Show there are infinitely many primes that are equivalent to $1 \pmod{8}$.
I am supposed to use $x^4 + 1$ to and how Euclid showed that there were infinetely many primes to prove this, but I am not sure how. I found that 17 and 41 work but I don't know how to show that there are infinetly many.
I also looked at Show there are infinitely many primes that are equivalent to 1 mod 8. but it didn't help much.
Show that if $p$ is a prime factor of $(2a)^4+1$, where $a$ is an integer, then $p$ is odd, and as $(2a)^4\equiv -1\pmod p$ then the multiplicative order of $2a$ modulo $p$ is $8$. Deduce that $p\equiv1\pmod 8$. Also observe that $p\nmid a$ so that one can avoid $p$ lying in any given finite set of primes.
As Angina Seng has already provided the standard technique for this particular problem, I will make some remarks so that you can solve other similar problems. This is a special case of Dirchlet's theorem which requires higher mathematics to prove. However, there are some special cases that can be proven in a Euclidean manner, meaning it is reminiscent of the classic proof of Euclid's theorem on the infinitude of primes. These proofs are elementary. See Keith Conrad's survey, which gives a biconditional criterion for when such proofs exist, which is that the residue squared must be congruent to $1.$ I am not aware of an easy way of always finding a suitable polynomial. In some cases, you will have to include $2$ in the product of the primes. A relatively elementary proof also exists for the residue $1$ case (your stated problem falls in this category) using cyclotomic polynomials.
