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!
 A: 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.
A: Generalization. Let $a,b$ be integers. If there exists consecutive integers $c,d$ such that $a-b=a^2c-b^2d$, then $|a-b|$ is a perfect square.
Proof. If $c=d+1$, we have $a-b=a^2(d+1)-b^2d=(a-b)(a+b)d+a^2$, so $$a^2=(a-b)(1-d(a+b))$$ Now let $g=\text{gcd}(a-b,1-d(a+b))$. We have $g^2|a^2$, so $g|a$. Now we have $g|b$, and we have $g|1$, so $g=1$.
Since $a-b$ and $1-d(a+b)$ are coprime and their multiple is a perfect square, we are done.
The case $c+1=d$ is handled similarly. $\blacksquare$.
A: Solve, 
$$x - y=5y^2 - 4x^2\tag1$$
hence,
$$x = \frac{-1\pm\sqrt{1+16y+80y^2}}{8}$$
Let,
$$1+16y+80y^2 = \big(\tfrac{2p}{q}y-1\big)^2$$
Expand and factor to get,
$$y = \frac{pq+4q^2}{p^2-20q^2}$$
with the relevant $x$ as,
$$x = \frac{pq+5q^2}{p^2-20q^2}$$
and $x,y$ will be integers if $p,q$ satisfy the Pell equation,
$$p^2-20q^2 = 1$$
One can then see that $x>y$ and,
$$x-y = \frac{q^2}{p^2-20q^2}=q^2$$
The first few $p,q$ are,
$$9, 2 \\161, 36 \\ 2889, 646$$
and these yield $x,y$,
$$38, 34 \\ 12276, 10980 \\ 3952874, 3535558$$
and so on, consistent with the numerical search made in the comments.
A: Use Bill's observation that

Theorem: If $a \mid b^2 $ and $ a = \pm c^2  \pm kb^2$, then $ a = \pm (c, b) ^2$
Proof: $ a = \gcd(a, b^2) = \gcd(\pm c^2 \pm kb^2, b^2) = \gcd(c^2 , b^2 ) = \gcd(c, b)^2$.

Now,

*

*$ x -y > 0$ by assumption.

*OP has showed that $x-y \mid y^2$.

*$x-y = 5y^2 - 4x^2 = -(2x)^2  + 5y^2$ .

Hence, $ x - y = (2x, y)^2$.
