Vtyjkyuk

We are given an equation:
$$2x^2-3xy-2y^2=7$$
And we have to find $x,y$ where $x,y ∈BbbZ$.

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 $x = frac3y±sqrt25y^2+564$ , $x$ to be an integer $sqrt25y^2+56$ 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 $y=1$, $sqrt25y^2+56=9$, so we get $x = frac3±94$ and when we have plus sign we get $x = frac3+94=3$. 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 $x = frac-3±94$, 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.

Hint:

$$2x^2-3xy-2y^2=2x^2-4xy+xy-2y^2=2xunderbrace(x-2y)+yunderbrace(x-2y)=?$$

Hint:

Clearly, if $x,y$ is a solution, $-x,-y$ will also be.

Let $25y^2+56=u^2iff(u-5y)(u+5y)=56$ where $uge0$

As $5y-u,5y+u$ have the same parity, both must be even

$impliesdfracu-5y2cdotdfracu+5y2=14$

Now $14=pm1cdotpm14,pm2cdotpm7$

If $y>0, u+5yge1+5impliesdfracu+5y2ge3$ i.e., $=7$ or $14$

If $y<0, u-5yge1+5$

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 $x=dfrac95,y=-dfrac135$, which are not integers.

2) $2x+y=7, x-2y=1.$ Thus $x=3,y=1$, which is a group of proper solution.

3) $2x+y=-1, x-2y=-7.$ $x=-dfrac95,y=dfrac135$, 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 $$x=-3,y=-1.$$

Note that $sqrt25y^2+56$ 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;
$$29^2=(28+1)^2=28^2+2cdot28+1=28^2+57>28^2+56,$$
so surely $z^2<29^2$ and $(5y)^2<28^2$. This only leaves a few values of $y$ to try.