If $ x,y ∈\Bbb{Z} $ find $x$ and $y$ given: $2x^2-3xy-2y^2=7$ We are given an equation: 
$$2x^2-3xy-2y^2=7$$
And we have to find $x,y$ where $x,y ∈\Bbb{Z}$.
After we subtract 7 from both sides, it's clear that this is quadratic equation in its standard form, where $a$ coefficient equals to 2, $b=-3y$ and $c=-2y^2-7$. Thus discriminant equals to $9y^2+16y^2+56=25y^2+56$.
Now $x = \frac{3y±\sqrt{25y^2+56}}{4}$ , $x$ to be an integer $\sqrt{25y^2+56}$ has to be an integer too. I substituted $y$ as nonnegative integer first, because the answer wouldn't differ anyway as it is squared and found that when $y=1$, $\sqrt{25y^2+56}=9$, so we get $x = \frac{3±9}{4}$ and when we have plus sign we get $x = \frac{3+9}{4}=3$. So there we have it, $x=3, y=1$ is on of the solution. But $y=-1$ will also work because $y$ is squared, again, we get $x = \frac{-3±9}{4}$, if we have minus sign we have $x=-3$. Thus another solution, leaving us with:
$$ x=3, y=1$$
$$ x=-3, y=-1$$
I checked other integers for $y$ but none of them lead to solution where $x$ is also an integer. But here's problem, how do I know for sure that these two solutions are the only solutions, I can't obviously keep substituting $y$ as integers, as there are infinitely many integers. So that's why I came here for help.
 A: Hint:
Clearly, if $x,y$ is a solution, $-x,-y$ will also be.
Let  $25y^2+56=u^2\iff(u-5y)(u+5y)=56$ where $u\ge0$
As $5y-u,5y+u$ have the same parity, both must be even
$\implies\dfrac{u-5y}2\cdot\dfrac{u+5y}2=14$
Now  $14=\pm1\cdot\pm14,\pm2\cdot\pm7$
If $y>0, u+5y\ge1+5\implies\dfrac{u+5y}2\ge3$ i.e., $=7$ or $14$
If $y<0, u-5y\ge1+5$
A: Notice $$2x^2-3xy-2y^2=(2x+y)(x-2y)=7.$$
Therefore, we have the four cases:
1)  $2x+y=1, x-2y=7.$ Thus $x=\dfrac{9}{5},y=-\dfrac{13}{5}$, which are not integers.
2)  $2x+y=7, x-2y=1.$ Thus $x=3,y=1$, which is a group of proper solution.
3) $2x+y=-1, x-2y=-7.$ $x=-\dfrac{9}{5},y=\dfrac{13}{5}$, which are not integers.
4)$2x+y=-7, x-2y=-1.$ Thus $x=-3,y=-1$, which is a second group of proper solution.
As a result, we have find two group of integer solution that $$x=3, y=1,$$ or $$x=-3,y=-1.$$
A: Hint:
$$2x^2-3xy-2y^2=2x^2-4xy+xy-2y^2=2x\underbrace{(x-2y)}+y\underbrace{(x-2y)}=?$$
A: Note that $\sqrt{25y^2+56}$ is an integer if and only if $(5y)^2+56=z^2$ for some integer $z$. This limits how big $5y$ can be. After all; 
$$29^2=(28+1)^2=28^2+2\cdot28+1=28^2+57>28^2+56,$$
so surely $z^2<29^2$ and $(5y)^2<28^2$. This only leaves a few values of $y$ to try.
