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 had scared that alligator. But that was the last they ever saw of Claude.

“Pretty soon after that, though, there was a shark scare at the bathing beaches along the Gulf coast. It seemed to be a lone shark that kept annoying women bathers, especially blondes; and they knew it was Claude Jackson. He was always hell after blondes.”

Fairchild ceased. The niece squealed and jumped up and came to him, patting his back. Jenny’s round ineffable eyes were upon him, utterly without thought. The Semitic man was slumped in his chair: he may have slept.

Major Ayers stared at Fairchild a long time. At last he said: “But why does the alligator one wear congress boots?”

Fairchild mused a moment. Then he said dramatically: “He’s got webbed feet.”

“Yes,” Major Ayers agreed. He mused in turn. “But this chap that got rich—” The niece squealed again. She sat beside Fairchild and regarded him with admiration.

“Go on, go on,” she said, “about the one that stole the money, you know.”

Fairchild looked at her kindly. Into the silence there came a thin saccharine strain. “There’s the victrola,” he said. “Let’s go up and start a dance.”

“The one who stole the money,” she insisted. “Please.” She put her hand on his shoulder.

“Some other time,” he promised, rising. “Let’s go up and dance now.” The Semitic man yet slumped in his chair, and Fairchild shook him. “Wake up, Julius. I’m safe now.”

The Semitic man opened his eyes and Major Ayers said: “How much did they gain with their fish ranching?”

“Not as much as they would have with a patent nicetasting laxative. All Americans don’t eat fish, you know. Come

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text: Page:The Pharsalia of Lucan; (IA cu31924026485809).pdf/66

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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 local dances and sat on their front porches of a Sunday evening in vain.

“Pretty soon he could outswim the old sheep, and having to dive so much after the young ones, he got to where he could stay under water longer and longer at a time. Sometimes he’d stay under for a half an hour or more. And pretty soon he got to where he’d stay in the water all day, only coming out to eat and sleep; and then they noticed that Claude’s skin was beginning to look funny and that he walked kind of peculiar, like his knees were stiff or something. Soon after that he quit coming out of the water at all, even to eat, so they’d bring his dinner down to the water and leave it, and after a while he’d swim up and get it. Sometimes they wouldn’t see Claude for days. But he was still catching those sheep, herding ’em into a pen old man Jackson had built in a shallow bayou and fenced off with hog wire, and his half of the money was growing in the bank. Occasionally half eaten pieces of sheep would float ashore, and old man Jackson decided alligators were getting ’em again. But he couldn’t put horns on ’em now because no one but Claude could catch ’em, and he hadn’t seen Claude in some time.

“It had been a couple of weeks since anybody had seen Claude, when one day there was a big commotion in the sheep pen. Old man Jackson and a couple of his other boys ran down there, and when they got there they could see the sheep jumping out of the water every which way, trying to get on land again; and after a while a big alligator rushed out from among ’em, and old man Jackson knew what had scared the sheep.

“And then, right behind the alligator he saw Claude. Claude’s eyes had kind of shifted around to the side of his head and his mouth had spread back a good way, and his teeth had got longer. And then old man Jackson knew what

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 still there, though, because they would occasionally hear ’em baa-ing at night way back in the swamp. They caught one occasionally on a trotline of shark hooks baited with ears of corn. But not many.

“Well, sir, the more old man Jackson thought about that swampful of sheep, the madder he got. He’d stamp around the house and swear he’d catch ’em if he had to buy a motor boat that would run fifty miles an hour, and a diving suit for himself and every one of his boys. He had one boy named Claude—Al’s brother, you know. Claude was kind of wild: hell after women, a gambler and a drunkard—a kind of handsome humorless fellow with lots of dash. And finally Claude made a trade with his father to have half of every sheep he could catch, and he got to work right away. He never bothered with boats or trotlines: he just took off his clothes and went right in the water and grappled for ’em.”

“Grappled for ’em?” Major Ayers repeated.

“Sure: run one down and hem him up under the bank and drag him out with his bare hands. That was Claude, all over. And then they found that this year’s lambs didn’t have any wool on ’em at all, and that its flesh was the best fish eating in Louisiana; being partly cornfed that way giving it a good flavor, you see. So that’s where old man Jackson quit the sheep business and went to fish ranching on a large scale. He knew he had a snap as long as Claude could catch ’em, so he made arrangements with the New Orleans markets right away, and they began to get rich.”

“By Jove,” Major Ayers said tensely, his mind taking fire.

“Claude liked the work. It was an adventurous kind of life that just suited him, so he quit everything and gave all his time to it. He quit drinking and gambling and running around at night, and there was a marked decrease in vice in that neighborhood, and the young girls pined for him at the

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text: Page:The Pharsalia of Lucan; (IA cu31924026485809).pdf/67


text:
User talk:Undekagon30

--Jan Kameníček (talk) 20:57, 10 February 2024 (UTC)

Translations
Hello. I have understood here that you want to translate some book to English and publish it in English Wikisource. This is possible, but often quite complicated, especially for new users. Per WS:Translations it is necessary to upload the publication in the original language (which must be in the public domain) to Commons and proofread it at the appropriate language Wikisource (Marathi Wikisource in our case) using the proofread extension. It is not enough if the Marathi Wikisource contains a copypasted text of the book, it explicitely must be proofread using the proofread extension. Only after this is done, it can be translated to English and hosted here. -- Jan Kameníček (talk) 21:08, 10 February 2024 (UTC)

 * Okay, and thank you for the information. Which format is acceptable as an original resouces. If I find a digital version at archives.org and somehow get the scanned pages linked here, is that adequate?
 * Or what else properties does it have to have? Because I already saw differences in spelling / wording in different digital versions online in some blogs. I also own a printed paper version given by a Marathi friend. I will go happily through the proofread process and try to find collaborators for it because I will learn myself a lot about the book. Undekagon30 (talk) 11:51, 11 February 2024 (UTC)
 * Great! The proofreading should definitely be based on scans of a printed version (often available at archive.org, or you can scan the book yourself if it is in the public domain), not on second-hand digital transcriptions from the web. The scans need to be uploaded to Commons so that they can be used to create the Index page using the proofreading extension. If there are more printed editions of the work, you can choose any. --Jan Kameníček (talk) 13:15, 11 February 2024 (UTC)
 * Okay, thank you for the clarifications. Will do so. Undekagon30 (talk) 07:57, 12 February 2024 (UTC)

title: User talk:Undekagon30


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Page:Mosquitos (Faulkner).pdf/252

 Or, lacking that, some other and more fortunate poet would divide a weekend or so with you: there seems to be a noblesse oblige among them,” the other answered. “Gentleman poets, that is,” he added.

“No,” said Fairchild, indefatigable, “I’d intersperse my book with photographs and art studies on ineffable morons in bathing suits or clutching imitation lace window curtains across their middles. That’s what I’d do.”

“That would damn it as Art,” Mark Frost objected.

“You're confusing Art with Studio Life, Mark,” Mrs. Wiseman told him. She forestalled him and accepted a cigarette. “I’m all out, myself. Sorry. Thanks.”

“Why not?” Mark Frost responded. “If studio life costs you enough, it becomes art. You have to have a good reason to give to your people back home in Ohio or Indiana or somewhere.”

“But everybody wasn’t born in the Ohio valley, thank God,” the Semitic man said. Fairchild stared at him, kind and puzzled, a trifle belligerent. “I speak for those of us who read books instead of write them,” he explained. “It’s bad enough to grow into the conviction after you reach the age of discretion that you are to spend the rest of your life writing books, but to have your very infancy darkened by the possibility that you may have to write the Great American Novel ”

“Oh,” Fairchild said. “Well, maybe you are like me, and prefer a live poet to the writings of any man.”

“Make it a dead poet, and I’ll agree.”

“Well” He settled his spectacles. “Listen to this”: Mark Frost groaned, rising, and departed. Fairchild read implacably:

“‘On rose and peach their droppings bled,

Love a sacrifice has lain,

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 Beneath his hand his mouth is slain,

Beneath his hand his mouth is dead—’

“No: wait.” He skipped back up the page. Mrs. Wiseman listened restively, her brother with his customary quizzical phlegm.

“ ‘The Raven bleak and Philomel

Amid the bleeding trees were fixed,

His hoarse cry and hers were mixed

And through the dark their droppings fell

“Upon the red erupted rose,

Upon the broken branch of peach

Blurred with scented mouths, that each

To another sing, and close—’”

He read the entire poem through. “What do you make of it?” he asked.

“Mostly words,” the Semitic man answered promptly, “a sort of cocktail of words. I imagine you get quite a jolt from it, if your taste is educated to cocktails.”

“Well, why not?” Mrs. Wiseman said with fierce protectiveness. “Only fools require ideas in verse.”

“Perhaps so,” her brother admitted. “But there’s no nourishment in electricity, as you poets nowadays seem to believe.”

“Well, what would you have them write about, then?” she demanded. “There’s only one possible subject to write anything about. What is there worth the effort and despair of writing about, except love and death?”

“That’s the feminine of it. You’d better let art alone and stick to artists, as is your nature.”

“But women have done some good things,” Fairchild objected. “I’ve read—”

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 “They bear geniuses. But do you think they care anything about the pictures and music their children produce? That they have any other emotion than a fierce tolerance of the vagaries of the child? Do you think Shakespeare’s mother was any prouder of him than, say, Tom o’ Bedlam’s?”

“Certainly she was,” Mrs. Wiseman said. “Shakespeare made money.”

“You made a bad choice for comparison,” Fairchild said.

“All artists are kind of insane. Don’t you think so?” he asked Mrs. Wiseman.

“Yes,” she snapped. “Almost as insane as the ones that sit around and talk about them.”

“Well—” Fairchild stared again at the page under his hand. He said slowly: “It’s a kind of dark thing. It’s kind of like somebody brings you to a dark door. Will you enter that room, or not?”

“But the old fellows got you into the room first,” the Semitic man said. “Then they asked you if you wanted to go out or not.”

“I don’t know. There are rooms, dark rooms, that they didn’t know anything about at all. Freud and these other—”

“Discovered them just in time to supply our shelterless literati with free sleeping quarters. But you and Eva just agreed that subject, substance, doesn’t signify in verse, that the best poetry is just words.”

“Yes infatuation with words,” Fairchild agreed. “That’s when you hammer out good poetry, great poetry. A kind of singing rhythm in the world that you get into without knowing it, like a swimmer gets into a current. WordsI had it once.”

“Shut up, Dawson,” Mrs. Wiseman said. “Julius can afford to be a fool.”

“Words,” repeated Fairchild. “But it’s gone out of me,

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text:
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RV 440 (ANSWER.])


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 she said. He didn’t move, and she added: “I haven’t got on a bathing suit. Go away a minute, David.”

But he didn’t move. He leaned over the rail, looking at her with a dumb and utter longing and after a while she slid quickly and easily into the tender, and still he remained motionless, making no move to help her as her grave simple body came swiftly aboard the yacht. “Be back in a minute,” she said over her shoulder and her startling white bathing suit sped across the deck and out of the ken of his dog’s eyes. The mist without thinning was filling with light: an imminence of dawn like a glory, a splendor of trumpets unheard.

Her minute was three minutes. She reappeared in her little colored linen dress, her dark coarse hair still damp, carrying her shoes and stockings in her hand. He hadn’t moved at all.

“Well, let’s get going,” she said. She looked at him impatiently. “Aren't you ready yet?” He stirred at last, watching her with the passive abjectness of a dog. “Come on,” she said sharply. “Haven’t you got the stuff for breakfast yet? What’s the matter with you, David? Snap out of your trance.” She examined him again, with a sober impersonality. “You didn‘t believe I was going to do it—is that it? Or are you backing out yourself? Come on, say so now, if you want to call it of.” She came nearer, examining his face with her grave opaque eyes. She extended her hand. “David?”

He took her hand slowly, looking at her, and she grasped his hand and shook his arm sharply. “Wake up. Say, you haven’t— Come on, let’s get some stuff for breakfast, and beat it. We haven’t got all day.”

He followed her and in the galley she switched on the light and chose a flat box of bacon and a loaf of bread, putting them on a table and delving again among boxes and lockers

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 and shelves. “Have you got matches? a knife?” she asked over her shoulder. “And—where are oranges? Let’s take some oranges. I love oranges, don’t you?” She turned her head to look at him. His hand was just touching her sleeve, so diffidently that she had not felt it. She turned suddenly, putting the oranges down, and put her arms about him, hard and firm and sexless, drawing his cheek down to her sober moist kiss. She could feel his hammering erratic heart against her breast, could hear it surging in the silence almost as though it were in her own body. His arms tightened and he moved his head, seeking her mouth, but she evaded him with a quick movement, without reproof.

“No, no, not that. Everybody does that.” She strained him against her hard body again, then released him. “Come on, now. Have you got everything?” She examined the shelves again, finding at last a small basket. It was filled with damp lettuce but she dumped the lettuce out and put her things in it. “You take my shoes. They’ll go in your pocket, won’t they?” She crumpled her limp blonde stockings into her slippers and gave them to him. Then she picked up the basket and snapped off the light.

Day was a nearer thing yet, though it was not quite come. Though the mist had not thinned, the yacht was visible from stem to stern, asleep like a gull with folded wings; and against the hull the water sighed a long awakening sigh. The shoreline was darker, a more palpable vagueness in the mist.

“Say,” she remarked, stopping suddenly, “how are we going to get ashore? I forgot that. We don’t want to take the tender.”

“Swim,” he suggested. Her dark damp head came just to his chin and she mused for a time in a sober consternation.

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226 lost his post as finance minister, but was compelled to flee for his life. He shared the fall of the Duke of Braganza, whose popularity was hateful to Alfonso’s successor. Don Isaac escaped to Castile in 1484, and, amid the friendly smiles of the cultured Jews of Toledo, set himself to resume the literary work he had been forced to lay aside while burdened with affairs of state. He began the compilation of commentaries on the historical books of the Bible, but he was not long left to his studies. Ferdinand and Isabella, under the very eyes of Torquemada and the Inquisition, entrusted the finances of their kingdom to the Jew Abarbanel during the years 1484 to 1492.

In the latter year, Abarbanel was driven from Spain in the general expulsion instigated by the Inquisition. He found a temporary asylum in Naples, where he also received a state appointment. But he was soon forced to flee again, this time to Corfu. “My wife, my sons, and my books


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Page:Thomas Patrick Hughes - Notes on Muhammadanism - 2ed. (1877).djvu/184

Muharram (lit. "that which is sacred") commences on the first of the month of that name, and is continued for ten days, the tenth day being called ʾA′shuráa. They are days of mátam, or lamentation, in commemoration of the martyrdom of ʾAli, and of Hasan,


text: Lays and Legends of Various Nations/Germany Part 3/Legends of Rubezahl/How Rubezahl Entertained a Party of Guests


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 164 and Husain, as observed by the Shíaʾhs; but the day ʾA′shuráa (the tenth) is also held sacred by the Sunnis, the observance of the month having been enjoined by Muhammad on account of its having been the month of creation.

The ceremonies of the Muharram differ much in different places; but the following are the main features of the festival as observed by the Shíaʾhs. A place is prepared which is called the ʾA′shúr-khána (the ten-day house), or Imám-Bára (the Imám place), in the centre of which is dug a pit, in which fires are kindled, and at night the people, young and old, fence across the fire with sticks and swords, and whilst dancing round it, call out, "Oh ʾAli! noble Hasan! noble Husain! bridegroom! alas friend! stay! stay! etc.; the cry being repeated in the most excited manner hundreds of times, until the whole assembly has reached the highest pitch of excitement. They then form


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Rh themselves in circles, and beat themselves with chains in the most frantic manner. The women repeat a funeral eulogium, and the Maulavís, the Rowzat-us-Shuhádáa, or the Book of Martyrs.

On the seventh day there are representations of the marriage ceremony of Qásim, and of the martyrdom of Husain; and on the eighth day a lance or spear is carried about the city to represent Husain's head, which was carried on the point of a javelin by order of Yazíd. In addition to these representations, there are the Tázías, Tábúts, or biers, of the tombs of Hasan and Husain, a horse-shoe in representation of Husain's swift horse, and the standards of Hasan, Husain, and Qásim, and other Muslim celebrities.

The Sunní Muhammadans do not usually take part in these ceremonies, but observe the tenth day, ʾA′shuráa, being the day on which God is said to have created Adam and Eve, heaven, hell, the tablet of decree, the pen, fate, life, and death.

Muhammad commanded his followers to observe the ʾA′shuráa by bathing, wearing new


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 166 clothes, applying surma to the eyes, fasting, prayers, making peace with one's enemies, associating with religious persons, relieving orphans, and giving of alms.

The fast of ʾA′shuráa is a Sunnat fast, i. e. not founded upon an injunction in the Qurán, but upon the example of Muhammad.


text: Notes on Muhammadanism/Chapter 33

title: Notes on Muhammadanism/Chapter 33


text: Lays and Legends of Various Nations/Germany Part 3/Legends of Rubezahl/Rubezahl Makes a Fool of a Noble


text: Lays and Legends of Various Nations/Germany Part 3/Legends of Rubezahl/Rubezahl Transforms Himself into an Ass


text: Lays and Legends of Various Nations/Germany Part 3/Legends of Rubezahl/Rubezahl Sells Pigs


text: Lays and Legends of Various Nations/Germany Part 3/Legends of Rubezahl/Rubezahl Gives Orders for a Cloak


text: Lays and Legends of Various Nations/Germany Part 3/Legends of Rubezahl/Rubezahl Turns Wood-Cutter


text: Lays and Legends of Various Nations/Germany Part 3/Legends of Rubezahl/Rubezahl Changes Himself into a Spear


text: Lays and Legends of Various Nations/Germany Part 3/Legends of Rubezahl/Rubezahl Changes Leaves into Ducats


text:
Translation:Disquisitiones Arithmeticae/Second Section

Recherches_arithmétiques

Article 13
Preliminary theorems about prime numbers, factorizations, etc.

The product of two positive numbers, each of which is smaller than a given prime, cannot be divisible by this prime.

Let $p$ be the prime, and let $a$  be a positive number $1,$ for if $b=1,$  then we would have $ab=a<p,$  so $ab$  would not be divisible by $p.$  Now, as a prime number, $p$  cannot be divisible by $b,$  and must instead fall between two multiples of $b, mb$  and $(m+1)b.$  Let $p-mb=b'.$  Then $b'$  will be positive and $<b.$  Since we have assumed $ab\equiv 0\pmod{p},$  it is also true that $mab\equiv 0$  (Art. 7), and hence subtracting from $ap\equiv 0$  we will have $a(p-mb)=ab'\equiv 0$ ; i.e. $b'$  is included among the numbers $b,$  $c,$  $d$  etc., even though $b$  was supposed to be the smallest of them. Q.E.A.

Article 14
If neither $a$ nor $b$  is divisible by a given prime number $p,$  then the product $ab$  will also not be divisible by $p$ .

Let $\alpha$ and $\beta$  be the minimal positive residues of the numbers $a$  and $b$  modulo $p.$  By hypothesis, neither of these will be zero. Now, if it were true that $ab\equiv 0 \pmod{p},$ then since $ab\equiv\alpha\beta,$  it would also be true that $\alpha\beta\equiv 0.$  But this is impossible, by the previous theorem.

The proof of this theorem has already been given by Euclid, Elements, vii, 32. However, we did not want to omit it, both because several modern authors have presented vague reasoning instead of a proof, or have neglected this theorem; and also because the nature of the method used here, which we will later apply to much more difficult problems, can be more easily understood from a simpler case.

Article 15
If none of the numbers $a, b, c, d, \dots$ is divisible by the prime number $p,$  then the product $abcd$  etc. will also not be divisible by $p$ .

According to the previous article, $ab$ is not divisible by $p,$  so the same is true for $abc,$  hence also for $abcd,$  etc.

Article 16
A composite number can only be resolved into prime factors in one way.

Proof. It is evident from first principles that any number can always be decomposed into prime factors; but it is often tacitly assumed that this decomposition is only possible in one way. Let us imagine a composite number $A=a^\alpha b^\beta c^\gamma$ etc., where $a,$  $b,$  $c$  etc. are unequal prime numbers, and suppose that $$A$$ can be resolved into prime factors in yet another way. It is first of all clear that in this second system of factors, primes other than $a,$ $b,$  $c$  etc. cannot enter, since these are the only primes which divide $A$. Similarly, none of the prime numbers $a,$ $b,$  $c$  etc. can be missing, because if that were the case, the missing primes would not divide $A$  (Art. 15); thus the two factorizations can only differ insofar as a prime is repeated more times in one than in the other. Let $p$ be a prime number, which is repeated $m$  times in one system and $n$  times in the other, with $m>n$. Now let the factor $p$ be removed from each system $n$  times, so that it remains in one system $m-n$  times, and has disappeared from the other. Then the number $\frac{A}{p^n}$ has two different factorizations, one of which is completely free of the factor $p$, while the other still contains it $m-n$  times, contrary to what we have just shown.

Article 17
Therefore, if a composite number $A$ is the product of $B,$  $C,$  $D$  etc., then it is clear that the numbers $B,$  $C,$  $D$  etc. cannot have prime factors different from those of $A,$  and that each of these factors must occur as many times altogether in the numbers $B,$  $C,$  $D$  etc. as it does in $A.$  From this, one obtains a criterion to determine whether one number $B$  divides another number $A.$  This will happen if $B$  does not involve any prime more times than $A;$  if this condition is not satisfied, then $B$  will not divide $A.$

With the help of the calculus of combinations, it is easy to see that if $A=a^\alpha b^\beta c^\gamma$ etc., where as above $a,$  $b,$  $c$  etc. denote distinct prime numbers, then $A$  has

$$(\alpha+1)(\beta+1)(\gamma+1)\dots$$

distinct divisors, including $1$ and $A.$

Article 18
It is therefore clear that if $A=a^\alpha b^\beta c^\gamma$ etc., $K=k^\kappa l^\lambda m^\mu$  etc., and if the primes $a,$  $b,$  $c$  etc., $k,$  $l,$  $m$  etc. are all distinct from one another, then $A$  and $K$  will have no common divisor other than $1,$  or in other words they will be relatively prime.

Given several numbers $A,$ $B,$  $C$  etc., their greatest common divisor is determined as follows. The numbers are resolved into their prime factors, and those that are common to all the numbers $A,$ $B,$  $C$  etc. are taken (if there are none, the given numbers would have no common divisor). Then one notes how many times each of these prime factors is contained in $A,$ $B,$  $C,$  etc., or equivalently the exponents of these factor in each of the numbers $A,$  $B,$  $C,$  etc. Then each factor is given the smallest of the exponents that it has in $A,$ $B,$  $C,$  etc. Finally, the product of the resulting powers is computed; this will be the greatest common divisor.

On the other hand, if the least common multiple of the numbers $A,$ $B,$  $C,$  etc. is desired, one can proceed as follows. Collect all the prime numbers that divide any of the numbers $A,$ $B,$  $C,$  etc., and assign to each of them the largest exponent that it has in the numbers $A,$  $B,$  $C,$  etc. The product of all these powers will be the required common multiple.

Ex. Let $A=504=2^3 3^2 7,$ $B=2880=2^6 3^2 5,$  $ C=864=2^5 3^3$. To find the greatest common divisor, the prime factors $2$ and $3$  which must be assigned exponents $3$  and $2;$  this gives the greatest common divisor $2^3 3^2 =72;$  the least common multiple will be $2^6 3^3 5. 7=60480.$

We omit the proofs because of their simplicity; after all, it is known how to solve these problems in an elementary way, even when the factorizations of the numbers $A,$ $B,$  $C,$  etc. are not given.

Article 19
If the numbers $a,$ $b,$  $c$  etc. are relatively prime to $k,$  then their product will also be relatively prime to $k.$

Indeed, because none of the numbers $a,$ $b,$  $c$  etc. has prime factors in common with $k,$  the product $abc$  etc. cannot have prime factors that do not belong to any of $a,$  $b,$  $c$  etc. Thus from the previous article, $abc$  etc. will be relatively prime to $k.$

If the numbers $a,$ $b,$  $c$  etc. are relatively prime to each other, and $k$  is divisible by each of them, then $k$  is also divisible by their product.

This is easily derived from Arts. 17 and 18. Let $p$ be any prime divisor of the product $abc$  etc., which is repeated $\pi$  times. Then it is clear that one of the numbers $a,$ $b,$  $c$  etc. will be divisible by $p^\pi,$  and consequently $k,$  which is divisible by this number, will also be divisible by $p^\pi$. The same reasoning holds for the remaining divisors of the product $abc$ etc.

Therefore, if two numbers $m,$ $n$  are congruent modulo several relatively prime moduli $a,$  $b,$  $c$  etc. then they will also be congruent modulo the product. Indeed, since $m-n$  is divisible by each of the numbers $a,$  $b,$  $c$  etc., it must also be divisible by their product.

Finally, if $a$ is relatively prime to $b,$  and $ak$  is divisible by $b,$  then $k$  will also be divisible by $b.$  Indeed, since $ak$  is divisible by $a$  and $b,$  it will be divisible by their product, i.e. $\frac{ak}{ab}=\frac{k}{b}$  will be an integer.

Article 20
If $A=a^\alpha b^\beta c^\gamma$ etc., where $a,$  $b,$  $c$  etc. are distinct prime numbers, and $A$  is equal to a perfect power $k^n,$  then all of the exponents $\alpha,$  $\beta,$  $\gamma, $  etc. will be divisible by $n.$

The number $k$ does not involve prime numbers other than $a,$  $b,$  $c$  etc. Let $\alpha'$  be the exponent of $a$  in $k.$   Then $k^n$  will contain this factor $n\alpha'$  times, so $n\alpha'=\alpha$  and $\frac{\alpha}{n}$  is an integer. It can be shown in a similar way that $\frac{\beta}{n}$ etc. are integers.

Article 21
If $a,$ $b,$  $c$  etc. are relatively prime to each other, and their product $abc$  etc. is a perfect power $k^n,$  then each number $a,$  $b,$  $c$  etc. will also be an $n$ th power.

Let $a=l^\lambda m^\mu p^\pi$ etc., where $l,$  $m,$  $p$  etc. are distinct prime numbers, none of which (by hypothesis) divide any of the numbers $b,$  $c$  etc. Then the product $abc$  etc. will include $l$  as a factor $\lambda$  times, $m$  as a factor $\mu$  times, etc.: hence (by the previous article) $\lambda,$  $\mu,$  $\pi$  will all be divisible by $n,$  and therefore

$$\sqrt[n]{a} = l^{\frac{\lambda}{n}} m^{\frac{\mu}{n}} p^{\frac{\pi}{n}}$$

will be an integer. The same will be true for $b,$ $c$  etc.

These facts about prime numbers having been put forth, we now turn to other things which will lead us more directly to our goal.

Article 22
If the numbers $a$ and $b$  are divisible by $k,$  and are congruent with respect to a modulus $m$  that is relatively prime to $k,$  then $\frac{a}{k}$  and $\frac{b}{k}$  are congruent with respect to that same modulus.

It is clear that $a-b$ is divisible by $k,$  and also by $m$  (by hypothesis); thus $\frac{a-b}{k}$  will be divisible by $m$  (Art. 19), i.e. $\frac{a}{k}\equiv\frac{b}{k}\pmod{m}.$

If the numbers $m$ and $k$  instead have a common divisor $e,$  then we will have $\frac{a}{k}\equiv\frac{b}{k}\pmod{\frac{m}{e}}$. For $\frac{k}{e}$ and $\frac{m}{e}$  are relatively prime, and $a-b$  is divisible by $k$  and by $m,$  so $\frac{a-b}{e}$  is divisible by $\frac{k}{e}$  and by $\frac{m}{e}$  and therefore by $\frac{mk}{e^2},$  i.e. $\frac{a-b}{k}$  is divisible by $\frac{m}{e},$  or equivalently $\frac{a}{k}\equiv\frac{b}{k}\pmod{\frac{m}{e}}.$

Article 23
If numbers $a$ and $m$  are relatively prime, and numbers $e$  and $f$  are incongruent modulo $m,$  then $ae$  and $af$  will also be incongruent modulo $m$ .

This is a restatement of the theorem of the preceding article.

Hence it is clear that if we multiply all of the integers from $0$ to $m-1$  by $a$, and reduce these products to their minimal residues modulo $m,$  the resulting residues will all be distinct. Since the number of these residues is $m,$ and none of them is $>m,$  it is also clear that every integer in the series from $0$  to $m-1$  will be found among them.

Article 24
The expression $ax+b,$ in which $a$  and $b$  are given numbers and $x$  is an indeterminate or variable number, can be made congruent to any given number modulo $m,$  provided that $m$  is relatively prime to $a.$

Let $c$ be the number to which it must be congruent, and let $e$  the minimal positive residue of $c-b.$  By the preceding article, it is possible to find a value $x<m$  such that the minimal residue of the product $ax$  modulo $m$  is equal to $e;$  let $v$  be this value. Then $av\equiv e\equiv c-b,$ thus $av+b\equiv c\pmod{m}$  Q.E.F.

Article 25
We will call any expression which takes the form of an equation, and relates two congruent quantities, a congruence; if it contains an unknown, to solve it means to find a value for this unknown which satisfies the congruence (a root). By this we understand what it means for a congruence to be solvable or unsolvable. It is easy to see that the same distinctions that apply to equations can also be applied congruences. Later we will see examples of transcendental congruences; as for algebraic congruences, they are divided according to the highest power of the unknown, into congruences of the first, second, and higher degree. One can even propose several congruences containing several unknowns, to which elimination must be applied.

Article 26
Solution of congruences of the first degree

From Art. 24 it follows that first degree congruences $ax+b\equiv c$ are always solvable, provided that the modulus is relatively prime to $a.$  In this case, let $v$  be the appropriate value for $x,$  or the root of the congruence. Then it is clear that all numbers congruent to $v$ modulo $m$, will also be roots (Art. 9). It is equally clear that all roots must be congruent to $v{:}$ for if $t$  is another root, then we have $av+b\equiv at+b,$  so $av\equiv at,$  and thus $v\equiv t$  (Art. 22). We conclude from this that the congruence $x\equiv v\pmod{m},$ gives the complete solution to the congruence $ax+b\equiv c.$

Since the congruence is solved by values of $x$ which are all congruent to each other, and in this context congruent numbers must be considered equivalent, we will consider these solutions as one and the same. Therefore, since our congruence $ax+b\equiv c,$ does not admit any other solutions, we say that it can only be solved in one way or has only one root. So e.g. the congruence $6x+5\equiv 13\pmod{11}$ does not admit any roots other than those which are $\equiv 5\pmod{11}.$  The situation is different for higher degree congruences, and even for first degree congruences where the coefficient of the unknown is not relatively prime to the modulus.

Article 27
It remains for us to add some details about how to solve this kind of congruence. First, observe that any congruence of the form $ax+t\equiv u,$ in which the modulus is assumed to be relatively prime to $a,$  can be reduced to the congruence $ax\equiv\pm 1.$  For if $x\equiv r$  satisfies the latter, then $x\equiv\pm r(u-t)$  will satisfy the former. But the congruence $ax\equiv\pm 1$, whose modulus is denoted by $b,$ is equivalent to the indeterminate equation $ax=by\pm 1,$  whose solution is known; therefore it will be sufficient for us to provide an algorithm by which this calculation can be carried out.

If quantities $A,$ $B,$  $C,$  $D,$  $E,$  etc. depend on others $\alpha,$  $\beta,$  $\gamma,$  $\delta$  etc. in such a way that

$$A=\alpha, \;\; B=\beta A+1, \;\; C=\gamma B+A, \;\; D=\delta C+B, \;\; E = \varepsilon D + C \text{ etc.}$$

then for the sake of brevity we write

$$A=[\alpha], \;\; B=[\alpha, \beta], \;\; C=[\alpha, \beta, \gamma], \;\; D=[\alpha, \beta, \gamma, \delta] \text{ etc.}$$

Now, consider the equation $ax=by\pm 1,$ where $a$  and $b$  are positive. Suppose, as is permitted, that $a$ is not less than $b.$  Then, according to the well-known algorithm for finding the greatest common divisor of two numbers, the following equations will be formed by division,

$$a=\alpha b+c, \;\; b=\beta c+d, \;\; c=\gamma d+e, \text{ etc.}$$

so that $\alpha,$ $\beta,$  $\gamma,$  $\delta$  etc. are positive integers, and $b,$  $c,$  $d,$  $e$  etc. are continually decreasing, until we reach $m=\mu n+1;$  which must always happen. Then we will have

$$a=[n, \mu, \dots\gamma, \beta, \alpha], \quad b=[n, \mu, \dots\gamma, \beta]$$

Setting

$$x=[\mu, \dots\gamma, \beta], \quad y=[\mu, \dots\gamma, \beta, \alpha]$$

we will have $ax=by+1,$ when the number of quantities $\alpha,$  $\beta,$  $\gamma\dots\mu,$  $n$  is even, and $ax=by-1,$  when it is odd.

Article 28
The general solution of these equations was first given by Euler, ''Comment. de Petersb. T. VII, p. 46''. The method he used consists of substituting other unknowns in place of $x$ and $y,$  and it is indeed well known. Lagrange treated the problem in a slightly different manner. He observed that if one decomposes the fraction $\frac{a}{b}$ into a continued fraction

$$\cfrac{1}{\alpha+{\cfrac{1}{\beta+{\cfrac{1}{\gamma+\cfrac{1}{\ddots+\cfrac{1}{\mu+\cfrac{1}{n}}}}}}}}$$

then after erasing the last part $\frac{1}{n},$ it reduces to an ordinary fraction $\frac{x}{y},$  and one has $ax=by\pm 1,$  so long as $a$  and $b$  are relatively prime. In fact, the two methods lead to the same algorithm. Lagrange’s investigations can be found in the Histoire de l’Académie de Berlin, Année 1767, p. 175, and along with others in Supplementis versioni gallicae Algebrea Eulerianae adiectis.

Article 29
The congruence $ax+t\equiv u,$ in which the modulus is not relatively prime to $a,$  is easily reduced to the previous case. Let $m$ be the modulus and let $\delta$  be the greatest common divisor of $a$  and $m.$  It is first of all clear that any value of $x$  which satisfies the proposed congruence modulo $m,$  will also satisfy it modulo $\delta$  (Art. 5). But since $\delta$ divides $a,$  we always have $ax\equiv 0\pmod{\delta}.$  Therefore, we must have $t\equiv u\pmod{\delta},$  i.e. $t-u$  must be divisible by $\delta,$  in order for the congruence to be solvable.

Let us therefore set $a=\delta e,$ $m=\delta f,$  $t-u=\delta k,$  so that $e$  and $f$  will be relatively prime. Then the proposed congruence $\delta ex+\delta k\equiv 0\pmod{\delta f},$ will be equivalent to $ex+k\equiv 0\pmod{f},$  i.e. any value of $x$  that satisfies the second one will also satisfy the first one, and vice versa. For it is clear that $ex+k$ will be divisible by $f$  whenever $\delta ex+\delta k$  is divisible by $\delta f,$  and vice versa. But we have already seen how to solve the congruence $ex+k\equiv 0\pmod{f},$ hence it is clear that if $\nu$  is one of the values of $x,$  then $x\equiv\nu\pmod{f}$  gives the complete solution of the proposed congruence.

Article 30
When the modulus is composite, it is often advantageous to use the following method:

Let the modulus $=mn,$ and let the proposed congruence $ax\equiv b.$  First solve the congruence modulo $m,$  and suppose that it is satisfied for $x\equiv\nu\pmod{\frac{m}{\delta}},$  where $\delta$  is the greatest common divisor of the numbers $m$  and $a.$  Now it is evident that any value of $x$  which satisfies the congruence $ax\equiv b\pmod{mn}$  will also satisfy the congruence $ax\equiv b\pmod{m},$  and thus it will be included in the form $\nu+\frac{m}{\delta}x',$  where $x'$  denotes an undetermined number. However, not all numbers of this form will satisfy the congruence modulo $mn.$ Rather, the values of $x'$  must be chosen in such a way that $\nu+\frac{m}{\delta}x',$  is a root of the congruence $ax\equiv b\pmod{mn};$  hence we will have $\frac{amx'}{\delta}+a\nu\equiv b\pmod{mn}$  or equivalently $\frac{am}{\delta}x'\equiv\frac{b-a\nu}{m}\pmod{n}.$  It follows from this that the solution of any congruence of the first degree modulo $mn$  can be reduced to the solution of two congruences, one modulo $m$  and the other modulo $n.$  It is easy to see that if $n$  is the product of two factors, the solution of the congruence modulo $n$  depends on the solution of two congruences with these factors as moduli; and generally, the resolution of a congruence modulo any composite modulus depends on the resolution of other congruences, whose moduli are the factors of the first one, and these moduli can always be chosen to be prime numbers, if it seems convenient.

Ex. consider the congruence $19x\equiv 1\pmod{140};$ if we first solve it modulo $2,$  we get $x\equiv 1\pmod{2}.$  By setting $x=1+2x',$  we get $38x'\equiv-18\pmod{140}$  or $19x'\equiv-9\pmod{70}.$  Solving this modulo $2,$  we get $x'\equiv 1\pmod{2},$  and setting $x'=1+2x,$  we get $38x\equiv-28\pmod{70}$  or $19x\equiv-14\pmod{35}.$  Solving this modulo $5$  gives $x\equiv 4\pmod{5},$  and substituting $x=4+5x,$  we have $95x\equiv-90\pmod{35}$  or $19x\equiv-18\pmod{7}.$   From this it follows that $x\equiv 2\pmod{7},$  and thus $x'=2+7x',$  from which we have $x=59+140x';$  therefore $x \equiv 59\pmod{140}$  is the complete solution of the proposed congruence.

Article 31
In the same way that the root of the equation $ax=b$ is expressed by $\frac{b}{a},$  we will also denote the root of the congruence $ax\equiv b$  by $\frac{b}{a},$  including the modulus of the congruence for the sake of clarity. So e.g. $\frac{19}{17}\pmod{12}$ represents an arbitrary number which is $\equiv 11\pmod{12}$. It is generally clear from the above that $\frac{b}{a}\pmod{c}$ does not have a real meaning in the case where $a$  and $c$  have a common divisor that does not divide $b$  (or, if one prefers, it is imaginary). However, except in this case, the expression $\frac{b}{a}\pmod{c}$ will always have real values, and indeed infinitely many, which will all be congruent modulo $c$  when $a$  is relatively prime to $c$, or more generally modulo $\frac{c}{\delta}$  where $\delta$  is the greatest common divisor of $a$  and $c.$

These expressions have almost the same arithmetic as ordinary fractions. Here we add some properties that can be easily deduced from what we have already shown:

 * 1) If $a \equiv \alpha$  and $b \equiv \beta$  modulo $c,$  then the expressions $\frac{a}{b} \pmod{c}$  and $\frac{\alpha}{\beta} \pmod{c}$  are equivalent.
 * 2) $\frac{a\delta}{b\delta} \pmod{c\delta}$  and $\frac{a}{b} \pmod{c}$  are equivalent.
 * 3) $\frac{ak}{bk} \pmod{c}$  and $\frac{a}{b} \pmod{c}$  are equivalent if $k$  is relatively prime to $c$

Many other similar propositions might be added here: but as they are not difficult, and unnecessary for what follows, we now move on to other things.

Article 32
On finding numbers congruent to given residues with respect to given moduli

A problem which will be of great use in what follows is to find all the numbers which form given remainders according to several given moduli. It is easy to solve this from what has already been presented. Let $A$ and $B$  be two given moduli, and let us seek a number $z$  which is congruent to the numbers $a$  and $b$  with respect to these moduli, respectively. Then all values of $z$ will necessarily be contained in the form $Ax+a,$  where $x$  is undetermined, but also satisfies $Ax+a\equiv b\pmod{B}.$  If the greatest common divisor of $A$  and $B$  is $\delta$, then the complete solution of this congruence will take the form $x\equiv v\pmod{\frac{B}{\delta}},$  or equivalently $x=v+\frac{kB}{\delta},$  where $k$  is an undetermined integer; thus the form $a+Av+\frac{kAB}{\delta}$  contains all values of $z,$  i.e. the complete solution of the problem will be $z\equiv a+Av\pmod{\frac{AB}{\delta}}.$  Now if there were a third modulus $C,$  relative to which the number $z$  must be $\equiv c,$  then it is clear how to proceed in the same way, since the two original conditions have already been merged into one. Namely, letting $\varepsilon$ be the greatest common divisor of the numbers $\frac{AB}{\delta}$  and $C,$  we will obtain the congruence $\frac{AB}{\delta}x+a+Av\equiv c\pmod{C},$  which can be solved by a congruence of the form $x\equiv w\pmod{\frac{C}{\varepsilon}}.$  Then the problem will be solved by the congruence $z\equiv\frac{AB}{\delta}w+a+Av\pmod{\frac{ABC}{\delta\varepsilon}}.$  One can proceed in the same way, regardless of the number of moduli. It is worth noting that $\frac{AB}{\delta}$ and $\frac{ABC}{\delta\varepsilon}$  are the least numbers divisible by $A,$  $B,$  and by $A,$  $B,$  $C,$  respectively. Thus one easily deduces, regardless of the number of moduli $A,$ $B,$  $C$  etc., that if $M$  denotes the smallest number divisible by all of them, then the complete solution will be of the form $z\equiv r\pmod{M}.$  Moreover, if one of the congruences is not solvable, it must be concluded that the problem is impossible. However, it is evident that this cannot happen if the numbers $A, B, C$ etc. are all relatively prime to each other.

Ex. Let $A=504,$ $B=35,$  $C=16,$  $a=17,$  $b=-4,$  $c=33$. Here the first two conditions, $z\equiv 17\pmod{504}$ and $\equiv-4\pmod{35},$  can be reduced to a single condition $z\equiv 521\pmod{2520}.$   Combining this with the third condition, $z\equiv 33\pmod{16},$  one obtains $z\equiv 3041\pmod{5040}.$

Article 33
When the numbers $A,$ $B,$  $C$  etc. are all relatively prime to each other, their product is the smallest number divisible by each of them; and in this case it is obvious that the congruences $z\equiv a\pmod{A},$  $z\equiv b\pmod{B}$  etc. together will be equivalent to a single congruence $z\equiv r\pmod{R},$  where $R$  denotes the product of the numbers $A,$  $B,$  $C$  etc. Conversely, it follows from this that a single condition $z\equiv r\pmod{R}$ can be decomposed into several; namely, if $R$  is decomposed into several mutually relatively prime factors $A,$  $B,$  $C$  etc., then the combined conditions $z\equiv r\pmod{A},$  $z\equiv r\pmod{B},$  $z\equiv r\pmod{C},$  etc. will be equivalent to the original one. This observation gives us not only the means to discover an impossibility when it exists, but also a more convenient and elegant method of calculation.

Article 34
Let the conditions proposed above be $z\equiv a\pmod{A},$ $z\equiv b\pmod{B},$  $z\equiv c\pmod{C}$. Resolve all of the moduli into mutually relatively prime factors, $A$ into $A'AA$  etc.; $B$  into $B'BB$  etc. etc., so that the numbers $A',$  $A$  etc. $B',$  $B$  etc. etc. are either primes or powers of primes. Of course, if one of the numbers $A,$ $B,$  $C$  is already prime itself, or a power of a prime number, then it need not be resolved into factors. Then it is clear from the above that, in place of the proposed conditions, one can substitute $z\equiv a\pmod{A'},$ $z\equiv a\pmod{A},$  $z\equiv a\pmod{A'}$  etc., $z\equiv b\pmod{B'},$  $z\equiv b\pmod{B''}$  etc. etc. Now, unless all the numbers $A,$ $B,$  $C$  etc. are mutually prime, e.g. if $A$  and $B$  are not relatively prime, then it is clear that the prime divisors of $A$  and $B,$  cannot all be different, and one of the divisors $A',$  $A,$  $A$  etc. must find its equal, its multiple, or its submultiple among the divisors $B',$  $B$  $B$  etc. In the first case, if $A'=B'$, then the conditions $z\equiv a\pmod{A'},$  $z\equiv b\pmod{B'},$  must be identical, or equivalently $a\equiv b\pmod{A' \text{ or } B'},$  then one or the other of these two conditions can be rejected. On the other hand, if $a\equiv b\pmod{A'}$ does not hold, the problem is impossible. In the second case, if $B'$ is a multiple of $A',$  then the condition $z\equiv a\pmod{A'}$  must be contained in $z\equiv b\pmod{B'},$  or in other words it can be used to deduce $z\equiv b\pmod{A'}.$  Under these conditions, the condition $z\equiv a\pmod{A'}$  can be rejected, if it does not contradict the other one (in which case the problem would be impossible). When all the superfluous conditions are thus rejected, it is clear that all the remaining moduli are mutually prime; one is then sure of the possibility of the problem, and one can proceed according to the method given above.

Article 35
If as above we set $z\equiv 17\pmod{504},$ $\equiv-4\pmod{35},$  and $\equiv 33\pmod{16},$  then these conditions can be reduced to the following: $z\equiv 17\pmod{8},$  $\equiv 17\pmod{9},$  $\equiv 17\pmod{7};$  $z\equiv-4\pmod{5},$  $\equiv-4\pmod{7};$  $z\equiv 33\pmod{16}.$  From these conditions, we can reject $z\equiv 17\pmod{8}$  and $z\equiv 17\pmod{7},$  since the former is contained in the condition $z\equiv 33\pmod{16},$  and the latter is equivalent to $z\equiv-4\pmod{7};$  we are then left with

$$z\equiv\left\{\begin{array}{rl} 17 &\pmod{9}\\-4 &\pmod{5}\\-4 &\pmod{7}\\ 33 &\pmod{16}\end{array}\right\}$$

from which we obtain

$$z\equiv 3041\pmod{5040}.$$

Now, it is clear that it would often be more convenient if, of the remaining conditions, those which had been derived from one and the same condition were combined into a single condition. This can be done easily; e.g. when some of the conditions $z\equiv a\pmod{A'},$ $z\equiv a\pmod{A''}$  etc. have been rejected, the rest can be restored as $z\equiv a$  with respect to the modulus formed by the product of the remaining moduli. Thus in our example the condition $z\equiv-4\pmod{35}$ can be recovered from the conditions $z\equiv-4\pmod{5}$  and $z\equiv-4\pmod{7}.$  Now, it does make some difference which of the superfluous conditions are rejected, as far as the brevity of the calculation is concerned. But this is not the place to discuss these details or other practical tricks, which are learned from practice much more easily than from precepts.

Article 36
When the moduli $A,$ $B,$  $C,$  $D$  etc. are mutually relatively prime, it is often preferable to use the following method. Let $\alpha$ be a number which is congruent to unity modulo $A,$  and which is divisible by the product of the other moduli; that is, $\alpha$  will be any value (the minimum is generally best) of the expression $\frac{1}{BCD\text{ etc.}}\pmod{A},$  multiplied by $BCD$  etc. (see Art. 32); similarly, let $\beta\equiv 1\pmod{B}$  and $\equiv 0\pmod{ACD\text{ etc.}},$  $\gamma\equiv 1\pmod{C}$  and $\equiv 0\pmod{ABD\text{ etc.}}$  etc. Then to find a number $z$  which is congruent to the numbers $a,$  $b,$  $c,$  $d$  etc. modulo $A,$  $B,$  $C,$  $D$  etc. respectively, we can set

$$z\equiv\alpha a+\beta b+\gamma c+\text{etc.}\pmod{ABCD\text{ etc.}}$$

Then we clearly have $\alpha a\equiv a\pmod{A},$ and the other terms are $\equiv 0\pmod{A};$  hence $z\equiv a\pmod{A}.$  The demonstration is the same for the other moduli. This solution is preferable to the first one when solving several problems of the same kind, for which the values of $A,$ $B,$  $C$  etc. are the same; for then the values of $\alpha,$  $\beta,$  $\gamma$  etc. remain constant. This can be applied to the chronological problem in which we must find the Julian year which has a given indiction, the golden number, and the solar cycle. Here $A=15,$ $B=19,$  $C=28;$  thus the value of the expression $\frac{1}{19. 28}\pmod{15},$ or $\frac{1}{532}\pmod{15},$  is $13,$  and we have $\alpha =6916.$  Similarly, we find $\beta =4200, \gamma =4845,$  hence the desired number will be the minimal residue of the number $6916a+4200b+4845c,$  where $a$  denotes the indiction, $b$  the golden number, and $c$  the solar cycle.

Article 37
Linear congruences involving several unknowns

Enough has been said about congruences of the first degree which contain only one unknown. It remains to deal with congruences that involve several unknowns, but since it would be too extensive to present everything rigorously in this chapter, and our goal is not to exhaust the subject here, but only to present what is most worthy of attention, we will limit our discussion to a few observations, reserving the complete exposition for another occasion.

1. Just as for equations, it can be seen that there must be as many congruences as there are unknowns for a solution to determined.

2. Let the following congruences be proposed

$$ \begin{aligned} ax+by+cz \dots &\equiv f\phantom{''} \pmod{m} & \quad. \quad. \quad. \quad. \quad \phantom{''}\,(A)\\ a'x+b'y+c'z \dots &\equiv f'\phantom{'} \phantom{\pmod{m}} & \quad. \quad. \quad. \quad. \quad \phantom{'}(A')\\ ax+by+cz \dots &\equiv f \phantom{\pmod{m}} & \quad. \quad. \quad. \quad. \quad (A'') \end{aligned} $$

and assume that the number of these congruences is the same as the number of unknowns $x,$ $y,$  $z$  etc.

Then determine numbers $\xi,$ $\xi',$  $\xi''$  etc. so that

$$ \begin{aligned} b\xi+b'\xi'+b'\xi''+\text{etc.} &=0, \\ c\xi+c'\xi'+c\xi+\text{etc.} &=0, \\ \text{ etc.} \qquad \qquad \end{aligned} $$

and so that these numbers are integers with no common divisor (which is always possible, by the theory of linear equations).

In a similar way, determine $\nu,$ $\nu',$  $\nu$  etc., $\zeta,$  $\zeta',$  $\zeta$  etc. etc. so that

$$\begin{aligned} a\nu+a'\nu'+a\nu+\text{etc.}&=0 \\ c\nu+c'\nu'+c\nu+\text{etc.}&=0 \\ \text{etc.} \qquad \qquad &\\ a\zeta+a'\zeta'+a\zeta+\text{etc.}&=0 \\ b\zeta+b'\zeta'+b\zeta+\text{etc.}&=0,\\ \text{etc.} \qquad \qquad & \end{aligned}$$

3. It is clear that if we multiply the congruences $A,$ $A',$  $A$  etc. first by $\xi,$  $\xi',$  $\xi$  etc. and then by $\nu,$  $\nu',$  $\nu''$  etc. etc. and add them, we will obtain the following:

$$\begin{aligned} (a\xi+a'\xi'+a\xi+\text{etc.})x & \equiv f\xi+f'\xi'+f\xi+\text{etc.} \\ (b\nu+b'\nu'+b\nu'+\text{etc.})y & \equiv f\nu+f'\nu'+f\nu''+\text{etc.} \\ (c\zeta+c'\zeta'+b\zeta+\text{etc.})z & \equiv f\zeta+f'\zeta'+f\zeta+\text{etc.} \end{aligned}$$

For the sake of brevity, we will write these congruences as

$$\Sigma(a\xi)x\equiv\Sigma(f\xi), \quad \Sigma(b\nu)y\equiv\Sigma(f\nu), \quad \Sigma(c\zeta)z\equiv\Sigma(f\zeta) \;\; \text{etc.}$$

4. There are now several cases to distinguish.

First, when the coefficients of the unknowns, $\Sigma(a\xi),$ $\Sigma(b\nu)$  etc. are all relatively prime to the modulus $m$, the above congruences can be solved by the methods already presented, and the complete solution of the problem will be given by congruences of the form $x\equiv p\pmod{m},$  $y\equiv q\pmod{m}$  etc. .  E.g. if we propose the congruences

$$x+3y+z\equiv 1, \quad 4x+y+5z\equiv 7, \quad 2x+2y+z\equiv 3\pmod{8},$$

we will find $\xi =9,$ $\xi'=1,$  $\xi''=-14,$  hence $-15x\equiv -26$  and therefore $x\equiv 6 \pmod{8};$  likewise, we will find $15y\equiv-4,$  $15z\equiv 1,$  and hence $y\equiv 4,$  $z\equiv 7\pmod{8}.$

5. Second, If the coefficients $\Sigma(a\xi),$ $\Sigma(a\xi),$  $\Sigma(b\nu)$  etc. are not all relatively prime to the modulus, let $\alpha,$  $\beta,$  $\gamma$  etc. be the greatest common divisors of $m$  and $\Sigma(a\xi),$  $\Sigma(b\nu),$  $\Sigma(c\zeta)$  etc. respectively. Then it is clear that the problem is impossible unless the numbers $\Sigma(f\xi),$ $\Sigma(f\nu),$  $\Sigma(f\zeta)$  etc. are divisible by $\alpha,$  $\beta,$  $\gamma$  etc. respectively. On the other hand, when these conditions hold, the problem will be completely solved by congruences of the form $ x\equiv p\pmod{\frac{m}{\alpha}}, \quad y\equiv q\pmod{\frac{m}{\beta}}, \quad z\equiv r\pmod{\frac{m}{\gamma}}$ etc., or if you prefer, there will be $\alpha$  different values for $x$  (i.e. the values $p, p+\frac{m}{\alpha}, \dots p+\frac{(\alpha-1)m}{\alpha},$  which are incongruent modulo $m$ ), $\beta$  values for $y,$  etc. that will satisfy the congruences. Clearly, all solutions to the problem, if there are any, must be found among those we have just indicated. However, it is not allowed to reverse this conclusion; for in general it is not true that all combinations of the $\alpha$  values of $x,$  with those of $y$  and those of $z$  etc. will satisfy the problem, but only some of them whose connection must be expressed by means of one or more conditional congruences. However, since the complete solution of this problem is not necessary for the following, we will not go any further on this subject and content ourselves to give the idea with an example.

Let the following congruences be proposed:

$$3x+5y+z\equiv 4, \quad 2x+3y+2z\equiv 7, \quad 5x+y+3z\equiv 6\pmod{12},$$

Then we will have

$$\begin{aligned} \xi &= 1,&\xi'&=-2,&\xi''&= 1, \\ \nu &= 1,&\nu'&= 1,&\nu''&=-1, \\ \zeta &=-13,&\zeta'&= 22,&\zeta''&=-1, \end{aligned}$$

hence $4x\equiv-4,$ $7y\equiv 5,$  $28z\equiv 96.$  From this we obtain four values of $x,$  namely, $x\equiv 2,$  $5,$  $8,$  $11;$  one value of $y,$  namely $y\equiv 11;$  and four values of $z,$  namely, $z\equiv 0, 3, 6, 9\pmod{12}.$  To find out which combinations of the values of $x$  and $z$  are permissible, substitute $2+3t,$  $11,$  $3u,$  for $x,$  $y,$  $z$  in the given congruences. After this, they become

$$57+9t+3u\equiv 0, \quad 30+6t+6u\equiv 0, \quad 15+15t+9u\equiv 0\pmod{12},$$

which are easily seen to be equivalent to

$$19t+3t+u\equiv 0,10+2t+2u\equiv 0,5+5t+3u\equiv 0\pmod{4}.$$

The first congruence clearly requires that $u\equiv t+1\pmod{4},$ and the rest will clearly be satisfies if this value is substituted in them. We conclude from this that the values $x\equiv 2,$ $5,$  $8,$  $11$  (which are obtained by successively setting $t=0,$  $1,$  $2,$  $3$ ) must be combined respectively with the values $z \equiv 3,$  $6,$  $9,$  $0,$  so there are four solutions altogether,

$$ \begin{array}{rrrrrl} x\equiv &2, &5, &8, &11 & \pmod{12}\\ y\equiv &11, &11, &11, &11\\ z\equiv &3, &6, &9, &0 \end{array} $$

To these investigations, which complete the task we had set ourselves in this section, we will add some propositions based on similar principles, which will be needed frequently in what follows.

Article 38
Various theorems

Determine how many numbers are less than and relatively prime to a given positive number $A.$

For the sake of brevity, let us denote the number we seek by the prefixed symbol $\phi$. So, the problem is to find $\phi A.$

I. When $A$ is prime, it is clear that all numbers from $1$  to $A-1,$  are relatively prime to $A;$  therefore, in this case, we have

$$\phi A=A-1.$$

II. When $A$ is a power of a prime number $p,$  say $p^m,$  then all numbers divisible by $p$  will not be prime with $A,$  and the others will be. Therefore, of the $p^m-1$ numbers, we must reject  $p,$  $2p,$  $3p$  $\dots$  $(p^{m-1}-1)p;$  so there remain $p^m-1-(p^{m-1}-1).$  Hence

$$\phi p^m =p^{m-1}. (p-1).$$

III. The other cases can be easily reduced to these, by means of the following proposition: If we factor $A$ into factors $M,$  $N,$  $P\dots$  etc. which are all relatively prime to each other, then

$$\phi A =\phi M. \phi N. \phi P \text{ etc.}$$

which can demonstrated as follows. Let $m,$ $m',$  $m''$  etc. be the numbers which are less than and relatively prime to $M$, whose multitude is therefore $\phi M$. Similarly, let the numbers less than and relatively prime to $N,$ $P$  etc. be $n,$  $n',$  $n$  etc.; $p,$  $p',$  $p$  etc. etc., whose multitudes are $\phi P,$  $\phi N$  etc. It is clear that any number relatively prime to $A$ will also be relatively prime to the factors $M,$  $N,$  $P$  etc. and vice versa (Art. 19). Furthermore, all numbers that are congruent to one of the numbers $m,$ $m',$  $m''$  etc. modulo $M$  will be relatively prime to $M,$  and the same applies for $N,$  $P$  etc. The question is therefore reduced to determining how many numbers are less than $A$ and congruent to one of the numbers $m,$  $m',$  $m$  etc. modulo $M,$  to one of the numbers $n,$  $n',$  $n$  etc. modulo $N$, etc. However, it follows from Art. 32 that all numbers with given residues modulo $M,$ $N,$  $P$  etc. must be congruent modulo the product $A,$  and therefore there can only be one number less than $A$  which is congruent to the given residues modulo $M,$  $N,$  $P$  etc. Therefore, the number we seek will be equal to the number of combinations of the individual numbers $m,$  $m',$  $m$  etc. with the individual numbers $n,$  $n',$  $n$  etc. and $p,$  $p',$  $p''$  etc. etc. It is evident from the theory of combinations that this number is $\phi(M). \phi(N). \phi(P)$ etc. Q.E.D.

IV. It is easy to see how to apply this proposition in the case at hand. We will decompose $A$ into prime factors; that is, we will reduce it to the form $a^\alpha b^\beta c^\gamma$  etc., with $a,$  $b,$  $c$  etc. being distinct prime numbers. Then we will have

$$\phi A=\phi a^\alpha\phi b^\beta\phi c^\gamma \text{ etc.} =a^{\alpha-1}(a-1) \cdot b^{\beta-1}(b-1) \cdot c^{\gamma-1}(c-1) \text{ etc.}$$

which can be put in the more elegant form

$$\textstyle \phi A=A \cdot \frac{a-1}{a} \cdot \frac{b-1}{b} \cdot \frac{c-1}{c} \text{ etc.}$$

Example: Take $A=60=2^2 \cdot 3 \cdot 5.$ Then we have $\phi A=60 \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5}=16.$  The numbers which are less than and relatively prime to $60$  are $1,$  $7,$  $11,$  $13,$  $17,$  $19,$  $23,$  $29,$  $31,$  $37,$  $41,$  $43,$  $47,$  $49,$  $53,$  $59.$

The first solution of this problem can be found in Euler’s commentary, Theoremata arithmetica nova methodo demonstrata, Comm. nov. Ac. Petrop. VIII p. 74. A demonstration was also provided in another dissertation Speculationes circa quasdam insignes proprietates numerorum Acta Petrop. VIII, p. 17.

Article 39
If the meaning of the symbol $\phi$ is determined in such a way that $\phi A$  expresses how many numbers are relatively prime to $A$  and not greater than $A,$  then we will have $\phi 1=1$  instead of $0.$  However, in all other cases, nothing will change. Adopting this definition, we have the following theorem:

If $a,$ $a',$  $a$  etc. are all of the divisors of $A,$  including 1 and $A,$  then

$$\phi a+\phi a'+\phi a''+\text{etc.} = A.$$

Ex. If $A=30,$ we have

$$\phi 1+\phi 2+\phi 3+\phi 5+\phi 6+\phi 10+\phi 15+\phi 30 = 1+1+2+4+2+4+8+8 = 30$$

Proof. If we multiply all numbers coprime with $a$ and not greater than $a$  by $\frac{A}{a},$  and similarly all numbers coprime with $a'$  by $\frac{A}{a'}$   etc., then we will have $\phi a+\phi a'+\phi a''+$  etc. numbers, all of which are not greater than $A.$  However,

 * 1) All of these numbers will be distinct. Indeed, it is clear that those coming from the same divisor of $A$  are all distinct. On the other hand, if two equal numbers resulted from different divisors $M$  and $N,$  and from numbers $\mu$  and $\nu$  respectively, that is, if $\frac{A}{M}\mu =\frac{A}{N}\nu,$  then we would have $N\mu =M\nu.$   Assuming that $M>N$  (which is allowed), then since $M$  is relatively prime to $\mu$  and divides $N\mu,$  it would follow that the larger number $M$  divides the smaller number $N,$  Q.E.A.
 * 2) Among these numbers, we will find all of those which make up the sequence $1, 2, 3, \dots, A.$ Indeed, let $t$  be any number that does not exceed $A,$  and let $\delta$  be the greatest common divisor of $A$  and $t.$  Then $\frac{A}{\delta}$  will be a divisor of $A$  which is relatively prime to $\frac{t}{\delta}$ . Then clearly the number $t$  will be among those that were produced by the divisor $\frac{A}{\delta}$.
 * 3) It follows from this that the multitude of these numbers is $A,$ and hence

$\phi a + \phi a' + \phi a'' + \text{etc.} = A.$ Q.E.D.

Article 40
If $\mu$ is the greatest common divisor of the numbers $A,$  $B,$  $C,$  $D$  etc., then numbers $a,$  $b,$  $c,$  $d$  etc. can be determined so that

$$aA + bB + cC + \text{etc.}= \mu.$$

Let us first consider only two numbers $A$ and $B,$  and let $\lambda$  be their greatest common divisor. Then the congruence $Ax \equiv \lambda \pmod{B}$ will be solvable (Art. 30). Let the root be $\equiv\alpha,$ and set $\frac{\lambda - A\alpha}{B} = \beta.$  Then we will have $\alpha A + \beta B = \lambda,$  as desired.

If there is a third number $C,$ let $\lambda'$  be the greatest common divisor of $\lambda$  and $C.$  Note that $\lambda'$  will also be the greatest common divisor of the three numbers $A,$  $B,$  $C$. Then let numbers $k$ and $\gamma$  be determined so that $k\lambda+\gamma C=\lambda',$  and we will have $k\alpha A\lambda+k\beta B+\gamma C=\lambda'.$

If there is a fourth number $D,$ let $\lambda''$  be the greatest common divisor of $\lambda'$  and $D$  (which, it is easily seen, will also be the greatest common divisor of the four numbers $A,$  $B,$  $C,$  and $D$ ). Then let $k'\lambda'+\delta D=\lambda,$ and we will have $k'k\alpha A+k'k\beta B+k'\gamma C +\delta D=\lambda.$

We would proceed in a similar way if there were more numbers.

In particular, if the numbers $A,$ $B,$  $C$  etc. have no common divisor, then it is clear that we can find $A,$  $B,$  $C$  etc. so that

$$aA+bB+cC+\text{etc.} = 1.$$

Article 41
If $p$ is a prime number, and if there are $p$  objects among which any number may be equal (provided that they are not all equal), then the number of permutations of these objects will be divisible by $p.$

For example, five objects $A, A, A, B, B$ can be arranged in ten different ways.

The demonstration of this theorem is easily deduced from the well-known theory of permutations. Indeed, suppose that, among these $p$ things, there are $a$  equal to $A,$  $b$  equal to $B,$  $c$  equal to $C$  etc. (where the numbers $a,$  $b,$  $c\dots$  can also represent unity), so that we have

$$a+b+c+\dots =p$$

Then the number of permutations will be

$$= \frac{1. 2. 3 \dots \dots p}{1. 2\dots a. 1. 2\dots b. 1. 2\dots c \text{ etc.}}$$

It is clear that the numerator of this fraction is divisible by the denominator, since the number of permutations is an integer: but the numerator is divisible by $p,$ whereas the denominator, which is composed of factors smaller than $p,$  is not divisible by $p$  (Art. 15). Therefore, the number of permutations will be divisible by $p$ (Art. 19).

That being said, we hope that the following demonstration will not be unwelcome.

When in two permutations the order of the objects only differs in that the item which occupied the first place in one, now occupies a different one in the other, while the other objects, starting from that one, follow the same order in each of the permutations, so that the last item in one is immediately before the first item in the other. Then we will call them similar permutations. So, for example, the permutations $ABAAB$ and $ABABA$  will be similar, since the things which in the former occupy the first place, second place, etc. occupy the third place, fourth place, etc. in the latter.

Now, as each permutation consists of $p$ objects, it is clear that one can find $p-1$  similar permutations, if one puts successively at the second place, third place, etc. the item which occupied the first place. Therefore, if none of these similar permutations are identical, it is evident that the total number of permutations will be equal to $p$ times the number of dissimilar permutations, and consequently will be divisible by $p.$   Let us suppose, then, that two similar permutations

$$PQ\dots TV\dots YZ, \qquad V\dots YZPQ\dots T$$

one of which has arisen from the other by the promotion of terms, are identical, in the sense that $P=V$ etc. Suppose that the term $P,$ which occupies the first place in the first permutation, occupies the ${n+1}^{st}$  in the second. Then in the latter series, the ${n+1}^{st}$ term will be equal to the first, the ${n+2}^{nd}$  will be equal to the second, etc., from which it follows that the ${2n+1}^{st}$  will again be equal to the first, and likewise for the ${3n+1}^{st},$  and in general the ${kn+m}^{th}$  will be equal to the ${m}^{th}$  (where, when $kn+m>p,$  it is necessary to imagine that either the series $V\dots YZPQ\dots T$  is repeated from the beginning, or that we subtract a multiple of $p$  from $kn+m$ ). Therefore, if we determine $k$ in such a way that $kn\equiv 1\pmod{p},$  which can always be done since $p$  is prime, it will follow that generally the $m^{th}$  term will be equal to the ${m+1}^{st},$  i.e. all the terms will be equal to each other, contrary to the hypothesis.

Article 42
If the coefficients $A,$ $B,$  $C \dots N;$  $a,$  $b,$  $c \dots n;$  of two functions of the form

$$\begin{aligned} x^m + Ax^{m-1} + Bx^{m-2} + Cx^{m-3}\dots + N & \quad.\quad.\quad.\quad.\quad(P) \\ x^{\mu} + ax^{\mu-1} + bx^{\mu-2} + cx^{\mu-3}\dots + n & \quad.\quad.\quad.\quad.\quad (Q) \end{aligned}$$

are all rational, but not all integers, and their product is

$$x^{m+\mu} + \mathfrak{A}x^{m+\mu-1} + \mathfrak{B}x^{m+\mu-2} \dots + \mathfrak{N},$$

then the coefficients $\mathfrak{A},$ $\mathfrak{B} \dots \mathfrak{N}$  cannot all be integers.

Indeed, let us simplify to their simplest form all fractions that may be found among the numbers $A,$ $B$  etc. $a,$  $b$  etc., and choose a prime number $p$  that divides one or more of the denominators of these fractions. Let us assume, as we may, that $p$ divides the denominator of a fractional coefficient in $(P).$  It is clear that if $(Q)$  is divided by $p,$  then at least one fractional coefficient in $\frac{(Q)}{p}$  will have a denominator divisible by $p$  (in particular, the coefficient of the first term will be $\frac{1}{p}$ ). It is then easy to see that in $(P)$ there is a term whose denominator contains a power of $p$  raised to a power which is greater than in all of the preceding terms, and not less than in all of the terms which follow. Let $Gx^g$ be such a term, and let $t$  be the exponent of $p$  in its denominator. Let the analogous term in $\frac{(Q)}{p}$ be $\Gamma x^\gamma,$  and let the exponent of $p$  in the denominator be $\tau.$  Then clearly $t+\tau$  will be at least $2.$  With this in mind, the term $x^{g + \gamma}$  of the product of $(P)$  and $(Q)$  will have a fractional coefficient whose denominator contains $p$  raised to the power $t+\tau-1,$  as will now be shown.

Let ${'}Gx^{g+1},$ ${}Gx^{g+2}$  etc. be the terms preceding $Gx^g$  in $(P),$  and let $G'x^{g-1},$  $Gx^{g-2}$  etc. be those which follow it. Similarly, let ${'}\Gamma x^{\gamma+1},$ ${}\Gamma x^{\gamma+2}$  etc. be the terms preceding $\Gamma x^\gamma$  in $\frac{(Q)}{p},$  and let $\Gamma'x^{\gamma-1},$  $\Gammax^{\gamma-2}$  be those which follow it. Then in the product of $(P)$ and $\frac{(Q)}{p},$  the coefficient of $x^{g+\gamma}$  will clearly be

$$\begin{aligned} = G\Gamma & {}+{'}G\Gamma'+{}G\Gamma+\dots.\\ & {}+{'}\Gamma G'+{}\Gamma G+\dots. \end{aligned}$$

The term $G\Gamma$ will be a fraction which, when reduced to its simplest form, will have a denominator divisible by $p^{t+\tau}.$  If the other terms are fractional, then their denominators will contain only powers of $p$  less than $p^{t+\tau},$  since each of them is the product of two factors, one of which contains only a power of $p$  smaller than $p^t$  or $p^\tau,$  and the other a power not greater than $p^t$  or $p^\tau.$  Thus $G\Gamma$  will be of the form $\frac{e}{fp^{t+\tau}},$  and the rest will be of the form $\frac{e'}{f'p^{t+\tau-\delta}},$  where $e,$  $f,$  and $f'$  are relatively prime to $p.$  Therefore the sum of these will be $= \frac{ef'+e'fp^\delta}{ff'p^{t+\tau}},$  a fraction whose numerator is not divisible by $p.$  Hence this fraction cannot be reduced in any way so that its denominator contains a a power of $p$  smaller than $p^{t+\tau}.$  Therefore, the coefficient of the term $x^{g+\gamma}$  in the product of $(P)$  by $(Q)$  will be

$$\frac{ef'+e'fp^\delta}{ff'p^{t+\tau-1}}$$

i.e. a fraction, whose denominator is divisible by at least the ${t+\tau-1}^{st}$ power of $p.$  Q.E.D.

Article 43
''A congruence of the $m^{th}$ degree,

$$Ax^m+Bx^{m-1}+Cx^{m-2}+\dots+Mx+N\equiv 0$$

whose modulus is a prime number $p$ that does not divide $A,$  cannot have more than $m$  solutions, or equivalently cannot have more than $m$  incongruent roots modulo $p$  (See Arts. 25, 26).

If anyone denies it, let us suppose that congruences of different degrees $m,$ $n$  etc. are given, which have more than $m,$  $n$  etc. roots. Let $m$ be the smallest of all of these, so that all congruences of degree lower than $m$  are consistent with our theorem. Since it has already been demonstrated above (Art. 26) that the theorem holds for congruences of the first degree, it is clear that $m$ will be $=2$  or greater. Let us therefore assume that the congruence

$$Ax^m+Bx^{m-1}+Cx^{m-2}+\dots+Mx+N\equiv 0$$

has at least $m+1$ roots $x\equiv\alpha,$  $x\equiv\beta,$  $x\equiv\gamma$  etc., and suppose that all of the numbers $\alpha,$  $\beta,$  $\gamma$  etc. are positive and less than $p,$, with the smallest being $\alpha$. Now let us substitute $y+\alpha$ for $x$  in the proposed congruence, with the result being

$$A'y^m+B'y^{m-1}+\dots+M'y+N'\equiv 0$$

Then it is clear that this congruence will be satisfied if we put $y\equiv 0$ or $\equiv\beta-\alpha,$  or $\equiv\gamma-\alpha$  etc., and that all of these roots will be distinct, with their multitude being $m+1.$  However, since $y\equiv 0$  is a root, it follows that $N'$  is divisible by $p.$  From this we obtain

$$y(A'y^{m-1}+B'y^{m-2}+\dots+M')\equiv 0\pmod{p}$$

a congruence that will be satisfied by substituting for $y$ any of the $m$  values $\beta-\alpha,$  $\gamma-\alpha$  etc. which are all $>0$  and $<p,$  and therefore in these cases

$$Ay^{m-1}+B'y^{m-2}+\dots+M'$$

will become $\equiv 0$ (Art. 22), i.e. the congruence

$$Ay^{m-1}+B'y^{m-2}+\dots+M'\equiv 0,$$

which is of degree $m-1,$ would have $m$  roots; which contradicts our theorem (for it is easy to  see that $A;=A,$  and therefore it is not divisible by $p,$  as required), even though we assumed that all congruences of a degree lower than $m$  satisfy it. Q.E.A.

Article 44
Although we have assumed here that the modulus $p$ does not divide the coefficient of the first term, the theorem is not restricted to this case. For if the first coefficient, and even some of the following ones, were divisible by $p,$ then these terms could be neglected without error, and the congruence would be reduced to a lower degree, such that the coefficient of the first term would no longer be divisible by $p.$  Indeed, not all coefficients can be divisible by $p$ ; in this case the congruence would be an identity, and the unknown would be completely indeterminate.

Lagrange was the first to propose and prove the theorem (Mem. de l'Ac de Berlin, Année 1768 p.192). It is also found in Legendre’s dissertation, ''Recherches d’Analyse indéterminée, Hist. de l'Acad. de Paris 1785 p. 466). Euler proved, in Nouveaux Commentaires Académiques. Pétersb. XVIII, p. 93, that the congruence $x^n-1\equiv 0$ could not have more than $n$  roots. Although this is only a particular case, his method can easily be applied to all congruences. He had already dealt with a still more limited case, Comm. nov. Ac. Petr. V. p.'' 6, but this method cannot be used in general. In Section VIII below, we will demonstrate this theorem in yet another way; but no matter how different all these methods may seem at first glance, experts who wish to compare them will easily verify that they all originate from the same principle. Furthermore, this theorem should only be considered here as a lemma, and the complete exposition does not belong to this section, so we will treat the case of a composite modulus elsewhere.


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is the "last Wednesday" of the month of Safar, and is a feast held in commemoration of Muhammad's having experienced some mitigation of his last illness and having bathed. It was the last time he performed the legal bathing, for he died on the twelfth day of the next month. In some parts of Islám it is customary, in the early morning of this day, to write seven verses of the Qurán, known as the Seven Saláms, and then wash off the ink and drink it as a charm against evil.

The A′khiri Chahár Shamba is not observed by the Wahhábís, not being enjoined in the Qurán and Hadís.


text: Notes on Muhammadanism/Chapter 34

title: Notes on Muhammadanism/Chapter 34


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Bára Wafát (i. e. Bára, "twelve," and Wafát, "death") is the twelfth day of the month, Rabí-ul-Awwal. It is observed in commemoration of Muhammad's death.

On this day, Fátihahs (i. e. the first chapter of the Qurán), are said for Muhammad; and both in private houses and in the mosques, the learned recite portions of the Traditions and other works in praise of the excellences of Muhammad. These customs are usually observed for the whole twelve days, although the twelfth day is held most sacred.

The Wahhábís do not observe the Bára Wafát, as its observance is not enjoined in the Qurán or Hadís.


text: Notes on Muhammadanism/Chapter 35

title: Notes on Muhammadanism/Chapter 35


text:
Author:Washington Atlee Burpee

As editor

 * How and What to Grow in a Kitchen Garden of One Acre (1898)

title: Author:Washington Atlee Burpee


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Our victory at Buna was the fruit of cooperation between ground and air forces. Either without the other would have failed. The Australian and American units of the U. S. 5th Air Force, under Lt. Gen. George C. Kenney, met all demands, strategic, tactical, and logistic. Though our squadrons were often outnumbered and always short of pilots, planes, and supplies, they were stronger than the forces which the Japanese could spare from their major effort in the Solomons. The skill and courage of Allied fliers, in combination with the superiority of our planes, won aerial mastery over the southeastern end of New Guinea. Our control of the air made large-scale reinforcement of the enemy troops in the area cost more than the Japanese were willing or able to pay in terms of losses. At the same time, we were able to bring in the soldiers and supplies to drive the enemy from Buna and Sanananda.

Our control of the air was won by constant patrolling, armed reconnaissance, and aggressive fighter operations. Yet we could not monopolize the air and considerable numbers of hostile planes from time to time succeeded in breaking through to bomb and strafe our troops and rear areas. For example, on 7 December, 3 Japanese navy dive bombers and 18 Zero-escorted high-level bombers attacked the 2d Field Hospital at Simemi, causing heavy casualties. As late as 27 December, when the enemy was withdrawing toward the sea for his last stand, 41 Zeros and dive bombers attempted to raid our positions but were intercepted by 8 of our P–38's. Two enemy planes were shot down, and the 3 bombs dropped did no damage. However, the

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 enemy could not maintain a continuous or effective aerial offensive and suffered severe losses in his occasional raids. He never completely severed our fragile lines of communication.

Strategic bombing and strafing of enemy airfields at Lae, Salamaua, and Rabaul by the 5th Air Force was severely limited by the small number of planes available. Between 1 November 1942 and 31 January 1943, only 13½ tons of explosives were dropped on ground targets and shipping outside the coastal region of southeastern New Guinea. At the same time, however, the 13th Air Force was engaging large numbers of Japanese planes in the Solomons and was also bombing Rabaul. These attacks had an effect out of proportion to their apparent weight, for they compelled the enemy to allocate part of his strength to defense of his bases and restricted his ability to interfere with Allied operations in the Buna-Sanananda area.

Tactical bombing and strafing of enemy forward areas played a relatively small part, for such operations early proved almost as dangerous to our own troops as to the enemy. Contact between our ground units and those of the enemy was exceedingly close and aerial observation was practically impossible. Enemy rear areas were constantly pounded by our B–25's and A–20's, yet during the 6 weeks of active operations only 163 tons of bombs were dropped and 144,790 rounds of ammunition fired on the Buna Mission and Old Strip area. Bombardiers in the combat zone usually had to aim at jungle-covered ground targets visible only at extremely low altitudes. In most instances, pilots had to report "results unobserved." Nevertheless, these raids had telling effect on Japanese forward supply lines and Japanese morale.

Tactical aerial reconnaissance and observation were likewise very difficult, but proved invaluable to our ground forces. A flight of Australian Wirraways, based at Dobodura late in November, gathered information contributing to a precision in artillery fire otherwise impossible because of the inaccuracy of available maps.

Of first importance and most novel was the contribution of the 5th Air Force to the transport and supply of the ground troops. Most of our infantrymen and supporting troops, a total of 14,900 were flown to the Buna area in the uncomfortable bucket seats of C–47's. The 19


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 small amount of artillery used in the operation was all airborne for at least part of the way. The air movement of the 128th Infantry from Australia to New Guinea was the first large-scale airborne troop movement by United States forces in a theater of operations. The bulk of the other units of the Buna Force, including the 127th and 163d Infantry Regiments, were flown by the 374th Troop Carrier Group directly to Dobodura and Popondetta, only 10 miles from the front lines. When they were wounded or when they fell ill of tropical diseases, the troops left as they had come; the planes which brought in reinforcements and supplies returned to Port Moresby laden with hospital cases, many of which were ferried immediately by air to Australia.

Supply of rations and equipment was an extraordinarily difficult problem throughout the operation. Land transportation across the mountains was almost impossible and very impractical, since there was only a rough and steep foot trail from Port Moresby to the front and a trip over this trail took 18 to 28 days. The distance could be flown by plane in 35 minutes. While units near the coast relied on supply by small boats, those inland had to depend on supply by aircraft.

The Air Force maintained a regular shuttle service across the treacherous mountains and over the coastal jungle to deliver food for the soldiers and native carriers, as well as the ammunition, the ¼-ton peeps used for transport on the corduroy trails, pick mattocks, shovels, axes, oil for the ordnance which was corroding in the steaming jungle, cots, blankets, surgical instruments, dressings, plasma, quinine and atabrine tablets for the sick and wounded, new shoes, and clothing. The only 105-mm howitzer used in the campaign was carried to the front by aircraft. Sixteen C–47's and several A–29's, based at Ward's Drome, Port Moresby, delivered a total of 2,450 tons to the landing strips at Dobodura, Popondetta, and Pongani and dropped 166 tons at small grounds designated along the jungle trails. Small arms, ammunition, mortar shells, and medical supplies were dropped with parachutes; food, clothing, and individual equipment other than arms, without parachutes.

The losses in dropping without parachutes averaged 50 per cent. For example, cans of bully beef broke open and spoiled. Dropping grounds were hard to find; smudge fires and panels often did not 20


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 identify the area sufficiently, and occasionally loads landed in enemy territory. Hungry men watched every incoming plane for good drops. "Convey our compliments to pilot and crew of Dutch Boy. They really laid delivery on the door step," said Maj. Herbert M. Smith, speaking for the 2d Battalion, 126th Infantry.

Pilots made their turn-around as quickly as possible. Two planes, Sleepy Sally and Eager Beaver, on 4 December made two trips from Port Moresby to Hariko, a distance of go miles, and dropped both loads at Hariko within 2 hours and 15 minutes.

Approximately half of the supplies were seaborne. On 14 November, 45 tons a week of rations for the men on the east flank were ordered shipped to Pongani via Milne Bay. Cargoes were transferred at Milne Bay from ocean freighters to smaller vessels of 50 to 500 tons, which crept around East Cape and through Ward Hunt Strait to Oro Bay. Enemy planes and surface craft made it dangerous for ships to remain at Oro Bay during daylight hours, so loads were transferred to smaller craft which made hazardous nightly runs through reef-studded coastal waters to an advance supply base. There the troops, partly infantry, partly units of the 107th Quartermaster Battalion, put out through the surf in native outrigger canoes and unloaded the cases of food and equipment.

This seaborne supply line remained always under threat of interruption. Attacks by enemy planes on 16–17 November put it out of operation for 3 weeks. The 22d Portable Hospital on the Alacrity, one of the four boats sunk in the first of these attacks, lost four men and all of its equipment and supplies. On 23 December, two enemy PT-type boats sank a ship off Hariko and machine-gunned the supply base. However, through most of the period, the Allied forces on the east flank were successfully supplied by the sea route.

The chain of supply stretched 1,700 miles from bases in Australia to the landing fields, dropping grounds, and coastal dumps. From such forward points, cargoes had to be transported to the men in and directly behind the combat lines. This last vital transport link was formed by a few peeps and some 700 fuzzy-headed native carriers, who delivered their 40 pounds apiece to dumps just outside the range of small-arms fire.

Requests for supplies flowed from the front to Lt. Col. Ralph T. Birkeness, Division Quartermaster at Port Moresby. All were marked 22


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 urgent. He had to assign priorities on the limited space of the planes and boats, determining whether the soldiers would receive bullets or rations, or food for the natives who were indispensable as carriers and would go home unless fed, or replacements for the ordnance watches "going to hell" in the damp climate of the Papuan jungle and needed for synchronization of combat efforts, or canister and fame throwers when weapons available in the front lines failed to solve tactical problems. On one occasion, General Harding wired from his advanced headquarters, "So many priorities on medical, engineer, anti-aircraft and other [supplies] are causing neglect of general ammunition and food for our fighting men."

Quartermaster officers in the field reported that the troops subsisted for almost a week on a daily diet of one-third of a "C" ration and one-sixth of a "D" ration, equivalent to about 1,000 calories a day. Even when shipping space was available, the problem was not solved. Foods in glass jars were packed in paper cartons which disintegrated in the rain. The jars, unprotected by cartons, often broke. There was a shortage of packing equipment. Parachutes used for the drops had to be salvaged for further use. However, as the operation progressed, both headquarters and officers at the front learned by trial and error what was needed, in what quantity, and how to use efficiently the available transportation.

The establishment and maintenance of communications across the Owen Stanley Mountains and in the Buna area involved many difficulties. Shortage of space on the air supply line from Port Moresby limited delivery of equipment. Radio sets corroded or shortcircuited in the hot, moist air even when protected from the heavy rains. The more powerful radios used for communication between the front and Port Moresby worked well, but the limited Signal Corps staff was swamped with the coding and decoding of messages. Portable radios were often ineffective because the dense growth of trees and under-brush limited their range.

Headquarters at Simemi was connected with Dobodura, Hariko, and Oro Bay by teletype. Telephone wires were laid to each regimental headquarters, to the four air strips, to aircraft warning stations, and to all artillery batteries. Where jungle and swamp were absolutely impenetrable, reels were mounted on rafts and field wire was

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 strung on trees as the rafts floated downstream. Altogether some 300 miles of wire were laid, all of it by hand and much of it under fire. On 23 December, 1st Lt. Philip S. Winson of the 32d Signal Company, while laying a battalion observation post telephone line, was with a platoon of the 126th Infantry which was isolated by enemy counter-attack. For organizing the defense of several captured enemy bunkers, he received a citation.

Lines were frequently broken by enemy patrols or bombing. Native carriers innocently cut lengths of wire to tie up their bundles. Repair parties were sniped at by the Japanese in daylight, and at night were fired on by our own men, who were suspicious of any movement in the dark. However, communications were maintained effectively throughout the operation; few complaints were heard. 25


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 Map No. 4

Cartographic Section, Diss. Unit by J. R. Hogan


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Heck v. Humphrey/Concurrence Thomas

, concurring.

The Court and correctly begin their analyses with the realization that "[t]his case lies at the intersection of  the Civil Rights Act of 1871, Rev. Stat. § 1979, as amended, 42 U.S.C. § 1983, and the federal habeas corpus statute, 28 U.S.C. § 2254." Ante, at 480; post, at 491. One need only read the respective opinions in this case to understand [p491] the difficulty of the task before the Court today. Both the Court and embark on a similar enterprise—harmonizing "[t]he broad language of § 1983," a "general" statute, with "the specific federal habeas corpus statute." Preiser v. Rodriguez, 411 U.S. 475, 489 (1973).

I write separately to note that it is we who have put § 1983 and the habeas statute on what appropriately terms a "collision course." Post, at 492. It has long been recognized that we have expanded the prerogative writ of habeas corpus and § 1983 far beyond the limited scope either was originally intended to have. Cf., e.g., Wright v. West, 505 U.S. 277, 285–286 (1992) (opinion of, J.) (habeas); Golden State Transit Corp. v. Los Angeles, 493 U.S. 103, 117 (1989) (, J., dissenting) (§ 1983). Expanding the two historic statutes brought them squarely into conflict in the context of suits by state prisoners, as we made clear in Preiser.

Given that the Court created the tension between the two statutes, it is proper for the Court to devise limitations aimed at ameliorating the conflict, provided that it does so in a principled fashion. Cf. Malley v. Briggs, 475 U.S. 335, 342 (1986). Because the Court today limits the scope of § 1983 in a manner consistent both with the federalism concerns undergirding the explicit exhaustion requirement of the habeas statute, ante, at 483, and with the state of the common law at the time § 1983 was enacted, ante, at 484–486, and n. 4, I join the Court's opinion.

title: Heck v. Humphrey/Concurrence Thomas


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text: How and What to Grow in a Kitchen Garden of One Acre (10th Ed)/Editor's Preface