If p and q are positive integers that satisfy
3p²+p=4q²+q, prove that p-q, 3p+3q+1 and 4p+4q+1 are squares of integers.

Rearrange the equation 3p²+p=4q²+q into:

3p^2 - 3q^2 + p - q = q^2

Factor the left side:

(p-q)(3p+3q+1) = q^2

From my previous post, p-q is square, therefore 3p+3q+1 must be square.