Prove a case of Dirichlet's Theorem: that there are infinitely many primes of the form $8k+1$

Question

Prove a case of Dirichlet's Theorem: that there are infinitely many primes of the form $8k+1$ using these steps.

(Dirichlet’s Theorem). Let $a$, $b$ be two positive integers. If $\gcd(a, b) = 1$, then there exists an infinite number of primes of the form $ak + b$.

The aim of this exercise is to prove Dirichlet's Theorem when $a = 8$ and $b = 1$. Let $x$ be an even integer and $p$ be a prime divisor of $x^4+1$.

1. Show that $\left(\frac{-1}{p}\right) = 1$.

2. Prove that $x$ and $p$ are coprime and deduce that $x$ is invertible modulo $p$.

3. Show that $\left(\frac2p\right) = 1$. Hint: You might find the following identity useful: $$ x^4+1 = (x^2+1)^2-2x^2 $$

4. Show that $p \equiv 1 \bmod 8$.

5. Deduce that there are infinitely many primes $p$ congruent to $1$ modulo $8$.

So far I've got $N = (2p_1p_2\ldots p_r)^4+1$

Let $p$ be a prime divisor of $N$. If $p|N$ then

$$p| (2p_1p_2\ldots p_r)^4+1$$

$$-1 = (2p_1p_2\ldots p_r)^4 \mod p$$

$$(-1/p) = 1$$

This is where I get stuck.

My lecturer has replied with:

'This is indeed the beginning of the correct answer. You have that N is of the form x^4+1, so you can use question 4) and deduce N=1 mod 8. Now, can it be equal to one of the p_i?'

I'm still unsure where to go from here?

Answer 1

Someone has already answered the question

Suppose that there are only finitely many such primes p1,…,pk≡1(mod8). Then consider the following number (2p1⋯pk)4+1, which is coprime with each pi, 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 ordp(2p1⋯pk)=8 which divides φ(p)=p−1 by Fermat's theorem. Therefore p is another prime ≡1(mod8).

Answer 2

First of all, note that all odd and prime divisors of n4+1 have the form 8k+1. Suppose that there are just finite prime numbers of form 8k+1. Construct the number a=(2p1…pn)4+1, where {pi}ni=1are all primes of form 8k+1. This number is odd, and has at least one prime divisor q. Then q has the form 8l+1. But then q=pj for some j. But it is not possible since q divides a and q divides (2p1…pn)4, i.e, q divides 1.

Answer 3

My proof goes like this:

Assume by way of contradiction that there are only finitely many of such prime numbers, say $p_1, p_2, \dots, p_r$. Consider $p=16(p_1p_2\cdot \cdot\cdot p_r)^4+1=(2p_1p_2\cdot \cdot \cdot p_r)^4+1$. Recall that the congruence $x^4 \equiv -1 \pmod {p}$ is solvable if and only if $p \equiv 1 \pmod{8}$. $p$ cannot be a prime of the form $8n+1$, but $p \equiv 1 \pmod{8}$, there is a contradiction. Thus, there must be infinitely many prime numbers of the form $8n+1$.

Answer 4

Maybe this will help fill in the details. Let us write $x = 2p_1\ldots p_r$, for some finite $r$, where $p_1 \ldots p_r$ are in Teri's answer, all the primes that are 1 mod 8. Then as Teri already noted there is a prime $p$ that divides $x^4+1$, and $p \not \in \{p_1,\ldots, p_r\}$. We claim that $p$ is of the form $8k+1$ for some integer $k$ next. As already noted by Teri, this will imply that $p_1,\ldots, p_r$ are not all the primes that are 1 mod 8, which will give you what you want.

Note that

$$x^4+1 \equiv_p 0 \Rightarrow x^4 \equiv_p -1 \Rightarrow x^8 \equiv_p 1;$$

That $x^4 \not \equiv_p 1$ and $x^8 \equiv_p 1$, together imply that $x$ has order precisely 8 in the group $\left(\mathbb{F}_p\right)^{\times}$, which implies that $|\left(\mathbb{F}_p\right)^{\times}| = p-1$ is divisible by 8, which implies that $p$ is of the form $8k+1$ for some integer $k$.