$(x_i+y_i)=(\sqrt{x_{i+1}}+\sqrt{y_{i+1}})^2$$

LEMMA 2
$$x_i$$ and $$y_i$$ have expressions

$$x_i=x^\prod_{j=0}^{j={i-2}}{(x^+y^)^{-1}}\quad\quad{(H)}$$

$$y_i=y^\prod_{j=0}^{j={i-2}}{(x^+y^)^{-1}}\quad\quad{(H')}$$

Proof of lemma 2
By induction

$$x=\sqrt{x_2}(\sqrt{x_2}+\sqrt{y_2})=\sqrt{x_2}(x+y)^{\frac{1}{2}}$$

$$x_2=\frac{x^{2}}{x+y}$$

also

$$y_2=\frac{y^{2}}{x+y}$$

it is verified for i=2. We suppose (H) and (H') true for i, then

$$x_i=\sqrt{x_{i+1}}(\sqrt{x_{i+1}}+\sqrt{y_{i+1}})=\sqrt{x_{i+1}}(x_i+y_i)^{\frac{1}{2}}$$

and

$$x_{i+1}=x_i^{2}{(x_i+y_i)}^{-1}$$

but (H) and (H'), then

$$x_{i+1}=x^\prod_{j=0}^{j=i-2}{(x^+y^)^{-2}}(x^{2^{i-1}}+y^{2^{i-1}})^{-1}\prod_{j=0}^{j={i-2}}{(x^+y^)}$$

$$=x^\prod_{j=0}^{j={i-1}}{(x^+y^)^{-1}}$$ and it is true for i+1, also for $$y_i$$ $$x_i$$ and $$y_i$$ can be written as it follows

$$x_i=x^{2^{i-1}}\prod_{j=0}^{j=i-2}{(x^+y^)^{-1}}$$

$$y_i=y^{2^{i-1}}\prod_{j=0}^{j=i-2}{(x^+y^)^{-1}}$$

$$\forall{i>1}$$

but, $$\forall{a,b}$$

$$a-b=(a^{2^{i-1}}-b^{2^{i-1}})\prod_{j=0}^{j={i-2}}{(a^{2^j}+b^{2^j})^{-1}}\quad\quad{(E)}$$

$$x-y=(x^-y^)\prod_{j=0}^{j={i-2}}{(x^+y^)^{-1}}\quad\quad{(E')}$$

$$\prod_{j=0}^{j={i-2}}{(x^+y^)}=\frac{(x^-y^)}{x-y}$$

the expressions of the sequences become, for

$$x\neq{y}$$ or

$$x_i\neq{y_i}$$

LEMMA 3
$$x_i=\frac{x^}{x^-y^}(x-y)$$

$$=U^n\frac{X^{n{2^{i-1}}}}{X^{n{2^{i-1}}}-Y^{n{2^{i-1}}}}(X^n-Y^n)$$ and

$$y_i=\frac{y^}{x^-y^}(x-y)$$

$$=U^n\frac{Y^{n{2^{i-1}}}}{X^{n{2^{i-1}}}-Y^{n{2^{i-1}}}}(X^n-Y^n)$$

$$u_i=x_i+y_i=U^n\frac{X^{n{2^{i-1}}}+Y^{n{2^{i-1}}}}{X^{n{2^{i-1}}}-Y^{n{2^{i-1}}}}(X^n-Y^n)$$

LEMMA 4
The equations (1) have (2) a constant

$$x_i-y_i=x-y$$

THE GHANOUCHI'S THEOREM
The only solution of equations

$$u=x+y,\quad((1))$$

and

$$\frac{1}{z}=\frac{1}{x}+\frac{1}{y},\quad{(2)}$$

is $$xy(x-y)=0$$

Proof of Ghanouchi's theorem
As

$$x_i=\frac{x^}{x^-y^}(x-y)$$ and

$$y_i=\frac{y^}{x^-y^}(x-y)$$ if

$$x>y\Rightarrow{\lim_{i\longrightarrow{\infty}}{(x_i)}=x-y,\quad\quad{\lim_{i\longrightarrow{\infty}}{(y_i)}=0}}$$ and $$y>x\Rightarrow{\lim_{i\longrightarrow{\infty}}{(y_i)}=y-x,\quad\quad{\lim_{i\longrightarrow{\infty}}{(x_i)}=0}}$$

We will give several proofs that $$xy(x-y)=0$$ is the solution with the series. We recapitulate

$$x_i>x_{i+1},\quad{y_i>y_{i+1}}\Rightarrow{x=y}$$

$$x_i=x_{i+1}=x_{i+1}+\sqrt{x_{i+1}y_{i+1}}\Rightarrow{xy=0}$$

$$y_i=y_{i+1}=y_{i+1}+\sqrt{x_{i+1}y_{i+1}}\Rightarrow{xy=0}$$

$$\Rightarrow{xy(x-y)=0}$$

x and y are not différent, the initial hypothesis is false there the only solution is

$$xy(x-y)=0$$

The proofs : we have

$$y_i^2-x_i^2=(y-x)(x_i+y_i)=(y_{i+1}-x_{i+1})(x_{i+1}+y_{i+1}+2\sqrt{x_{i+1}y_{i+1}})$$

$$=y_{i+1}^2-x_{i+1}^2+2\sqrt{x_{i+1}y_{i+1}}(y_{i+1}-x_{i+1})$$

$$\sqrt{y_{i+1}}=Z$$ is solution of the following equation

$$Z^4+2\sqrt{x_{i+1}}Z^3-2\sqrt{x_{i+1}^3}Z-x_{i+1}^2+x_i^2-y_i^2=0$$

Also $$\sqrt{x_{i+1}}=Z'$$ is solution of

$${Z'}^4+2\sqrt{y_{i+1}}{Z'}^3-2\sqrt{y_{i+1}^3}Z'-y_{i+1}^2+y_i^2-x_i^2=0$$

And

$$Z^4+2Z'Z^3-2{Z'}^3Z-{Z'}^4+x_i^2-y_i^2=0$$

Also

$${Z'}^4+2Z{Z'}^3-2Z^3Z'-Z^4+y_i^2-x_i^2=0$$

Let

$$Z=u-\frac{Z'}{2}$$

But

$$(u-\frac{Z'}{2})^4+2Z'(u-\frac{Z'}{2})^3-2{Z'}^3(u-\frac{Z'}{2})-{Z'}^4+x_i^2-y_i^2=0$$

Hence

$$u^4+\frac{{Z'}^4}{16}-2u^3{Z'}+\frac{3}{2}u^2{Z'}^2+$$

$$-\frac{1}{2}u{Z'}^3+2Z'u^3-3u^2{Z'}^2+\frac{3}{2}u{Z'}^3-\frac{1}{4}{Z'}^4+$$

$$-2{Z'}^3u+{Z'}^4-{Z'}^4+x_i^2-y_i^2=0$$

$$=u^4-(\frac{3}{2}{Z'}^2)u^2-{Z'}^3u-\frac{3}{16}{Z'}^4+x_i^2-y_i^2=0$$

We deduce

$$(u^2-\frac{3}{4}{Z'}^2)^2-\frac{3}{4}{Z'}^4-{Z'}^3u+x_i^2-y_i^2=0$$

Also $$Z'=v-\frac{Z}{2}$$ leads to

$$(v-\frac{Z}{2})^4+2Z(v-\frac{Z}{2})^3-2{Z}^3(v-\frac{Z}{2})-{Z}^4+y_i^2-x_i^2=0$$

Hence

$$v^4+\frac{{Z}^4}{16}-2v^3{Z}+\frac{3}{2}v^2{Z}^2+$$

$$-\frac{1}{2}v{Z}^3+2Zv^3-3v^2{Z}^2+\frac{3}{2}v{Z}^3-\frac{1}{4}{Z}^4+$$

$$-2{Z}^3v+{Z}^4-{Z}^4+y_i^2-x_i^2=0$$

$$=v^4-(\frac{3}{2}{Z}^2)v^2-{Z}^3v-\frac{3}{16}{Z}^4+y_i^2-x_i^2=0$$

We deduce

$$(v^2-\frac{3}{4}{Z}^2)^2-\frac{3}{4}{Z}^4-{Z}^3v+y_i^2-x_i^2=0$$

If we add

$$(u^2-\frac{3}{4}{Z'}^2)^2+(v^2-\frac{3}{4}Z^2)^2-\frac{3}{4}(Z^4+{Z'}^4)-{Z'}^3u-Z^3v=0$$

$$=(Z^2+\frac{{Z'}^2}{4}+ZZ'-\frac{3}{4}Z^2)^2+({Z'}^2+\frac{Z^2}{4}+Z'Z-\frac{3}{4}{Z'}^2)^2-\frac{3}{4}(Z^4+{Z'}^4)-Z{Z'}^3-Z^3Z'-\frac{{Z'}^4+Z^4}{2}=0$$

$$=(\frac{Z^2+{Z'}^2}{4}+ZZ')^2+(\frac{Z^2+{Z'}^2}{4}+Z'Z)^2-\frac{5}{4}(Z^4+{Z'}^4)-ZZ'(Z^2+{Z'}^2)=0$$

$$=\frac{Z^4+{Z'}^4+2Z^2{Z'}^2}{8}+2Z^2{Z'}^2+ZZ'(Z^2+{Z'}^2)-\frac{5}{4}(Z^4+{Z'}^4)-ZZ'(Z^2+{Z'}^2)=0$$

$$=-\frac{9}{8}(Z^4+{Z'}^4)+\frac{9}{4}Z^2{Z'}^2=0$$

$$=-(Z^2-{Z'}^2)\frac{9}{8}=0$$

The solution is

$$Z^2-{Z'}^2=y-x=0$$

Another proof : we have

$$\sqrt{x_{i-1}}+\sqrt{x_i}=\frac{x_{i-1}-x_i}{\sqrt{x_{i-1}}-\sqrt{x_i}}=\frac{\sqrt{x_iy_i}}{\sqrt{x_{i-1}}-\sqrt{x_i}}$$

$$=\frac{1}{\sqrt{\frac{x_{i-1}}{x_iy_i}}-\sqrt{\frac{x_i}{x_iy_i}}}=\frac{1}{\sqrt{\frac{\sqrt{x_{i-1}+y_{i-1}}}{\sqrt{x_i}y_i}}-\frac{1}{\sqrt{y_i}}}$$

$$=\frac{1}{\frac{\sqrt[4]{x_{i-1}+y_{i-1}}-\sqrt[4]{x_i}}{\sqrt[4]{x_i}\sqrt{y_i}}}=\frac{\sqrt[4]{x_i}\sqrt{y_i}}{\sqrt[4]{x_{i-1}+y_{i-1}}-\frac{\sqrt{x_{i-1}}}{\sqrt[4]{x_{i-1}+y_{i-1}}}}$$

$$=\frac{\sqrt[4]{x_i}\sqrt{y_i}(\sqrt[4]{x_{i-1}+y_{i-1}})}{\sqrt{x_{i-1}+y_{i-1}}-\sqrt{x_{i-1}}}\leq{\frac{\sqrt[4]{x_i}\sqrt{y_i}\sqrt[4]{x_{i-1}+y_{i-1}}}{\sqrt[4]{y_{i+1}}}}$$

Because

$$\sqrt{x_{i-1}+y_{i-1}}-\sqrt{x_{i-1}}=\frac{y_{i-1}}{\sqrt{x_{i-1}+y_{i-1}}+\sqrt{x_{i-1}}}\geq{\sqrt[4]{y_{i+1}}}$$

$$y_{i-1}=\sqrt{y_i}\sqrt{x_{i-1}+y_{i-1}}\geq{\sqrt[4]{y_{i+1}}(\sqrt{x_{i-1}+y_{i-1}}+\sqrt{x_{i-1}})}=\frac{\sqrt{y_i}}{\sqrt[4]{x_i+y_i}}(\sqrt{x_{i-1}+y_{i-1}}+\sqrt{x_{i-1}})$$

And

$$\sqrt[4]{x_i+y_i}\sqrt{x_{i-1}+y_{i-1}}\geq{\sqrt{x_{i-1}+y_{i-1}}+\sqrt{x_{i-1}}}$$

Because

$$\sqrt[4]{x_i+y_i}=\frac{\sqrt[4]{x_{i-1}^2+y_{i-1}^2}}{\sqrt[4]{x_{i-1}+y_{i-1}}}\geq{1+\sqrt{\frac{x_{i-1}}{x_{i-1}+y_{i-1}}}}$$

And

$$\sqrt[4]{x_{i-1}^2+y_{i-1}^2}\geq{\sqrt[4]{x_{i-1}+y_{i-1}}+\sqrt[4]{x_{i-1}+y_{i-1}}\frac{\sqrt{x_{i-1}}}{\sqrt{x_{i-1}+y_{i-1}}}}$$

We deduce

$$\sqrt{x_{i-1}}+\sqrt{x_i}\leq{\frac{\sqrt[4]{x_i}\sqrt{y_i}}{\sqrt[4]{y_{i+1}}}}=\frac{\sqrt[4]{x_i}\sqrt[4]{y_{i+1}}(\sqrt[4]{x_i+y_i})}{\sqrt[4]{y_{i+1}}}=\sqrt[4]{x_i}(\sqrt[4]{x_i+y_i})$$

In the infinity

$$0<\sqrt{x-y}\leq{\lim_{i\longrightarrow{\infty}}{(\sqrt{x_{i-1}}+\sqrt{x_i})}=2\sqrt{x-y}}\leq{\lim_{i\longrightarrow{\infty}}{(\sqrt[4]{x_i}\sqrt[4]{x_i+y_i})}=\sqrt{x-y}}$$

Therefore

$$\sqrt{x-y}=2\sqrt{x-y}=0$$

Another proof : let

$$x_i=a_ix_{i+1}=x_{i+1}+\sqrt{x_{i+1}y_{i+1}}\Rightarrow{a_i=1+\sqrt{\frac{y_{i+1}}{x_{i+1}}}=1+\frac{y_i}{x_i}}$$

Also

$$y_i=b_iy_{i+1}=y_{i+1}+\sqrt{x_{i+1}y_{i+1}}\Rightarrow{b_i=1+\sqrt{\frac{x_{i+1}}{y_{i+1}}}=1+\frac{x_i}{y_i}}$$

We deduce

$$a_i=1+\frac{a_i}{b_i}=1+\frac{y_i}{x_i}$$

And $$b_i=1+\frac{b_i}{a_i}=1+\frac{x_i}{y_i}$$

Thus $$a_ib_i=a_i+b_i$$ And

$$x_i=c_iy_{i+1}=x_{i+1}+\sqrt{x_{i+1}y_{i+1}}\Rightarrow{c_i=b_i(b_i-1)}$$

Also

$$y_i=d_ix_{i+1}=y_{i+1}+\sqrt{x_{i+1}y_{i+1}}\Rightarrow{d_i=a_i(a_i-1)}$$

We have

$$x_i=b_i\sqrt{x_{i+1}y_{i+1}},\quad{y_i=a_i\sqrt{x_{i+1}y_{i+1}}}$$

Let also $$x_i=i{\alpha}_i,\quad{y_i=i{\beta}_i}$$

And

$${\alpha}_i=a'_i{\alpha}_{i+1},\quad{{\beta}_i=b'_i{\beta}_{i+1}}$$ We deduce

$$\frac{a_i}{b_i}=\frac{y_i}{x_i}=\frac{{\beta}_i}{{\alpha}_i}=\frac{a'_i}{b'_i}$$ Because

$$i{\alpha}_i=x_i=x_{i+1}+\sqrt{x_{i+1}y_{i+1}}=(i+1)({\alpha}_{i+1}+\sqrt{{\alpha}_{i+1}{\beta}_{i+1}})$$

$$i{\beta}_i=y_i=y_{i+1}+\sqrt{x_{i+1}y_{i+1}}=(i+1)({\beta}_{i+1}+\sqrt{{\alpha}_{i+1}{\beta}_{i+1}})$$ Or

$$\frac{{\alpha}_i}{{\beta}_i}=\sqrt{\frac{{\alpha}_{i+1}}{{\beta}_{i+1}}}=\frac{a'_i{\alpha}_{i+1}}{b'_i{\beta}_{i+1}}$$ And

$$b'_i=\frac{i+1}{i}b_i\neq{2},\quad{a'_i=\frac{i+1}{i}a_i\neq{2}}$$ But

$$b'_i-2=\frac{{\alpha}_i}{\sqrt{{\alpha}_{i+1}{\beta}_{i+1}}}-2=a'_i\frac{{\alpha}_i}{{\beta}_i}-2$$

$$=(a'_i-2)\frac{x_i}{y_i}+2(\frac{x_i}{y_i}-1)$$ And

$$a'_i-2=\frac{{\beta}_i}{\sqrt{{\alpha}_{i+1}{\beta}_{i+1}}}-2=b'_i\frac{{\beta}_i}{{\alpha}_i}-2$$

$$=(b'_i-2)\frac{y_i}{x_i}+2(\frac{y_i}{x_i}-1)$$ Thus

$$(a'_i-2)^2\frac{x_i}{y_i}+2(\frac{x_i}{y_i}-1)(a'_i-2)$$

$$=(b'_i-2)^2\frac{y_i}{x_i}+2(\frac{y_i}{x_i}-1)(b'_i-2)$$

$$=((a'_i-2)\frac{x_i}{y_i}+2(\frac{x_i}{y_i}-1))^2\frac{y_i}{x_i}+2(\frac{y_i}{x_i}-1)((a'_i-2)\frac{x_i}{y_i}+2(\frac{y_i}{x_i}-1))$$

$$=(a'_i-2)^2\frac{x_i}{y_i}+4(\frac{x_i}{y_i}-1)^2\frac{y_i}{x_i}+4(\frac{x_i}{y_i}-1)(a'_i-2)+2(1-\frac{x_i}{y_i})(a'_i-2)+4(\frac{y_i}{x_i}-1)^2$$

$$=(a'_i-2)^2\frac{x_i}{y_i}+4(\frac{x_i}{y_i}-1)^2\frac{y_i}{x_i}+2(\frac{x_i}{y_i}-1)(a'_i-2)+4(\frac{y_i}{x_i}-1)^2$$ Hence

$$4(\frac{x_i}{y_i}-1)^2\frac{y_i}{x_i}+4(\frac{y_i}{x_i}-1)^2=0$$

$$=4(\frac{(x-y)^2}{y_i^2})\frac{y_i}{x_i}+4(\frac{(x-y)^2}{x_i^2})=0$$

$$=4(x-y)^2(\frac{1}{x_iy_i}+\frac{1}{x_i^2})=0$$

$$\Rightarrow{(x-y)^2=0}$$

Another proof : We have

$$c_id_i=a_ib_i$$

Let

$$a_{i-1}b_{i-1}=f_{i}a_ib_i$$

We have

$$f_{i}=\frac{a_{i-1}b_{i-1}}{a_ib_i}=\frac{x_{i-1}y_{i-1}x_{i+1}y_{i+1}}{x_i^2y_i^2}$$

$$=\frac{\sqrt{x_iy_i}(\sqrt{x_i}+\sqrt{y_i})^2x_{i+1}y_{i+1}}{x_{i+1}y_{i+1}(x_i+y_i)^2}\leq{1}$$

$$a_ib_i=\frac{a_i}{b_i}+\frac{b_i}{a_i}+2=(\sqrt{\frac{a_i}{b_i}}+\sqrt{\frac{b_i}{a_i}})^2=(a_{i-1}b_{i-1}-2)^2$$

And

$$a_ib_i=e_i=(f_{i}a_ib_i-2)^2$$

It means

$$e_i^2f_i^2-(4f_i+1)e_i+4=0$$

And

$$e_i=a_ib_i=\frac{4f_i+1+\sqrt{8f_i+1}}{2f_i^2}$$

But

$$\sqrt{a_ib_i}=f_ia_ib_i-2\Rightarrow{f_ia_ib_i-\sqrt{a_ib_i}-2=0}$$

Thus

$$\sqrt{e_i}=\sqrt{a_ib_i}=\frac{1+\sqrt{1+8f_i}}{2f_i}$$

And

$$\sqrt{e_i}-2=\frac{1-f_i+\sqrt{1+8f_i}-3f_i}{2f_i}=\frac{1-f_i+\frac{1-f_i^2+8f_i(1-f_i)}{\sqrt{1+8f_i}+3f_i}}{2f_i}$$

$$=(1-f_i)\frac{1+\frac{1+9f_i}{\sqrt{8f_i1}+3f_i}}{2f_i}$$

And

$$e_i-4=a_ib_i-4=(\sqrt{e_i}-2)(\sqrt{e_i}+2)=(\sqrt{e_i}-2)f_ia_ib_i=(1-f_i)f_ia_ib_i$$

$$\frac{1-f_i^2+4f_i(1-f_i)+\frac{1-f_i^2+8f_i(1-f_i)}{\sqrt{8f_i+1}+3f_i^2}}{2f_i^2}$$

$$=(1-f_i)\frac{1+5f_i+\frac{1+9f_i}{\sqrt{8f_i+1}+3f_i^2}}{2f_i^2}$$

$$(1-f_i)(f_ia_ib_i-\frac{1+5f_i+\frac{1+9f_i}{\sqrt{1+8f_i}+3f_i^2}}{2f_i^2})=0$$

Or

$$(1-f_i)(a_ib_i-\frac{1+5f_i+\frac{1+9f_i}{\sqrt{1+f_i}+3f_i^2}}{2f_i^3})=0$$

$$=(1-f_i)(\frac{f_i(1+4f_i)+f_i\sqrt{8f_i+1}-1-5f_i-\frac{1+9f_i}{\sqrt{1+8f_i}+3f_i^2}}{2f_i^3})=0$$

The expression between the parenthesis is not equal to zero, we deduce

$$f_i=1$$

Else

$$\sqrt{a_ib_i}=\frac{2}{f_i\sqrt{a_ib_i}-1}$$

$$\sqrt{a_ib_i}-2=2(\frac{2-f_i\sqrt{a_ib_i}}{f_i\sqrt{a_ib_i}})=2(\frac{2(1-f_i)+f_i(2-\sqrt{a_ib_i})}{f_i\sqrt{a_ib_i}})$$

We deduce

$$(\sqrt{a_ib_i}-2)(1+\frac{2}{\sqrt{a_ib_i}})=(\sqrt{a_ib_i}-2)(\frac{\sqrt{a_ib_i}+2}{\sqrt{a_ib_i}})$$

$$=(\sqrt{a_ib_i}-2)(\frac{f_ia_ib_i}{\sqrt{a_ib_i}})=(\sqrt{a_ib_i}-2)f_i\sqrt{a_ib_i}=4\frac{1-f_i}{f_i\sqrt{a_ib_i}}$$

Thus

$$(\frac{1+\sqrt{1+8f_i}}{2f_i}-2)f_i\sqrt{a_ib_i}=(1-f_i)(\frac{1+\frac{1+9f_i}{\sqrt{1+8f_i}+3f_i}}{2f_i})=(1-f_i)\frac{4}{f_i\sqrt{a_ib_i}}$$

The expressions between the parenthesis are not equal, therefore

$$f_i=1$$

Else

$$a_i-4=(\sqrt{e_i}-2)(\sqrt{e_i}+2)=(1-f_i)(\frac{1+\frac{1+9f_i}{\sqrt{1+8f_i}+3f_i}}{2f_i})(\frac{1+8f_i+\sqrt{1+8f_i}}{2f_i})$$

Hence

$$(1-f_i)(\frac{1+5f_i+\frac{1+9f_i}{\sqrt{1+8f_i}+3f_i^2}}{2f_i^2}-(\frac{1+\frac{1+9f_i}{\sqrt{1+8f_i}+3f_i}}{2f_i})(\frac{1+8f_i+\sqrt{8f_i+1}}{2f_i}))=0$$

Or

$$(1-f_i)(2+10f_i+2\frac{1+9f_i}{\sqrt{1+8f_i}+3f_i^2}-(1+8f_i+\frac{(1+9f_i)\sqrt{8f_i+1}}{\sqrt{8f_i+}+3f_i}+\sqrt{8f_i+1}+\frac{(1+8f_i)(1+9f_i)}{\sqrt{8f_i+1}+3f_i}))=0$$

The expression between the parenthesis is not equal to zero, we deduce

$$f_i=1\Rightarrow{x_iy_i(a_i-b_i-4)=(x-y)^2=0}$$

And, else

$$2f_i^2e_i=4f_i+1+\sqrt{8f_i+1}$$

$$(2f_i^2e_i-4f_i-1)^2=8f_i+1=4f_i^4e_i^2+16f_i^2-16f_i^3e_i-4f_i^2e_i+8f_i+1$$

$$f_i^2e_i^2+4-4f_ie_i-e_i=0$$

$$(f_i^2e_i-4f_i-1)e_i=-4$$

$$e_i=\frac{4}{1+4f_i-f_i^2e_i}$$

$$e_i-4=\frac{-16f_i+4f_i^2e_i}{1+4f_i-f_i^2e_i}$$

$$=\frac{-16f_i+16f_i^2+4f_i^2(e_i-4)}{1+4f_i-f_i^2e_i}$$

$$(e_i-4)(1-\frac{4f_i^2}{1+4f_i-f_i^2e_i})$$

$$=(f_i-1)\frac{16f_i}{1+4f_i-f_i^2e_i}=4f_ie_i(f_i-1)$$

$$=(e_i-4)(1-f_i^2e_i)$$

$$e_i=\frac{4f_i+1+\sqrt{8f_i+1}}{2f_i^2}$$

$$e_i-4=\frac{4f_i(1-f_i)+1-f_i^2+\sqrt{8f_i+1}-3f_i^2}{2f_i^2}$$

$$=\frac{4f_i(1-f_i)+(1-f_i)(1+f_i)+\frac{8f_i-8f_i^4+1-f_i^4}{\sqrt{8f_i+1}+3f_i^2}}{2f_i^2}$$

$$=(1-f_i)(\frac{5f_i+1+\frac{9f_i^3+9f_i^2+9f_i+1}{\sqrt{8f_i+1}+3f_i^2}}{2f_i^2})$$

$$(f_i-1)(4f_ie_i+(1-f_i^2e_i)(\frac{5f_i+1+\frac{9f_i^3+9f_i^2+9f_i+1}{\sqrt{8f_i+1}+3f_i^2}}{2f_i^2}))=0$$

$$=(f_i-1)(\frac{4(4f_i+1+\sqrt{8f_i+1})}{2f_i}+(1-\frac{4f_i+1+\sqrt{8f_i+1}}{2})(\frac{5f_i+1+\frac{9f_i^3+9f_i^2+9f_i+1}{\sqrt{8f_i+1}+3f_i^2}}{2f_i^2}))=0$$

$$=(f_i-1)(\frac{5f_i+1+\frac{9f_i^3+9f_i^2+9f_i+1}{\sqrt{8f_i+1}+3f_i^2}}{2f_i^2}-(\frac{4f_i+1+\sqrt{8f_i+1}}{2})(\frac{-3f_i+1+\frac{9f_i^3+9f_i^2+9f_i+1}{\sqrt{8f_i+1}+3f_i^2}}{2f_i^2}))=0$$

And the expression between the parenthesis is not equal to zero, it means that

$$f_i=1\Rightarrow{a_ib_i=e_i=4}$$

$$a_i^2b_i^2-4a_ib_i=(a_i-b_i)^2=0=(y_i-x_i)^2x_{i+1}y_{i+1}$$

And

$$(a_i-b_i)\sqrt{x_{i+1}y_{i+1}}=y-x=0$$

Another proof :

$$x_i$$ and $$y_i$$ are particular cases of $$v_i$$ and $$w_i$$ which follow

$$v_i=\frac{x^{i}}{x^{i}-y^{i}}(x-y),\quad{w_i=\frac{y^{i}}{x^{i}-y^{i}}(x-y)}$$

$$x_i=v_{2^{i-1}},\quad{y_i=w_{2^{i-1}}}$$

Also

$$e_i=\frac{v_i+w_i}{w_{i}},\quad{f_i=\frac{v_i+w_i}{v_i}}$$

$$e_i=\frac{e_i+f_i}{f_i},\quad{f_i=\frac{e_i+f_i}{e_i}}$$

$$b_i=e_{2^{i-1}},\quad{a_i=f_{2^{i-1}}}$$

So, we have

$$v_i-w_i=x-y$$

And

$$e_if_i=e_i+f_i$$

But

$$b_1b_2...b_i=1+\frac{e_1}{f_1}+...+\frac{e_{2^{i}-1}}{f_{2^{i}-1}}$$

Let

$$1+\frac{e_1}{f_1}+\frac{e_2}{f_2}+...+\frac{e_i}{f_i}=1+\frac{v_1}{w_1}+\frac{v_2}{w_2}+...+\frac{v_i}{w_i}$$

$$=1+\frac{x}{y}+\frac{x^{2}}{y^{2}}+...+\frac{x^{i}}{y^{i}}=\frac{1-\frac{x^{(i+1)}}{y^{(i+1)}}}{1-\frac{x}{y}}=\frac{2-e_{i+1}}{2-e_1}$$

Thus

$$e_{i+1}-2=\frac{e_{i+1}-f_{i+1}}{f_{i+1}}=\frac{v_{i+1}-w_{i+1}}{w_{i+1}}=\frac{x-y}{w_{i+1}}$$

$$=(e_1-2)(1+(e_1-1)+(e_2-1)+...+(e_i-1))=(e_1-2)(1+i+(e_1-2)+(e_2-2)+...+(e_i-2))$$

$$=(\frac{e_1-f_1}{f_1})(1+i+\frac{e_1-f_1}{f_1}+\frac{e_2-f_2}{f_2}+...+\frac{e_i-f_i}{f_i})$$

$$=(\frac{v_1-w_1}{w_1})(1+i+\frac{v_1-w_1}{w_1}+\frac{v_2-w_2}{w_2}+...+\frac{v_i-w_i}{w_i})$$

Or

$$(x-y)(\frac{1}{w_{i+1}}-\frac{1}{w_1}(1+i+\frac{x-y}{w_1}+\frac{x-y}{w_2}+...+\frac{x-y}{w_i}))=0$$

And

$$\frac{1}{w_{i+1}}-(1+i)\frac{1}{w_1}$$

is not equal to zero for $$x-y=0$$, and it is the case, of course, of

$$-\frac{1}{w_1}(x-y)(\frac{1}{w_1}+...+\frac{1}{w_i})$$

therefore the expression between the parenthesis is not equal to zero. We gave the fourth proof that the only solution is

$$xy(x-y)=0$$

And there are others (see the series).

The Ghanouchi's series
As seen $$\sqrt{x_iy_i}=y_{i-1}-y_i=x_{i-1}-x_i$$ we deduce $$x_i-x_{i+1}=\sqrt{x_{i+1}y_{i+1}}$$

$$x_{i-1}-x_i=\sqrt{x_iy_i}$$

$$x_1-x_2=x-x_2=\sqrt{x_2y_2}$$

then

$$\sum_{j=2}^{j=i+1}{(\sqrt{x_jy_j})}=x-x_2+x_2-x_3+...+x_i-x_{i+1}=x-x_{i+1}$$ and

$$\sum_{j=2}^{j=\infty}{(\sqrt{x_jy_j})}=\lim_{i\longrightarrow{\infty}}{(x-x_{i+1})}$$

if $$x>y$$

$$\sum_{j=2}^{j=\infty}{(\sqrt{x_jy_j})}=\lim_{i\longrightarrow{\infty}}{(x-x_{i+1})}=x-(x-y)=y$$

if $$x<y$$

$$\sum_{j=2}^{j=\infty}{(\sqrt{x_jy_j})}=\lim_{i\longrightarrow{\infty}}{(x-x_{i+1})}=x$$

The applications of the Ghanouchi's series
$$\sum_{j=2}^{j=i}{((-1)^j\sqrt{x_jy_j})}=x-x_2-(x_2-x_3)+(x_3-x_4)-...+(-1)^i(x_{i-1}-x_i)$$

$$=x-2x_2+2x_3-...+2(-1)^{i-1}x_{i-1}+(-1)^{i+1}x_i$$ $$=2\sum_{j=2}^{j=i-1}{((-1)^{j+1}x_j)}+x+(-1)^{i+1}x_i$$

$$=2\sum_{j=1}^{j=i}{((-1)^{j+1}x_j)}-x-(-1)^{i+1}x_i$$

$$=\sum_{j=2}^{j=i-1}{((-1)^{j+1}x_j)}+\sum_{j=1}^{j=i}{((-1)^{j+1}x_j)}$$

$$=2\sum_{j=2}^{j=i-1}{((-1)^{j+1}y_j)}+y+(-1)^{i+1}y_i$$

$$=2\sum_{j=1}^{j=i}{((-1)^{j+1}y_j)}-y-(-1)^{i+1}y_i$$

$$=\sum_{j=2}^{j=i-1}{((-1)^{j+1}y_j)}+\sum_{j=1}^{j=i}{((-1)^{j+1}y_j)}$$ or $$2\sum_{j=1}^{j=i}{((-1)^{j+1}x_j)}=\sum_{j=2}^{j=i}{((-1)^j\sqrt{x_jy_j})}+x+(-1)^{i+1}x_i$$ and $$2\sum_{j=1}^{j=i}{((-1)^{j+1}y_j)}=\sum_{j=2}^{j=i}{((-1)^j\sqrt{x_jy_j})}+y+(-1)^{i+1}y_i$$ As we do not know the limit of $$(-1)^{i+1}x_i$$, then $$\sum_{j=1}^{j=\infty}{((-1)^{j}x_j)}$$

can be not convergent. But $$\sum_{j=2}^{j=\infty}{((-1)^j\sqrt{x_jy_j})}$$

is convergent.

Also, knowing that $$y_i$$ tends to zero in the infinity, we can say $$\sum_{j=1}^{j=\infty}{((-1)^{j}y_j)}$$

is convergent. The limit of $$\sum_{j=2}^{j=i}{((-1)^j\sqrt{x_jy_j})}=2\sum_{j=1}^{j=i}{((-1)^{j+1}y_j)}-y-(-1)^{i+1}y_i$$ $$=2\sum_{j=1}^{j=i}{((-1)^{j+1}x_j)}-x-(-1)^{i+1}x_i$$

exists and the series are convergent. It means that for x and y integers and for conditions on the exponents like $$n\geq{3}$$ for Fermat equation : $$\lim_{i\longrightarrow{\infty}}{(x_i)}=x-y=0$$

It is confirmed by the fact that the général term of the series tends to zero. Let us prove it. We give two proofs. We must remark that we prove firstly that the following series are convergent, we do not present the proof, here. Let

$$\sum_{k=1}^{k=2m}{((-1)^{k+1}x_{k}e^{-\frac{k}{\sqrt{2m}}})}$$

$$=xe^{-\frac{1}{\sqrt{2m}}}-x_2e^{-\frac{2}{\sqrt{2m}}}+x_3e^{-\frac{3}{\sqrt{2m}}}-...+(-1)^{2m+1}x_{2m}e^{-\frac{2m}{\sqrt{2m}}}$$

$$=xe^{-\frac{2}{\sqrt{2m}}}+x(e^{-\frac{1}{\sqrt{2m}}}-e^{-\frac{2}{\sqrt{2m}}})-x_2e^{-\frac{2}{\sqrt{2m}}}+x_3e^{-\frac{4}{\sqrt{2m}}}+x_3(e^{-\frac{3}{\sqrt{2m}}}-e^{-\frac{4}{\sqrt{2m}}})-x_4e^{-\frac{4}{\sqrt{2m}}}+...-x_{2m}e^{-\frac{2m}{\sqrt{2m}}}$$

$$=xe^{-\frac{2}{\sqrt{2m}}}(e^{\frac{1}{\sqrt{2m}}}-1)+x_3e^{-\frac{4}{\sqrt{2m}}}(e^{\frac{1}{\sqrt{2m}}}-1)+...+x_{2m-1}e^{-\frac{2m}{\sqrt{2m}}}(e^{\frac{1}{\sqrt{2m}}}-1)+$$

$$+(x-x_2)e^{-\frac{2}{\sqrt{2m}}}+(x_3-x_4)e^{-\frac{4}{\sqrt{2m}}}+...+(x_{2m-1}-x_{2m})e^{-\frac{2m}{\sqrt{2m}}}$$

$$=(e^{\frac{1}{\sqrt{2m}}}-1)(xe^{-\frac{2}{\sqrt{2m}}}+x_3e^{-\frac{4}{\sqrt{2m}}}+...+x_{2m-1}e^{-\sqrt{2m}})+(\sqrt{x_2y_2}e^{-\frac{2}{\sqrt{2m}}}+\sqrt{x_4y_4}e^{-\frac{4}{\sqrt{2m}}}+...+\sqrt{x_{2m}y_{2m}}e^{-\frac{2m}{\sqrt{2m}}})$$ $$=(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k}{\sqrt{2m}}})}+\sum_{k=1}^{k=m}{(\sqrt{x_{2k}y_{2k}}e^{-\frac{2k}{\sqrt{2m}}})}$$ Also $$\sum_{k=1}^{k=2m}{((-1)^{k+1}y_{k}e^{-\frac{k}{\sqrt{2m}}})}$$ $$=ye^{-\frac{1}{\sqrt{2m}}}-y_2e^{-\frac{2}{\sqrt{2m}}}+y_3e^{-\frac{3}{\sqrt{2m}}}-...+(-1)^{2m+1}y_{2m}e^{-\frac{2m}{\sqrt{2m}}}$$ $$=(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k}{\sqrt{2m}}})}+\sum_{k=1}^{k=m}{(\sqrt{x_{2k}y_{2k}}e^{-\frac{2k}{\sqrt{2m}}})}$$ But $$(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k}{\sqrt{2m}}})}=S$$ We have $$(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k+1}{\sqrt{2m}}})}<S<(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k-1}{\sqrt{2m}}})}$$ $$(e^{\frac{1}{\sqrt{2m}}}-1)e^{-\frac{3}{\sqrt{2m}}}\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})}<S<(e^{\frac{1}{\sqrt{2m}}}-1)e^{-\frac{1}{\sqrt{2m}}}\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})}$$ Thus $$\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})})}=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)y+(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=2}^{k=m}{(y_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})})}$$ $$=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=2}^{k=m}{(y_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})})}=\lim_{m\longrightarrow{\infty}}{(S)}$$ We have $$(e^{\frac{1}{\sqrt2m}}-1)\sum_{k=p}^{k=m}{(y_{2k-1}e^{-\frac{2k-p}{\sqrt{2m}}})}=A$$ And $$(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(y_{2k-1}e^{-\frac{2k-p+1}{\sqrt{2m}}})}<A<(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(y_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})}$$ Or $$(e^{\frac{1}{\sqrt{2m}}}-1)e^{-\frac{2}{\sqrt{2m}}}\sum_{k=p}^{k=m}{(y_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})}<A<(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(y_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})}$$ Hance $$\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(y_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})})}$$ $$=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)y_{2p-1}e^{-\frac{p-1}{\sqrt{2m}}}+(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p+1}^{k=m}{(y_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})})}$$ $$=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p+1}^{k=m}{(y_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})})}=\lim_{m\longrightarrow{\infty}}{(A)}=\lim_{m\longrightarrow{\infty}}{(S)}$$ And $$p+1=m\Rightarrow{\lim_{m\longrightarrow{\infty}}{(A)}=\lim_{m\longrightarrow{\infty}}{(S)}=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)y_{2m-1}e^{-\frac{m}{\sqrt{2m}}})}=0}$$ Consequently $$\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k}{\sqrt{2m}}})})}=0$$ Also $$(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k}{\sqrt{2m}}})}=S$$ We have $$(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k+1}{\sqrt{2m}}})}<S<(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k-1}{\sqrt{2m}}})}$$ $$(e^{\frac{1}{\sqrt{2m}}}-1)e^{-\frac{3}{\sqrt{2m}}}\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})}<S<(e^{\frac{1}{\sqrt{2m}}}-1)e^{-\frac{1}{\sqrt{2m}}}\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})}$$ Thus $$\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})})}=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)x+(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=2}^{k=m}{(x_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})})}$$ $$=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=2}^{k=m}{(x_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})})}=\lim_{m\longrightarrow{\infty}}{(S)}$$ And $$(e^{\frac{1}{\sqrt2m}}-1)\sum_{k=p}^{k=m}{(x_{2k-1}e^{-\frac{2k-p}{\sqrt{2m}}})}=A$$ And $$(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(x_{2k-1}e^{-\frac{2k-p+1}{\sqrt{2m}}})}<A<(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(x_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})}$$ Or $$(e^{\frac{1}{\sqrt{2m}}}-1)e^{-\frac{2}{\sqrt{2m}}}\sum_{k=p}^{k=m}{(x_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})}<A<(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(x_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})}$$ Hence $$\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(x_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})})}$$ $$=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)x_{2p-1}e^{-\frac{p-1}{\sqrt{2m}}}+(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p+1}^{k=m}{(x_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})})}$$ $$=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p+1}^{k=m}{(x_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})})}=\lim_{m\longrightarrow{\infty}}{(A)}=\lim_{m\longrightarrow{\infty}}{(S)}$$ Or $$p+1=m\Rightarrow{\lim_{m\longrightarrow{\infty}}{(A)}=\lim_{m\longrightarrow{\infty}}{(S)}=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)x_{2m-1}e^{-\frac{m}{\sqrt{2m}}})}=0}$$ Consequently $$\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k}{\sqrt{2m}}})})}=0$$

We deduce

$$0<\lim_{m\longrightarrow{\infty}}{(\sum_{k=1}^{k=2m}{((-1)^{k+1}x_{k}e^{-\frac{k}{\sqrt{2m}}})})}$$

$$=\lim_{m\longrightarrow{\infty}}{(\sum_{k=1}^{k=2m}{((-1)^{k+1}x_{k}e^{-\frac{k}{\sqrt{2m}}})})}$$

$$=\lim_{m\longrightarrow{\infty}}{(\sum_{k=1}^{k=m}{(\sqrt{x_{2k}y_{2k}}e^{-\frac{2k}{\sqrt{2m}}})})}<\lim_{m\longrightarrow{\infty}}{(\sum_{k=1}^{k=m}{(\sqrt{x_{2k}y_{2k}})})}$$

$$<\lim_{m\longrightarrow{\infty}}{(\sum_{k=1}^{k=2m}{(\sqrt{x_{k}y_{k}})})}=y$$

Thus

$$\lim_{m\longrightarrow{\infty}}{(\sum_{k=1}^{k=2m}{((-1)^{k+1}(x_k-y_k)e^{-\frac{k}{\sqrt{2m}}})})}=0$$

$$=\lim_{m\longrightarrow{\infty}}{((x-y)\sum_{k=1}^{k=2m}{(e^{-\frac{k}{\sqrt{2m}}})})}=\lim_{m\longrightarrow{\infty}}{((x-y)e^{-\frac{1}{\sqrt{2m}}}\frac{1-e^{-\sqrt{2m}}}{1+e^{-\frac{1}{\sqrt{2m}}}})}=\frac{x-y}{2}=0$$

And

$$x-y=0$$

the only solution is $$xy(x-y)=0$$, $$xy=0$$ if at least one of the sequences $$x_i$$ or $$y_i$$ is constant. And the second proof. Let

$$x_i(t)=\sum_{k=2}^{k=i-1}{(x_ke^{-kt})}=\sum_{k=2}^{k=i}{(x_ke^{-kt})}-x_ie^{-it}={x'}_i(t)-x_ie^{-it}$$

$$y_i(t)=\sum_{k=2}^{k=i-1}{(y_ke^{-kt})}=\sum_{k=2}^{k=i}{(y_ke^{-kt})}-y_ie^{-it}={y'}_i(t)-y_ie^{-it}$$

$$u_i(t)=\sum_{k=2}^{k=i}{(\sqrt{x_ky_k}e^{-kt})}=\sum_{k=2}^{k=i-1}{((x_{k-1}-x_k)e^{-kt})}$$

$$=xe^{-2t}+x_2(-e^{-2t}+e^{-3t})+x_3(-e^{-3t}+e^{-4t})+...+x_{i-1}(-e^{-(i-1)t}+e^{-it})-x_ie^{-it}$$

$$=xe^{-2t}+x_2e^{-2t}(-1+e^{-t})+x_3e^{-3t}(-1+e^{-t})+...+x_{i-1}e^{-(i-1)t}(-1+e^{-t})-x_ie^{-it}$$

$$=xe^{-2t}-x_ie^{-it}+(-1+e^{-t})(x_2e^{-2t}+x_3e^{-3t}+...+x_ie^{-(i-1)t})$$

$$=xe^{-2t}-x_ie^{-it}+(-1+e^{-t})x_i(t)=xe^{-2t}-x_ie^{-it}+(-1+e^{-t}){x'}_i(t)-(-1+e^{-t})x_ie^{-it}=xe^{-2t}-x_ie^{-(i+1)t}+(-1+e^{-t}){x'}_i(t)$$

let

$$t=\frac{1}{\sqrt{i}}\Rightarrow{\lim_{i\longrightarrow{\infty}}{(u_i(\frac{1}{\sqrt{i}}))}=\lim_{i\longrightarrow{\infty}}{(\sum_{k=2}^{k=i}{(\sqrt{x_ky_k}e^{-\frac{k}{\sqrt{i}}})})}}=\lim_{i\longrightarrow{\infty}}{(\sum_{k=2}^{k=i}{(\sqrt{x_ky_k})})}=y$$

$$=x+\lim_{i\longrightarrow{\infty}}{((-1+e^{-\frac{1}{\sqrt{i}}})\sum_{k=2}^{k=i-1}{(x_ke^{-\frac{k}{\sqrt{i}}})}-x_ie^{-\sqrt{i}})}=x$$

$$=y+\lim_{i\longrightarrow{\infty}}{((-1+e^{-\frac{1}{\sqrt{i}}})\sum_{k=2}^{k=i-1}{(y_ke^{-\frac{k}{\sqrt{i}}})}-y_ie^{-\sqrt{i}})}=y$$

We Recapitulate

$$x_i>x_{i+1},\quad{y_i>y_{i+1}}\Rightarrow{x=y}$$

$$x_i=x_{i+1}=x_{i+1}+\sqrt{x_{i+1}y_{i+1}}\Rightarrow{xy=0}$$

$$y_i=y_{i+1}=y_{i+1}+\sqrt{x_{i+1}y_{i+1}}\Rightarrow{xy=0}$$

$$\Rightarrow{xy(x-y)=0}$$ is the only solution of (1) and (2). This result is paradoxal, we remark that we have not put any condition on n, because there are solutions for $$n\leq{2}$$. The answer is related to Matiasevic theorem which claims that dos not exist an algorithm to prove theorems related to diophantine equations and we gave one : The approach must conduct then to an impossibility. We confirm Matiasevic theorem and prove it because our algorithm is available for n=1. The approach is more important than it appears, it is an answer to problems more general than Fermat theorem or Beal or Fermat-Catalan conjectures. We will try to prove them. The series become

$$\sum_{k=2}^{k={\infty}}{((-1)^k\sqrt{x_ky_k})}=\sum_{k=2}^{k={\infty}}{((-1)^kx_k)}=\sum_{k=2}^{k={\infty}}{((-1)^ky_k)}$$

$$=x_2-x_3+x_4-x_5+...$$

$$=2\sum_{k=1}^{k={\infty}}{((-1)^{k+1}x_k)}-x=x-2x_2+2x_3-2x_4+...$$

$$=2\sum_{k=1}^{k={\infty}}{((-1)^{k+1}y_k)}-y=y-2y_2+2y_3-2y_4+...$$

$$\Rightarrow{3(x_2-x_3+x_4-x_5+...)=3\sum_{k=2}^{k={\infty}}{((-1)^kx_k)}=3\sum_{k=2}^{k={\infty}}{((-1)^ky_k)}=x=y}$$

$$\Rightarrow{x_2-x_3+x_4-x_5+...=\sum_{k=2}^{k={\infty}}{((-1)^kx_k)}=\sum_{k=2}^{k={\infty}}{((-1)^ky_k)}=\frac{x}{3}=\frac{y}{3}}$$

and

$$\sum_{k=2}^{k={\infty}}{(\sqrt{x_ky_k})}=\sum_{k=2}^{k={\infty}}{(x_k)}=\sum_{k=2}^{k={i}}{(y_k)}$$

$$=x_2+x_3+x_4+x_5+...=y_2+y_3+y_4+y_5+...=y=x$$

and

$$x_i=x^{2^{i-1}}\prod_{j=0}^{j=i-2}{(x^+y^)^{-1}}=y_i$$

$$=x^{2^{i-1}}\prod_{j=0}^{j=i-2}{(2x^{2^j})^{-1}}=\frac{x^{2^{i-1}}}{2^{i-1}x^{2^{i-1}-1}}$$

$$=x_i=\frac{x}{2^{i-1}}=y_i=\frac{y}{2^{i-1}}$$ This development is in fact a test of impossibility. The sequences and series are a consequence of Fermat equation and of other diophantine equations (as we will see). The question now is : why are there solutions for n=2 ? The answer is in the formulas, as seen. It is important to note that for n=1, there are trivial solutions. But, for n=1, lemma 3 allows to write

$$x_i=\frac{x^}{x^-y^}(x-y)=U\frac{X^}{X^-Y^}(X-Y)$$

$$=U\frac{X^{2{2^{i-2}}}}{X^{2{2^{i-2}}}-Y^{2{2^{i-2}}}}(X-Y)$$ and

$$y_i=\frac{y^}{x^-y^}(x-y)=U\frac{Y^}{X^-Y^}(X-Y)$$

$$=U\frac{Y^{2{2^{i-2}}}}{X^{2{2^{i-2}}}-Y^{2{2^{i-2}}}}(X-Y)$$

and

$$u_i=x_i+y_i=U\frac{X^{2{2^{i-2}}}+Y^{2{2^{i-2}}}}{X^{2{2^{i-2}}}-Y^{2{2^{i-2}}}}(X-Y)=\frac{X^{2{2^{i-2}}}+Y^{2{2^{i-2}}}}{X^{2{2^{i-2}}}-Y^{2{2^{i-2}}}}(X^2-Y^2)$$

It is the expression of $${u'}_{i-1}$$, the $$u_{i-1}$$ exponent 2.

The case n=1 conducts to the case n=2 and as there are solutions for n=1, it will be the same for n=2 !

For n=4

$$u_i=x_i+y_i=\frac{X^{4{2^{i-3}}}+Y^{4{2^{i-3}}}}{X^{4{2^{i-3}}}-Y^{4{2^{i-3}}}}U(X-Y)$$

the case n=4 is different, in this formula the exponent i-3 does not guarantee the existence of the sequences if i=2.

Then, the case n=2 is the only exception. The only solution for n>2 is xy(x-y)=0, there is no solution.

Another application is Beal equation. It is

$$U^c=X^a+Y^b$$

$$GCD(X,Y)=1$$

We pose $$u=U^{2c}$$

$$x=U^cX^a$$

$$y=U^cY^b$$

$$z=X^aY^b$$

and

$$u=U^{2c}=U^c(X^a+Y^b)=x+y$$

and

$$\frac{1}{z}=\frac{1}{X^aY^b}=\frac{U^{2c}}{U^cX^aU^cY^b}=\frac{u}{xy}=\frac{x+y}{xy}=\frac{1}{x}+\frac{1}{y}$$

it is lemma 1 and, with

$$U^c=X^a+Y^b$$

the solutions are

$$xy(x-y)=U^{2c}X^aY^b(U^cX^a-U^cY^b)=0$$

or impossible solutions for a>2 et b>2 et c>2.

Another application is the following equation.

$$U^n=X_1^{n_1}+X_2^{n_2}+...+X_i^{n_i}$$

$$GCD(X_j,X_k)=1;\forall{j,k,j\neq{k}}$$

We conjecture and prove that there is no solutions for n>i(i-1) and $$n_j>i(i-1)$$, $$\forall{j\in\{1,2,...,i\}}$$. We can not know when there are solutions as proved by Matiasevic when one of the exponent is less or equal to i(i-1).

LEMMA 6
The solution $$\forall{k\geq{2}}$$ is

$$X_k=0$$

Proof of lemma 6
We pose

$$u=U^{2n}$$

$$x=U^nX_k^{n_k}$$

$$y=U^n(U^n-X_k^{n_k})$$

$$z=X_k^{n_k}(U^n-X_k^{n_k})$$

or

$$u=x+y$$

and

$$\frac{1}{z}=\frac{1}{x}+\frac{1}{y}$$

it is lemma 1. Its solution is

$$U=X_k=0;\forall{k\in{\{1,2,...,i\}}}$$

But, why are not they solutions for n>i(i-1) and $$n_j>i(i-1)$$ ?

We will generalize the definition of the sequences.

We will define general sequences. Our goal is to prove that if $$X_i$$,

($$i\geq{2}$$), $$n_i$$, $$U$$, $$n$$, $$PGCD(X_1,X_2,...,X_i)=1$$

are positive integers, then

$$X_a=X_b=0;\forall{a,b=1,2,...,i}$$

$$n_k>{i(i-1)},\forall{k=1,2,...,i},n>i(i-1)$$

for the equation

$$X_1^{n_1}+X_2^{n_2}+...+X_i^{n_i}=U^n\quad\quad{(e)}$$

When $$n\leq{i(i-1)},n_k\leq{i(i-1)},k=1,2,...,i$$ there are solutions, for example : $$i=2$$ has $$3^2+4^2=5^2$$ and $$i=3$$ has $$3^3+4^3+5^3=6^3$$

and

$$95800^4+217519^4+414560^4=422481^4$$

and $$i=4$$ has $$27^5+84^5+110^5+133^5=144^5$$

etc...

We suppose (e) verified and that $$GCD(X_k)=1$$

$$k\in\{1,2,...,i\};i\geq{2}$$, soit

$$x_k=U^{(i-1)n}X_k^{n_k}$$

$$k\in{\{1,2,...,i\}}$$

and

$$u=U^{in}$$

and

$$v=X_1^{n_1}X_2^{n_2}...X_i^{n_i}$$

LEMMA 7
$$x_k$$, $$k=1,2,...i$$, u, v verify

$$x_1+x_2+...+x_i=U^{(i-1)n}(X_1^{n_1}+X_2^{n_2}+...+X_i^{n_i})=U^{in}=u\quad\quad{(3)}$$

and

$$\frac{1}{v}=\frac{1}{X_1^{n_1}X_2^{n_2}...X_i^{n_i}}=\frac{U^{{i(i-1)}n}}{U^{(i-1)n}X_1^{n_1}U^{(i-1)n}X_2^{n_2}...U^{(i-1)n}X_i^{n_i}}=\frac{u^{(i-1)}}{x_1x_2...x_i}\quad\quad{(4)}$$

We pose

$$x_{k,0}=x_k$$

$$u_0=u$$

$$v_0=v$$

and

$$x_{k,1}=x_k^{i}(x_1+x_2+...+x_i)^{-(i-1)}$$

$$k=1,2,...,i$$

it implies

$$u=x_1+x_2+...+x_i=(x_{1,1}^{\frac{1}{i}}+x_{2,1}^{\frac{1}{i}}+...+x_{i,1}^{\frac{1}{i}})^i>u_1>1$$

and

$$x_{k,0}=x_k=x_{k,1}^{\frac{1}{i}}(x_1+x_2+...+x_i)^{\frac{(i-1)}{i}}$$

$$=x_{k,1}^{\frac{1}{i}}(x_{1,1}^{\frac{1}{i}}+x_{2,1}^{\frac{1}{i}}+...+x_{i,1}^{\frac{1}{i}})^{(i-1)}>x_{k,1}>0$$

and

$$v=\frac{x_{1,0}x_{2,0}...x_{i,0}}{u^{(i-1)}}=x_{1,1}^{\frac{1}{i}}x_{2,1}^{\frac{1}{i}}...x_{i,1}^{\frac{1}{i}}>v_1=\frac{x_{1,1}x_{2,1}...x_{i,1}}{u_1^{(i-1)}}>0$$

The reasoning is available until infinity. Then

$$u_j=x_{1,j}+x_{2,j}+...+x_{i,j}=(x_{1,{j+1}}^{\frac{1}{i}}+x_{2,{j+1}}^{\frac{1}{i}}+...+x_{i,{j+1}}^{\frac{1}{i}})^i>u_{j+1}>0$$

and

$$x_{k,j}=x_{k,{j+1}}^{\frac{1}{i}}(x_{1,{j+1}}^{\frac{1}{i}}+x_{2,{j+1}}^{\frac{1}{i}}+...+x_{i,{j+1}}^{\frac{1}{i}})^{(i-1)}>x_{k,{j+1}}>0$$

and

$$k=1,2,...,i$$

and

$$v_j=\frac{x_{1,j}x_{2,j}...x_{i,j}}{u_j^{(i-1)}}=x_{1,{j+1}}^{\frac{1}{i}}x_{2,{j+1}}^{\frac{1}{i}}...x_{i,{j+1}}^{\frac{1}{i}}>v_{j+1}=\frac{x_{1,{j+1}}x_{2,{j+1}}...x_{i,{j+1}}}{u_{j+1}^{(i-1)}}>0$$

$$x_{k,j}, v_j, u_j$$ are positive $$\forall{j>1}$$, $$\forall{k=1,2,...,i}$$.

LEMMA 8
(P) is the expression :

$$x_{k,j}=x_k^(\prod_{l=0}^{l={j-1}}{x_1^+x_2^+...+x_i^})^{-(i-1)}$$

Proof of lemma 8
for j=1, it is verified because of the definition of $$x_{k,1}, u_1, v_1$$,

we suppose (P) true and the expression of $$u_j$$ implies, with (P), that

$$x_{k,j}=x_{k,{j+1}}^{\frac{1}{i}}(x_{1,{j+1}}^{\frac{1}{i}}+x_{2,{j+1}}^{\frac{1}{3}}+...+x_{i,{j+1}}^{\frac{1}{i}})^{(i-1)}$$

or

$$x_{k,{j+1}}^{\frac{1}{i}}=x_{k,j}(x_{1,{j+1}}^{\frac{1}{i}}+x_{2,{j+1}}^{\frac{1}{i}}+...+x_{i,{j+1}}^{\frac{1}{i}})^{-(i-1)}$$

and

$$x_{k,{j+1}}=x_{k,j}^{i}(x_{1,{j+1}}^{\frac{}{i}}+x_{2,{j+1}}^{\frac{1}{i}}+...+x_{i,{j+1}}^{\frac{1}{i}})^=x_{k,j}^{i}(x_{1,j}+x_{2,j}+...+x_{i,j})^{-(i-1)}$$

$$=x_k^\prod_{l=0}^{l={j-1}}{(x_1^+x_2^+...+x_i^)^{-{i(i-1)}}(x_1^+x_2^+...+x_i^)^{-(i-1)}}\prod_{l=0}^{l={j-1}}{(x_1^+x_2^+...+x_i^)^{(i-1)^2}}$$

$$=x_k^\prod_{l=0}^{l=j}{(x_1^+x_2^+...+x_i^)^{-(i-1)}}$$

then

$$x_{k,j}=x_k^\prod_{l=0}^{l={j-1}}{(x_1^+x_2^+...+x_i^)^{-(i-1)}}$$ Why are not they solutions for $$n>i(i-1), n_k>i(i-1)$$ ?

We suppose

$$n=n_k=i(i-1)$$

the formula becomes

$$x_{k,j}=x_k^(\prod_{l=0}^{l={j-1}}{x_1^+x_2^+...+x_i^})^{-(i-1)}$$

$$=U^{n{i^j}}X_k^{n_k{i^j}}(\prod_{l=0}^{l={j-1}}{(U^{n{i^l}}X_1^{n_1{i^l}}+U^{n{i^l}}X_2^{n_2{i^l}}+...+U^{n{i^l}}X_i^{n_i{i^l}})})^{-(i-1)}$$

$$=U^{i(i-1){i^j}}X_k^{i(i-1){i^j}}(\prod_{l=0}^{l={j-1}}{(U^{i(i-1){i^l}}X_1^{i(i-1){i^l}}+...+U^{i(i-1){i^l}}X_i^{i(i-1){i^l}})})^{-(i-1)}$$

$$=U^{(i-1)i^{j+1}}X_k^{(i-1)i^{j+1}}(\prod_{l=0}^{l={j-1}}{(U^{(i-1){i^{l+1}}}x_1^{(i-1){i^{l+1}}}+...+U^{(i-1){i^{l+1}}}X_i^{(i-1){i^{l+1}}})})^{-(i-1)}$$

It is the expression of $$x_{k,j+1}$$ of the exponent i-1.

If we suppose that exist solutions for the exponent i-1, there will exist solutions for an exposant not greater than i(i-1).

Some times, we must make attention to the initial change of the data. For example, let the following equationd

$$kU^n=X^n+Y^n$$

for some k integers like 7, there are solutions, for others like 2, there is no solution. It is too easy toi pose

$$u=(kU^n)^2$$

$$x=kU^nX^n$$

$$y=kU^nY^n$$

$$z=X^nY^n$$

the lemma 1 is satisfied

$$u=kU^n(X^n+Y^n)=x+y$$

$$\frac{1}{z}=\frac{k^2U^{2n}}{kU^nX^nkU^nY^n}=\frac{u}{xy}=\frac{1}{x}+\frac{1}{y}$$

The correct solution is to pose

$$u=U^{2n}$$

$$x=U^nX^n$$

$$y=U^nY^n$$

$$z=X^nY^n$$

Like this

$$u=U^{2n}=\frac{kU^n(X^n+y^n)}{k^2}=\frac{x+y}{k}$$

and

$$\frac{1}{z}=\frac{U^{2n}}{U^nX^nU^nY^n}=\frac{u}{xy}=\frac{x+y}{kxy}=\frac{1}{kx}+\frac{1}{ky}$$

Conclusion
The conclusion is that Ghanouchi's sequences, series and numbers have several applications in all diophantine equations, we saw some of them and there are many others like Pilai equation, Smarandache equation, the Catalan equation, etc...

title: Diophantine equations


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