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.

Take the given identity 3p²+p=4q²+q and multiply both sides by 16 and add 1:

48p^2+16p+1 = 64q^2+16q+1

Expressing each side in terms of polynomial squares gives:

(8p+1)^2 - 16p^2 = (8q+1)^2

Applying a common parameterization for the Pythagorean theorem gives:

8p+1 = x^2+y^2

2xy = 8p^2

8q+1 = x^2-y^2

From there take the sum and difference of 8p+1 and 8q-1:

(8p+1) + (8q+1) = (x^2+y^2) + (x^2-y^2)

2(4p+4q+1) = 2x^2

4p+4q+1 = x^2

(8p+1) - (8q+1) = (x^2+y^2) - (x^2-y^2)

8(p-q) = 2y^2

4(p-q) = y^2

This is sufficent to prove that if 3p²+p=4q²+q then p-q and 4p+4q+1 are both squares