Find all integers $x,y$ such that $x^5-x^3-x^2+1=y^2$ Find all integers $x,y$ such that $x^5-x^3-x^2+1=y^2$
I think we need to factor this out and I've managed to factor it to $(x-1)(x+1)^2(x^2-x+1)=y^2$, but I'm not sure what to do here. Have I done something wrong? Am I on the right path? Please help, thanks in advance.
 A: There is a much simpler way to finish the above answer.
We first show that there is no integer $x$ satisfying $x \ge 2$ that is in a solution. Indeed, let us assume that $x$ is an integer satisfying $x \ge 2$ that is in an solution $(x,y)$ to this equation. Then as noted, $x^2+x+1$ is a square for such an $x$. [To elaborate, factoring the LHS of the equation $x^5-x^3-x^2+1 = y^2$ yields
$$(x-1)^2(x+1)(x^2+x+1) = y^2.$$
So for $(x,y)$ to be an integral solution, the LHS to be an integral square, and thus $(x+1)(x^2+x+1)$ has to be an integral square. However, as $x^2+x+1 = x(x+1)+1$, it follows that $x^2+x+1$ and $x+1$ are positive [in particular, non-zero] and relatively prime to each other for any integral $x \ge 2$ i.e., they share no integral factors. So, the only way the product $(x+1)(x^2+x+1)$ can be an integral square is if each of $x+1$, $x^2+x+1$ is itself an integral square. Thus, for $(x,y)$ to be an integral solution, each of $x+1$, $x^2+x+1$ has to be an integral square, and thus in particular indeed, $x^2+x+1$ has to be an integral square.] However, it is impossible for $x^2+x+1$ to be an integral square as
$$(x+1)^2 =x+2x+1>x^2+x+1>x^2.$$
Then one can check that there is no negative integer $x$ satisfying $|x|\ge 2$ that can be a solution, as the LHS of the equation $x^5-x^3-x^2+1 = y^2$ is negative for such $x$.
Thus if $x$ is in a solution, it follows that $|x| \le 1$ and thus $x \in \{-1,0,1\}$. However, note that $x=-1,y=0$, $x=0,y \in \{-1,1\}$, and $x=1,y=0$ are solutions. These from the above paragraph are in fact all the solutions.
