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Let $n \in \mathbb{N}$. Any odd prime factor $p$ of $n^2 +1$ has the form $p = 4k+1$ for some integer $k \geq 0$.

Proof using contradiction. I am a little bit stuck, can someone help me get started on this question

Arthur
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albert
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  • Have a look at http://math.stackexchange.com/questions/552398/let-n-suppose-that-p-is-an-odd-prime-number-that-divides-n21-show-that-if-p/552414#comment1175532_552414 – Arthur Nov 05 '13 at 05:31
  • That is quite different from this. – albert Nov 05 '13 at 05:33
  • Assume there is a prime divisor $p = 4k + 3$ of $n^2 + 1$. Then by the result of the question I linked we have $n^{p-1} \equiv -1$ mod $p$. You should be able to derive a contradiction from that. – Arthur Nov 05 '13 at 05:34
  • So, dude, you know what the Legendre symbol is? – Will Jagy Nov 05 '13 at 05:36
  • this question is already asked and answer is in the comments... http://math.stackexchange.com/questions/549783/any-odd-prime-factor-p-of-n2-1-has-the-form-p-4k-1-for-some-integer –  Nov 05 '13 at 06:22

1 Answers1

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Hint: Suppose, towards a contradiction, that $p$ does not have the form $4k+1$, where $k \geq 0$. Then by the Division Algorithm, we know that for some $k \geq 0$, we have either $p=4k$ or $p=4k+2$ or $p=4k+3$. But it can't be $4k$ or $4k+2$, since $p$ is odd. So we may assume that $p=4k+3$ for some $k \geq 0$. Use this to get a contradiction.

Adriano
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