Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / C / C2 / vi / Solution

Suppose there are only finitely many primes $p_1,\dots ,p_r\equiv 3(\mathrm{mod}8)$, and let $N$ be their product. This is odd, so $N^2+2\equiv 3(\mathrm{mod}8)$. No listed prime divides this number, since reduction modulo any $p_i$ gives two. Every prime divisor $q$ is odd and does not divide $N$, and satisfies $N^2\equiv -2(\mathrm{mod}q)$. Thus $(-2/q)=1$, forcing $q\equiv 1$ or $3(\mathrm{mod}8)$ by (v).

A product of prime factors all congruent to one modulo eight could not be three modulo eight. More precisely, the product of factors congruent to three is three modulo eight exactly when their total exponent is odd. Thus at least one prime divisor $q\equiv 3(\mathrm{mod}8)$ occurs. It is not in the alleged complete list, a contradiction. Hence there are infinitely many primes congruent to three modulo eight.