Prove that there are infinitely many primes of the form $8k- 1$ without using Dirichlet’s Theorem.
Hint: Suppose there are a finite number of primes $p_1, p_2, \ldots , p_k$ of this form and consider the integer $(4p_1p_2\ldots p_k)^2− 2$ and then use Proposition (7.15).
We assume, for the purpose of contradiction, there are finite primes $p_1, p_2, \ldots, p_k$ of the form $8k-1$, where $k$ is a positive integer.
Let's consider a constructed number $N=(p_1p_2 \ldots p_k)^2-2$, noting that $p_i \equiv -1 \pmod 8$,
$$ N \equiv (p_1p_2 \ldots p_k)^2 -2 \equiv (-1 \times -1 \ldots -1)^2 -2 \equiv 1-2 \equiv -1 \pmod 8 $$
This means $N$ is of the form $8k-1$.
We have two cases for $N$, prime or composite.
Case $N$ prime
If $N$ is prime, then we have found another prime of the form $8k-1$ that is larger than any of the $p_1, p_2, \ldots, p_k$ of the form $8k-1$. This contradicts the assumption there are only $k$ primes of form $8k-1$.
Case $N$ composite
If $N$ is composite, then it has a prime factor, which we can call $q$. This means
$$ (p_1p_2 \ldots p_k)^2 \equiv 2 \pmod q $$
By Proposition (7.15) this means $q \equiv \pm 1 \pmod 8$.
Let's consider each of these two options.
- If $q \equiv -1 \pmod 8$, that is $q=8t-1$ for some integer $t$, then $q$ must be one of the known $p_1, p_2, \ldots, p_k$. This gives us $q \mid (p_1p_2 \ldots p_k)^2-2 \implies q \mid 2$. This is a contradiction, and invalidates the assumption there are only $k$ primes of form $8k-1$.
- If all the odd prime factors of $N$ are congruent to $1 \pmod 8$ that would mean $N \equiv 1 \pmod 8$. But we have seen above that $N \equiv -1 \pmod 8$. This means some (an odd number) of the prime factors must be congruent to $-1 \pmod 8$, and we have just shown how this invalidates the assumption there are only $k$ primes of the form $8k-1$.
For all the possible cases, we have shown the assumption of a finite number of primes of the form $8k-1$ leads to a contradiction. And so there are an infinite number of primes of the form $8k-1$.