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Suppose that $p_1,p_2,...,p_r$ are the only primes congruent to $1\ (\text{mod}\ 4)$. Prove that $4p_1^2p_2^2...p_r^2+1$ is divisible only by the primes congruent to $3\ (\text{mod}\ 4)$. So far all I have been able to do is prove there are infinitely many primes congruent to $3\ (\text{mod}\ 4)$

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  • $\begingroup$ $$(4p_1^2p_2^2...p_r^2+1,p_r)=1$$ $\endgroup$ – lab bhattacharjee Nov 9 '17 at 6:12
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Are you assuming that there are only finitely many primes $\equiv1\pmod4$ for sake of contradiction?

If $p$ is a prime factor of $4m^2+1$ with $m$ an integer, then $p$ is odd, and $(2m)^2\equiv-1\pmod p$, that is $-1$ is a quadratic residue modulo $p$. Do you know a criterion for when $-1$ is a quadratic residue?

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