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Show there are infinitely many primes that are equivalent to $1 \pmod{8}$.

Hello there! I have been trying to do this problem for a pretty long time with no avail.

I noticed that this is really similar to Euclid's proof that there are infinitely many primes. However, I can't find a way to use that here. Can anyone help me?

Frank
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Suppose that there are only finitely many such primes $p_1,\ldots,p_k \equiv 1 \pmod 8$. Then consider the following number $$ (2p_1\cdots p_k)^4+1, $$ which is coprime with each $p_i$, and has remainder $1$ modulo $8$. Since it is odd and greater than $1$, it has to be divisible by an odd prime $p$. Then $$ \mathrm{ord}_p(2p_1\cdots p_k)=8 $$ which divides $\varphi(p)=p-1$ by Fermat's theorem. Therefore $p$ is another prime $\equiv 1\pmod{8}$.

Paolo Leonetti
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    Note that when you said that $(2p_1 \cdots p_k)^4+1>1$, you assumed that $k \geq 1$. In order to complete your proof, you must show that this assumption was valid. In other words, you must show that there is at least one prime which is congruent to $1 \pmod{8}$. An example of such a prime that you could use would be 17. – Aoden Teo Masa Toshi Aug 20 '17 at 09:14
  • This is obvious, come on. – Paolo Leonetti Aug 20 '17 at 10:31
  • Why does $\mathrm{ord}_p(2p_1\cdots p_k)=8$? –  Aug 11 '20 at 23:19
  • Because we have $$(2p_1\cdots p_k)^4+1\equiv 0\pmod p$$ which gives us $$(2p_1\cdots p_k)^4\equiv -1\pmod p.$$ We square this getting $$\mathrm{ord}_p(2p_1\cdots p_k)=8.$$ Any smaller 'order' would not result in it being congruent to $1$, hence not being an order. – Mike Smith Aug 12 '20 at 21:36