TITLE: Given positive integers $ x , y$ and that $3x^2 + x = 4y^2 +y$ Prove that $x-y$ is a perfect square
QUESTION [5 upvotes]: This is some old Olympiad problem (I vaguely remember its Iranian )
I factorised it as
$ (x-y)(3x + 3y +1) = y^2 $
Then I fruitlessly tried to prove 
$\gcd(x-y, 3x + 3y +1 ) = \gcd (6x+1,6y+1) = 1$
for proving that $x-y$ is a perfect square.
How do I continue my method or are there other ways to approach this intuitively?

REPLY [8 votes]: Defining $z = x-y$, from $y^2 = (x-y)(3x + 3y +1)$, we have 
$$ y^2 = 3z^2 + 6yz+ z$$ which is equivalent to $$(y - 3z)^2 = z(12z+1).$$ 
Since $\gcd(z, 12z+1)=1 $, $z$ and $12z+1$ both must be perfect squares.