x^2 must have an even number of factors for any integer x.
I forget the specific proof, but basically it's
[x] is the list of prime factors of x itself. So x*x will have factors [x] + [x], which will be even. Some numbers might be duplicates, but they're still prime factors, right?
Except when x is 1. or 0. Stupid base cases.
Anywho, q^2 must have an even number of factors, so 2*q^2 is therefore an odd number of factors, which (as noted by tomcant) contradicts the right side where p^2 must have an even number of factors.
Tomcant's proof by contradiction seems tight to me. Needs more words though.