 Solution

Suppose only finitely many primes $q_1,\dots ,q_r$ are congruent to $7$ modulo $8$, and setThenIf an odd prime $q$ divides $M$, then $n^2\equiv 2(\mathrm{mod}q)$, so part (a) givesSince their product $M$ is $7$ modulo $8$, at least one prime divisor $q$ is $7$ modulo $8$ to an odd power. Yet no listed $q_i$ divides $M$, because $n\equiv 0(\mathrm{mod}{q}_i)$ givesThis produces a new prime congruent to $7$ modulo $8$, a contradiction. HenceThis is the Euclid proof for infinitely many primes congruent to seven modulo eight.