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Show that if $n$ is an integer, then every prime factor of $4n^2+1$ is congruent to $1 \pmod 4$. (Hint: if $p\mid 4n^2+1$), then what can you say about $(-1\mid p)$?

Approach:

I went over all the cases for n and concluded that $4n^2+1 \equiv 1\pmod 4$. We know that $4n^2+1$ can be represented as the product of prime numbers, so let $4n^2+1=q_1\cdots q_\ell$. This implies that $q_i \mid 4n^2+1$

b) Show that there are infinitely many primes congruent to 1(mod 4)

Approach

The common way to do this is by contradiction. Assume there are finitely many primes congruent to 1 mod 4. Call them $p_1,....,p_k$. Now let's build a number n congruent to 1 mod 4 with these primes. n=$4p_1*....*p_{k}+1 \equiv 1(mod\text{ } 4)$. By definition we can represent n as a product of primes. Let $n=q_1,...,q_l$. If $q_j=p_j$ then $q_j | 4p_1*...*p_k$ and $q_j | 1$ which can't be possible, so contradiction.

I am just wondering if there is a way to do the latter by using what was just done.

Martin Sleziak
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TheMathNoob
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    I'd say so. If $p$ is a prime divisors of $4n^2+1$, $-1$ is a quadratic residue $\pmod{p}$, because $(2n)^2\equiv -1\pmod{p}$, hence $p\equiv 1\pmod{4}$ by computing the Legendre symbol $\left(\frac{-1}{p}\right)$. – Jack D'Aurizio Jul 09 '16 at 23:38
  • , so you used here the lemma if $(\frac{-1}{p})=1$ then $p \equiv 1(mod 4)$ from quadratic reciprocity right? – TheMathNoob Jul 09 '16 at 23:53
  • If $p$ divides $4n^2+1$, then $(2n)^2\equiv -1\pmod{p}$, so $-1$ is a QR of $p$, so $p\equiv 1\pmod{4}$. – André Nicolas Jul 09 '16 at 23:55
  • @TheMathNoob: quadratic reciprocity is not really needed here, $\left(\frac{-1}{p}\right)=1$ implies $p\equiv 1!!\pmod{4}$ since $$\left(\frac{-1}{p}\right) = (-1)^{\frac{p-1}{2}}.$$ – Jack D'Aurizio Jul 09 '16 at 23:57
  • Yes that's what I was thinking, it comes from euler's criterion. – TheMathNoob Jul 09 '16 at 23:57
  • I am going to post another question on same thread hence it's the part of the same question, so I think it's related – TheMathNoob Jul 10 '16 at 00:00

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