When trying to solve a bigger problem, I came across the problem of characterising all primes $p,q$ such that $p^2+1=2q$. That is, are there necessary and sufficient conditions for primes $p,q$ to satisfy the equation? Even better, is there a paramatrisation of solutions (though, this is probably unlikely)? I know the solutions $(p,q)=(3,5),(5,13),(11,61),(19,181),(29,421)$ but I don't know that these are the only ones (indeed, it seems likely there are many more).
Some incomplete thoughts: I immediately made the factorisation $(p+1)(p-1)=2(q-1)$, and because $p$ is obviously odd, let $p=2p_1+1$. This leads us to the equation $2p_1(p_1+1)=q-1$. If we substituted $q=2q_1+1$ we have $p_1(p_1+1)=q_1$, so $q$ is twice of the product of two consecutive integers, plus one. It follows that $q$ is $1$ mod $4$, and $p^2$ is $1$ mod $8$. The same conclusion can be obtained simply by noting that $-1$ must be a quadratic residue mod $2q$, hence $1$ is a $4$th power residue mod $2q$. By Lagrange's Theorem we have $4\mid\phi(2q)=q-1$, so $q \equiv1$ mod $4$. Any further thoughts are appreciated!