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I want to prove:

There are infinitely many primes $p$ with $p\equiv 1 \mod 4$.

My lecture notes say this can be proven by considering the numbers $(n!)^2+1$ and using that they satisfy $(n!)^2+1\equiv 1\mod 4$.

I have two questions:

  • How to conclude the claim from this fact?
  • Why is the fact true?
MyNameIs
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    $-1$ is a quadratic residue modulo any prime factor $p$ of $(n!)^2+1$. Therefore $p\equiv1\pmod4$. If there were only finitely many such primes what would happen if $n$ were the largest of that lot? – Jyrki Lahtonen Jul 19 '16 at 12:21
  • see this question – n1000 Jul 19 '16 at 12:49
  • Why is $p\equiv 1\mod 4$ if $p$ is a prime factor of $(n!)^2+1$, @JyrkiLahtonen ? – MyNameIs Jul 20 '16 at 10:05
  • Is the following argument OK? $p$ divides $(n!)^2+1$, hence $(n!)^2\equiv -1 \mod p$ and therefore $\left(\frac{-1}{p} \right)=1$. This implies, by Euler's lemma, that $p\equiv 1 \mod 4$. – MyNameIs Jul 20 '16 at 10:10
  • So we found a prime $p$ which is of the form $4k+1$. Now we repeat the procedure for $n'=(p!)^2+1$. This gives us a prime $p'$ of the form $4l+1$. This prime must be bigger than $p$ because otherwise it would divide $1$. – MyNameIs Jul 20 '16 at 10:12
  • That's the idea. Alternatively you can do a contrapositive argument. If there were only finitely many primes of this type, there would be a largest such prime... – Jyrki Lahtonen Jul 20 '16 at 10:25

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