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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 Lagrange's theorem (number theory) ： ウィキペディア英語版 Lagrange's theorem (number theory) In number theory, Lagrange's theorem is a statement named after Joseph-Louis Lagrange about how frequently a polynomial over the integers may evaluate to a multiple of a fixed prime. More precisely, it states that if ''p'' is a prime number and $\backslash textstyle\; f(x)\; \backslash in\; \backslash mathbb()$ is a polynomial with integer coefficients, then either: * every coefficient of ''f(x)'' is divisible by ''p'', or * $f(x)\; \backslash equiv\_p\; 0$ has at most deg ''f(x)'' incongruent solutions. Solutions are "incongruent" if they do not differ by a multiple of ''p''. If the modulus is not prime, then it is possible for there to be more than deg ''f(x)'' solutions. ==A proof of Lagrange's theorem== The two key ideas are the following. Let $\backslash textstyle\; g(x)\; \backslash in\; (\backslash mathbb/p)()$ be the polynomial obtained from $f(x)$ by taking the coefficients $\backslash mod\; p$. Now (i) $f(k)$ is divisible by $p$ if and only if $g(k)\; =\; 0$; (ii) $g(k)$ has no more roots than its degree. More rigorously, start by noting that $g(x)\; =\; 0$ if and only if each coefficient of $f(x)$ is divisible by $p$. Assume $g(x)$ is not 0; its degree is thus well-defined. It's easy to see $\backslash textstyle\; \backslash deg\; g(x)\; \backslash leq\; \backslash deg\; f(x)$. To prove (i), first note that we can compute $g(k)$ either directly, i.e. by plugging in (the residue class of) $k$ and performing arithmetic in $\backslash textstyle\; \backslash mathbb/p$, or by reducing $f(k)\; \backslash mod\; p$. Hence $g(k)=0$ if and only if $f(k)\; \backslash equiv\_p\; 0$, i.e. if and only if $f(k)$ is divisible by $p$. To prove (ii), note that $\backslash textstyle\; \backslash mathbb/p$ is a field, which is a standard fact; a quick proof is to note that since $p$ is prime, $\backslash textstyle\; \backslash mathbb/p$ is a finite integral domain, hence is a field. Another standard fact is that a non-zero polynomial over a field has at most as many roots as its degree; this follows from the division algorithm. Finally, note that two solutions $\backslash textstyle\; f(k\_1),\; f(k\_2)\; \backslash equiv\_p\; 0$ are incongruent if and only if $\backslash textstyle\; k\_1\; \backslash not\; \backslash equiv\_p\; k\_2$. Putting it all together: the number of incongruent solutions by (i) is the same as the number of roots of $g(x)$, which by (ii) is at most $\backslash deg\; g(x)$, which is at most $\backslash deg\; f(x)$. 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「Lagrange's theorem (number theory)」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.