Problem 3

By Step 5, $n\le\frac{p-h}2$ and $p\mid n^2+1$, so by Step 1's computation $p$ is a prime factor of $n^2+1$ with $p>2n+\sqrt{2n}$. There are infinitely many primes $p\equiv1\pmod8$ (a special case of Dirichlet's theorem, or shown directly via cyclotomic polynomials). The corresponding positive integers $n=p-x$ cannot take only finitely many values: if they did, the numbers $n^2+1$ would have only finitely many prime divisors, contradicting the infinitely many distinct primes $p$ constructed above. Hence infinitely many distinct positive integers $n$ satisfy the required property. $\blacksquare$