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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?

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    $\begingroup$ If an odd prime $p$ divides $n^4+1$, what does that tell you about $p$? $\endgroup$ – Daniel Fischer Aug 16 '15 at 21:41
  • $\begingroup$ Prof. Leo Morse observed that way to show there"s no largest prime is to produce a strictly increasing sequence of natural numbers that are pairwise co-prime.For example.the Fermat numbers. $\endgroup$ – DanielWainfleet Aug 16 '15 at 23:39
  • $\begingroup$ See also Dirichlet's theorem on arithmetic progressions. $\endgroup$ – Lucian Aug 17 '15 at 4:24
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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 $\equiv 1\pmod{8}$.

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  • $\begingroup$ 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. $\endgroup$ – Aoden Teo Masa Toshi Aug 20 '17 at 9:14
  • $\begingroup$ This is obvious, come on. $\endgroup$ – Paolo Leonetti Aug 20 '17 at 10:31

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