Equation solving; three expressions as squares If positive integer $x,y$ satisfy,
$$2x^2+x=3y^2+y$$ then $x-y,2(x+y)+1,3(x+y)+1$ are perfect squares.
I somehow managed to prove $x-y$ is a perfect square but couldn't prove the others.

$2x^2+x=3y^2+y\Longrightarrow (x-y)(3x+3y+1)=x^2$. Now observe if a prime $p$ divides both the factors, $x-y$ and $3x+3y+1$ on the LHS, then $p^2$ divides $x^2$ and so $p$ divides $x$. 
Now, $p$ divides $x$ and $p$ divides $x-y$ implies $p$ divides $y$. But this is a contradiction as $p$ divides $3x+3y+1$. So we can conclude that the two factors on LHS are co prime. Now if two co prime numbers multiply to yield a perfect square, then each of them should be perfect squares.
Now observe again that the given equation is equivalent to $(x-y)(2x+2y+1)=y^2$. A similar reasoning follows.
