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Is it necessary for a number of the form $4k^2+1$ to have at least one prime factor of the form $4n+1$?

I was trying to prove that there are infinitely many primes of the form $4n+1$, but to prove it, I need to prove the above statement true.

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  • $\begingroup$ Even more : EVERY prime factor of such a number must be of the form $4n+1$ $\endgroup$
    – Peter
    Jan 13, 2019 at 15:53
  • $\begingroup$ @Peter I would really appreciate if you provide a proof for that. $\endgroup$ Jan 13, 2019 at 15:55
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    $\begingroup$ Considering quadratic residues, the proof is trivial. Every prime factor $p$ must be odd, and since the number is of the form $n^2+1$ , $-1$ is a so-called quadratic residue mod $p$ , whenever $p$ divides $4k^2+1$ , we can conclude immediately that $p$ must be congruent to $1$ modulo $4$. $\endgroup$
    – Peter
    Jan 13, 2019 at 16:05
  • $\begingroup$ Cf. here $\endgroup$ Jan 13, 2019 at 16:11
  • $\begingroup$ Also there are infinitely many primes of the form $4n+1$ trivially by Dirichlet's theorem on arithmetic progressions. $\endgroup$
    – Tejas Rao
    Jan 16, 2019 at 17:12

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