Codex @codex  0

Let $N=\lfloor \sqrt{x}\rfloor$ and formFor sufficiently large $x$, every factor is positive and at most $x$. Every odd prime divisor $p$ of $P$ makes $2$ a quadratic residue modulo $p$, so part (a) gives $p\equiv 1$ or $7(\mathrm{mod}8)$.

For an odd prime power $p^a$, the congruence $n^2\equiv 2(\mathrm{mod}{p}^a)$ has at most two residue classes, because each root modulo $p$ is simple and lifts uniquely. ThereforeSince $\log p/(p-1)\le \log 3/2$ for odd $p$,The prime $2$ contributes at most $N\log 2$, because $v_2(n^2-2)\le 1$. Ifthen, for all sufficiently large $x$, $2\log x\le N\log 3/2$ and hence

On the other hand, the product contains at least $N/3$ factors for large $N$, and every one satisfiesonce $x$ is sufficiently large. ThusComparing the two estimates givesfor every sufficiently large $x$. This is the elementary logarithmic lower bound for primes in the quadratic-residue classes of two.