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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Proofs of Fermat's theorem on sums of two squares ： ウィキペディア英語版 Proofs of Fermat's theorem on sums of two squares Fermat's theorem on sums of two squares asserts that an odd prime number ''p'' can be expressed as : $p\; =\; x^2\; +\; y^2$ with integer ''x'' and ''y'' if and only if ''p'' is congruent to 1 (mod 4). The statement was announced by Fermat in 1640, but he supplied no proof. The "only if" clause is easy: a perfect square is congruent to 0 or 1 modulo 4, hence a sum of two squares is congruent to 0, 1, or 2. An odd prime number is congruent to either 1 or 3 modulo 4, and the second possibility has just been ruled out. The first proof that such a representation exists was given by Leonhard Euler in 1747 and was complicated. Since then, many different proofs have been found. Among them, the proof using Minkowski's theorem about convex sets〔See Goldman's book, §22.5〕 and Don Zagier's short proof based on involutions have appeared. ==Euler's proof by infinite descent== Euler succeeded in proving Fermat's theorem on sums of two squares in 1749, when he was forty-two years old. He communicated this in a letter to Goldbach dated 12 April 1749.〔(Euler à Goldbach, lettre CXXV )〕 The proof relies on infinite descent, and is only briefly sketched in the letter. The full proof consists in five steps and is published in two papers. The first four steps are Propositions 1 to 4 of the first paper〔De numerus qui sunt aggregata quorum quadratorum. (Novi commentarii academiae scientiarum Petropolitanae 4 (1752/3), 1758, 3-40) ()〕 and do not correspond exactly to the four steps below. The fifth step below is from the second paper.〔Demonstratio theorematis FERMATIANI omnem numerum primum formae 4n+1 esse summam duorum quadratorum. (Novi commentarii academiae scientiarum Petropolitanae 5 (1754/5), 1760, 3-13) ()〕 〔The summary is taken from Edwards book, pages 45-48; italics in the original.〕 1. ''The product of two numbers, each of which is a sum of two squares, is itself a sum of two squares.'' ::This is a well known property, based on the identity :::$(a^2+b^2)(p^2+q^2)\; =\; (ap+bq)^2\; +\; (aq-bp)^2\backslash ,$ ::due to Diophantus of Alexandria. 2. ''If a number which is a sum of two squares is divisible by a prime which is a sum of two squares, then the quotient is a sum of two squares.'' (This is Euler's first Proposition). ::Indeed, suppose for example that $a^2\; +\; b^2$ is divisible by $p^2+q^2$ and that this latter is a prime. Then $p^2\; +\; q^2$ divides :::$(pb-aq)(pb+aq)\; =\; p^2b^2\; -\; a^2q^2\; =\; p^2(a^2+b^2)\; -\; a^2(p^2+q^2).$ ::Since $p^2+q^2$ is a prime, it divides one of the two factors. Suppose that it divides $pb-aq$. Since :::$(a^2+b^2)(p^2+q^2)\; =\; (ap+bq)^2\; +\; (aq-bp)^2\backslash ,$ ::(Diophantus identity) it follows that $p^2+q^2$ must divide $(ap+bq)^2$. So the equation can be divided by the square of $p^2+q^2$. Dividing the expression by $(p^2+q^2)^2$ yields: :::$\backslash frac\; =\; \backslash left(\backslash frac\backslash right)^2\; +\; \backslash left(\backslash frac\backslash right)^2$ ::and thus expresses the quotient as a sum of two squares, as claimed. ::If $p^2+q^2$ divides $pb+aq$, a similar argument holds by using :::$(a^2+b^2)(q^2+p^2)\; =\; (aq+bp)^2\; +\; (ap-bq)^2\backslash ,$ ::(Diophantus identity). 3. ''If a number which can be written as a sum of two squares is divisible by a number which is not a sum of two squares, then the quotient has a factor which is not a sum of two squares.'' (This is Euler's second Proposition). ::Suppose $x$ divides $a^2+b^2$ and that the quotient, factored into its prime factors is $p\_1p\_2\backslash cdots\; p\_n$. Then $a^2+b^2\; =\; x\; p\_1p\_2\backslash cdots\; p\_n$. If all factors $p\_i$ can be written as sums of two squares, then we can divide $a^2+b^2$ successively by $p\_1$, $p\_2$, etc., and applying the previous step we deduce that each quotient is a sum of two squares. This until we get to $x$, concluding that $x$ would have to be the sum of two squares. So, by contraposition, if $x$ is not the sum of two squares, then at least one of the primes $p\_i$ is not the sum of two squares. 4. ''If $a$ and $b$ are relatively prime then every factor of $a^2\; +\; b^2$ is a sum of two squares.'' (This is Euler's Proposition 4. The proof sketched below includes the proof of his Proposition 3). ::This is the step that uses infinite descent. Let $x$ be a factor of $a^2+b^2$. We can write :::$a\; =\; mx\; \backslash pm\; c,\backslash qquad\; b\; =\; nx\; \backslash pm\; d$ ::where $c$ and $d$ are at most half of $x$ in absolute value. This gives: :::$a^2\; +\; b^2\; =\; m^2x^2\backslash pm\; 2mxc\; +\; c^2\; +\; n^2x^2\; \backslash pm\; 2nxd\; +\; d^2\; =\; Ax\; +\; (c^2+d^2).$ ::Therefore, $c^2+d^2$ must be divisible by $x$, say $c^2+d^2\; =\; yx$. If $c$ and $d$ are not relatively prime, then their gcd must be relatively prime to $x$ (else the common factor of their gcd and $x$ would also be a common factor of $a$ and $b$ which we assume are relatively prime). Thus the square of the gcd divides $y$ (as it divides $c^2+d^2$), giving us an expression of the form $e^2+f^2\; =\; zx$ for relatively prime $e$ and $f$, and with $z$ no more than half of $x$, since :::$zx\; =\; e^2\; +\; f^2\; \backslash leq\; c^2+d^2\; \backslash leq\; \backslash left(\backslash frac\backslash right)^2\; +\; \backslash left(\backslash frac\backslash right)^2\; =\; \backslash fracx^2.$ ::If $c$ and $d$ are relatively prime, then we can use them directly instead of switching to $e$ and $f$. ::If $x$ is not the sum of two squares, then by the third step there must be a factor of $z$ which is not the sum of two squares; call it $w$. This gives an infinite descent, going from $x$ to a smaller number $w$, both not the sums of two squares but dividing a sum of two squares. Since an infinite descent is impossible, we conclude that $x$ must be expressible as a sum of two squares, as claimed. 5. ''Every prime of the form $4n+1$ is a sum of two squares.'' (This is the main result of Euler's second paper). ::If $p=4n+1$, then by Fermat's Little Theorem each of the numbers $1,\; 2^,\; 3^,\backslash dots,\; (4n)^$ is congruent to one modulo $p$. The differences $2^-1,\; 3^-2^,\backslash dots,(4n)^-(4n-1)^$ are therefore all divisible by $p$. Each of these differences can be factored as :::$a^-b^\; =\; \backslash left(a^+b^\backslash right)\backslash left(a^-b^\backslash right).$ ::Since $p$ is prime, it must divide one of the two factors. If in any of the $4n-1$ cases it divides the first factor, then by the previous step we conclude that $p$ is itself a sum of two squares (since $a$ and $b$ differ by $1$, they are relatively prime). So it is enough to show that $p$ cannot always divide the second factor. If it divides all $4n-1$ differences $2^-1,\; 3^-2^,\backslash dots,(4n)^-(4n-1)^$, then it would divide all $4n-2$ differences of successive terms, all $4n-3$ differences of the differences, and so forth. Since the $k$th differences of the sequence $1^k,\; 2^k,\; 3^k,\backslash dots$ are all equal to $k!$ (Finite difference), the $2n$th differences would all be constant and equal to $(2n)!$, which is certainly not divisible by $p$. Therefore, $p$ cannot divide all the second factors which proves that $p$ is indeed the sum of two squares. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Proofs of Fermat's theorem on sums of two squares」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.