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prove that the prime factors of $n^{2}+4$ , $ n \in \mathbb{N}$ are congruent to $ 1 \ or \ 5 \ (mod \ 8)$. $$$$ I can that the statement holds for n=1,2,3 , so can i use induction principle ? If there is any other way to solve this? Thanks

Jyrki Lahtonen
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mmath
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    Have you heard about quadratic residues? – Seewoo Lee Mar 05 '17 at 11:22
  • Odd prime factors, maybe? – Oscar Lanzi Mar 05 '17 at 11:29
  • If you seek yo prove it directly, that is a proof that constructs the result for all of them. If it is true for infinite $n$ then you require induction or contradiction. For contradiction it would be like assuming there exists an $n$ such that the premis isnt true or factors not congruent to 1 or 5. Then deducing a contradiction would show that they must at least be congruent to 1 or 5. Just to explain that if it helps. Induction would also work. – marshal craft Mar 05 '17 at 12:10

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If $p$ is a prime factor of $n^2+4$ then $n^2\equiv-4\pmod p$, and $-4$ is a quadratic residue modulo $p$. Because $4$ is a square this is equivalent to $-1$ being a quadratic residue modulo $p$. It is well known that this is so if and only if $p\equiv1\pmod4$ (with the obvious exception $p=2$).

Therefore $p$ is congruent to either $1$ or $5$ modulo $8$, or $p=2$.

Jyrki Lahtonen
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  • In case you have never seen it, the fact that $-1$ being a quadratic residue modulo $p$ implies $p\equiv1\pmod4$ is explained e.g. here. Probably also earlier, that is one of the most often used tricks in elementary number theory. – Jyrki Lahtonen Mar 05 '17 at 11:40
  • correct to match one part of the answer – mmath Mar 05 '17 at 11:40
  • @mmath: Sorry, what do you mean by part of the answer? I don't think there is anything missing :-) – Jyrki Lahtonen Mar 05 '17 at 11:42
  • ok, my apologise . everything is there – mmath Mar 05 '17 at 11:45
  • That's ok. I was just worried for a moment. May be I misread of something :-) – Jyrki Lahtonen Mar 05 '17 at 11:46
  • @JyrkiLahtonen: the link you included states that $p \equiv 3 \mod 4$ implies that $-1$ is not a quadratic residue modulo $p$... Shouldn't your first comment say that since $-1$ is quadratic residue, then $p - 3$ is not a multiple of $4$? – Student Mar 05 '17 at 12:18
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    Correct, @Student. Sorry about that. I edited the comment (one of the diamond superpowers is the ability to edit old comments). – Jyrki Lahtonen Mar 05 '17 at 12:29