numbers

If q =2, then p=5 is a solution.

If q is an odd prime then let p = 6m±1

Then 36m^2 \pm 12m = 6q^2 \Rightarrow 2m(3m \pm 2) = q^2 which means q is even, a contradiction.

So (p,q) = (5,2) is the unique solution.

edited: as pointed out by anant sir. also apologies as I didnt notice this was in 9th/10th forum.picked it up from home page

Imagining none to be evn, we can just write p² as 4k+1, so that we are left with
3q²=2k, which directly gives q=2.