Certain primes $p$ which seems to be $\equiv 1\pmod{10}$

Question

$A=\{5,61,181,1741,36721,60901,135721,431521,531481,552301,685621,2834581,3567121,3674761,3696481,4503001,6121501,6811741,9456901,11002741,11524801,12495001,15629641,16068781,18611101,20218441,20231161,25625641,25999261,28042561,33874681,35540881,38377561,42698041\}$

is a set of the smallest primes $p$ such that $\frac{p^2+1}{2}$ is prime and such that there is a prime $q$ with $p=\frac{q^2+1}{2}$. It seems that for all $p\in A$ except $p=5$, $p\equiv 1\pmod{10}$. Can this be proved?

Answer 1

If $q$ is an odd prime $\equiv 3,7\pmod {10}$ then $q^2+1\equiv 0 \pmod {10}$, so $q^2+1\equiv 0\pmod 5$. Hence those are impossible.

So, if $q>5$ we must have $q\equiv 1, 9\pmod {10}$ and the conclusion follows at once.