$-1$ is a quadratic residue modulo $8n+2$; what can be stated about $n$?

Question

When $a$ exists for which $a^2 \equiv -1 \mod {8n+2}$, then what can be stated about $n$? I am not looking for theoretical statements about the prime factorization of $8n+2$; I need explicit statements about $n$, when possible. In case that is not possible, then a statement about $8n+2$ is OK.

Answer 1

In general the numbers which have $-1$ as a quadratic residue are characterized precisely as those numbers not divisible by $4$ or by any prime of the form $4k+3$. Since $8n+2$ does not fall in the former category, it is necessary and sufficient that it be $2$ times the product of primes of type $4k+1$.

This latter property is equivalent to being expressible as the sum of two co-prime odd squares, so $8n + 2 = a^2 + b^2$ for some $(a,b)=1$.

Finally, we get a necessary condition on $n$ by noting that $n = (a^2-1)/8 + (b^2-1)/8$, so $n$ is the sum of two triangular numbers. This would be sufficient, except there is no direct way to express the coprimality of the underlying values of $a$ and $b$. The best we could say is that "$n$ is the sum of two triangular numbers $A,B$ such that $8A+1$ and $8B+1$ have no common factor".