<s>[INST] 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.
 [/INST] 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 $(x,y)=(0,0),(1,-1),(-1,1)$
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