Find all solutions to the equation $x^2 + y^2 + xy = (xy)^2$ Can anyone help me find the number of solutions to the equation:

$$  x^2 + y^2 + xy = (xy)^2 $$

Let me give a brief account of what I've tried to proceed with:
Case 1: One of $x$ and $y$ is odd. This results into a contradiction where the parity of LHS and RHS differs.
Case 2: Both of $x$ and $y$ are even. I have shown that no solution apart from $x=y=0$ exists and I've proved that using Infinite Descent.
Now I'm stuck in case $3$ where both $x$ and $y$ are odd.
 A: You can write your equation (leaving the trivial $x=0,y=0$) as $$\frac xy+ \frac yx +1 =xy$$ now let $x/y=k$ and rearrange: $$k^2(1-y^2)+k+1=0$$ which has rational solutions only if $4y^2-3=p^2$. This implies that $y=\pm 1$. Hence the only integer solutions are $(-1,1)$ and $(1,-1)$.
A: I'm assuming you're looking for integer solutions, even though you didn't say os. 
If you rewrite the LHS as $(x+y)^2 - xy$, your equations becomes
$$(x+y)^2 = xy(xy + 1)$$ 
For positive $x$ and $y$:
Suppose (by swapping names if necessary) that $x \ge y$. Then the left hand side is no more than $4x^2$, while the right hand side is at least $y^2 x^2$.
We have
$$
LHS = (x+y)^2 \le (x+x)^2 = 4x^2
$$
and
$$
RHS = xy(xy + 1) > (xy)^2 = y^2 x^2.
$$
If $y \ge 2$, then 
$$
RHS > y^2 x^2 \ge 2^2 x^2 = 4 x^2
$$
but RHS and LHS are equal, so we have a number that's both $ \le 4x^2$ and $> 4x^2$, and that's a contraduction. 
That means that $y$ is at most one. That should get you on your way. 
For the case where one of $x$ and $y$ is negative, you still have some work to do. 
A: First, write it as $(x^2-1)(y^2-1)=xy+1$.
In the narrow ranges where $|x|\leq 1$ or $|y|\leq 1$, you can solve yourself.
Then, when $x>1$ and $x>1$ you have $(x^2-1)(y^2-1)>=2(x^2+y^2-2)=2(x-y)^2+4xy-4\geq 4xy-4$. 
But $xy\geq 4$, so $4xy -4> xy +3\cdot 4 -4 \geq xy-1$.
The case when $x<-1$ and $y<-1$ is the same.
There are no cases when $x<-1$ and $y>1$ because then $xy+1<0$.
Finally, if $x=0$, $y=0$. If $x=-1$ then $y=1$, and if $x=1,$ $y=-1$.
The key is that in most cases, it is "obvious" that $(x^2-1)(y^2-1)$ is a lot bigger than $xy$.

Another approach. Let $d=xy$. Then assume $d\neq 0$ and you have:
$$x^2+\frac{d^2}{x^2}=d^2-d$$
or
$$x^4-(d^2-d)x^2 + d^2=0$$
So by the quadratic formula:
$$x^2=\frac{d^2-d \pm \sqrt{d^2(d-3)(d+1)}}{2}$$
When is $(d-3)(d+1)=(d-1)^2-4$ a perfect square? The only case of perfect squares that differ by four is $0$ and $4$. That means $d=3$ or $d=-1$.
The only cases left to handle then are $d=0$ and $d=3$.
A: We need $$x^2(1-y^2)+xy+y^2=0$$
The discriminant is $$y^2-4y^2(1-y^2)=y^2(4y^2-3)$$ 
So, $4y^2-3$ needs to be perfect square
$4y^2-3=a^2\iff(2y-a)(2y+a)=3$
$\implies2y-a,2y+a=\pm3,\pm1$
A: Your answer is three, and in my opinion, a basic algebraic approach most clearly shows the solution.  This equation is comprised of symmetric polynomials $x^2+y^2$ and $xy$, so finding one solution, $(x,y)=(p,q) \ | \ p \neq q$ is equivalent to finding two $(x,y)=(p,q),(q,p)$.  If we let algebra drive:
$x^2 + y^2 + xy = (xy)^2 \to \\
(x + y)^2 = (xy)^2+(xy) \to \\
(2x + 2y)^2 = 4(xy)^2+4(xy) \to \\
(2x + 2y)^2 = 4(xy)^2+4(xy) +1-1\to \\
(2x + 2y)^2 = (2xy+1)^2-1 \implies$
$$(2xy+1)^2-(2x+2y)^2=1$$
Then we arrive at a difference of two squares being $1$.  It can only imply that
$$
\begin{cases}
2xy+1&=(\pm 1)\\
2x+2y&=0
\end{cases}
$$
From the second $x=-y$.  Putting this into the first yields $$-2y^2+1=(\pm 1) \to \\
y^2=\frac{1-(\pm 1)}{2}$$
So $(x,y)=(0,0),(1,-1),(-1,1)$
A: Note that $$t^2-1\geq |t|+1>0$$ for all real numbers $t$ such that $|t|\geq 2$  (this is because $\big(|t|-2\big)\big(|t|+1\big)\geq 0$).  Therefore, if $|x|\geq 2$ and $|y|\geq 2$, then
$$\left(x^2-1\right)\left(y^2-1\right)\geq \big(|x|+1\big)\big(|y|+1\big)>|xy|+1\geq xy+1\,.$$
This means $x^2+y^2+xy<(xy)^2$.  Thus, any solution $(x,y)\in\mathbb{Z}\times\mathbb{Z}$ to $x^2+y^2+xy=(xy)^2$ must come from the case where $|x|\leq 1$ or $|y|\leq 1$.  
Clearly, $x=0$ if and only if $y=0$.  We now exclude the solution $(x,y)=(0,0)$, so we are left with the case $|x|=|y|=1$.   This means $$xy+1=\left(x^2-1\right)\left(y^2-1\right)=0\,,$$ so $xy=-1$.  This implies $(x,y)=(-1,+1),(+1,-1)$.
