For primes $p,q$ one has $2p^2=q^2 + 1$. $p=5$ and $q=7 $ are solutions. Any others?

Question

Given primes $p, q$ and $2p^2 = q^2 + 1$ one sees that $p = 5$ and $q = 7$ are solutions. Are they unique? Does this in any way relate to the fact that $7$ is the only solution to a prime $r$ in $r^3 + r^2 + r + 1 = n^2$?

Answer 1

How about $2\cdot 29^2=41^2+1?$ Ignoring the requirement of primes, the solutions recur with $p_{i+1}=3p_i+2q_i, q_{i+1}=4p_i+3q_i$. You can compute those in a spreadsheet and look for pairs which are both prime. My solution is the next one from yours. I didn't find any more until Excel went into scientific notation.

Answer 2

At the risk of having Servaes downvote this answer as incoherent, from his comment I decided to look at the units in $\mathbb Z[\sqrt 2]$. Compute the sequence $(1 + \sqrt 2)^{2n + 1}$ to get units of norm $-1$ in the form $a + b \sqrt 2$, with $a$ and $b \in \mathbb Z$.

Then you're simply looking for $a$ and $b$ to both be prime. And so in that way from $7 + 5 \sqrt 2$ (corresponding to $n = 1$) I get $41 + 29 \sqrt 2$. The next one is $239 + 169 \sqrt 2$... oh, so close!