If $x-y = 5y^2 - 4x^2$, prove that $x-y$ is perfect square Firstly, merry christmas!
I've got stuck at a problem. 

If x, y are nonzero natural 
  numbers with $x>y$ such that
  $$x-y = 5y^2 - 4x^2,$$
  prove that $x - y$ is perfect square.

What I've thought so far:
$$x - y = 4y^2 - 4x^2 + y^2$$
$$x - y = 4(y-x)(y+x) + y^2$$
$$x - y + 4(x-y)(x+y) = y^2$$
$$(x-y)(4x+4y+1) = y^2$$
So $4x+4y+1$ is a divisor of $y^2$.
I also take into consideration that $y^2$ modulo $4$ is $0$ or $1$ (I don't know if this can help.)
So how do I prove that $4x+4y+1$ is a perfect square (this would involve $x-y$ - a perfect square)? While taking examples, I couldn't find any perfect square with a divisor that is $M_4 + 1$ and is not perfect square. 
If there are any mistakes or another way, please tell me.
Some help would be apreciated. Thanks!

Let $z=x-y$. Then $(x-5z)^2=z(20z+1)$ and so $z(20z+1)$ is a perfect square. Since $\gcd(z,20z+1)=1$ both $z$ and $20z+1$ must be perfect squares.
