<s>[INST] Find all integer solutions: $x^4+x^3+x^2+x=y^2$ Find all integer solutions of the following equation:
$$x^4+x^3+x^2+x=y^2$$
 [/INST] EDIT : The case (1) in my answer has a big mistake. The case (2) is true.
$(x,y)=(0,0)$ is one of the solutions. In the following, suppose that $(x,y)\not=(0,0).$
We have
$$x(x+1)(x^2+1)=y^2.$$
First, we know that $x$ and $x+1$ are coprime, and that $x$ and $x^2+1$ are coprime, too. This leads that there is an integer $k$ such that $x=k^2.$ (see the right hand side)
In the following, we break up the answer into two cases.
(1) The case that $x+1$ and $x^2+1$ are not coprime.
Letting $$x+1=ps,x^2+1=pt\ \ (\text{$p$ is a prime number, $s,t\in\mathbb Z$})$$
we have
$$(ps-1)^2+1=pt\iff p(t-ps^2+2s)=2\Rightarrow p=2.$$
So, we know we only have the following possibilities :
$$x+1=0,\pm1, \pm2^k\Rightarrow x=-1,0,-2,-1\pm2^k.$$
However, we have $x\not=-1, x\not=0, x\not=-2$, because $x$ has to be a square of an integer (and $x\not=0$). So, we have $$x=-1\pm2^k.$$ In the same argument, we have $$x^2=-1\pm2^l.$$ Note that $k,l\ge1\in\mathbb Z.$
Hence, we have
$$(-1\pm2^k)^2=-1\pm2^l\iff 2^{2k}\mp2^k+2=\pm2^l\iff 2^{2k-1}\mp2^{k-1}+1=\pm2^{l^1}.$$
Now, if $k-1\ge1$ and $l-1\ge1$, then we know that the left hand side is odd, and the right hand side is even, which is a contradiction. 
So, we have "$k=1$ or $l=1$".
This leads $x=\pm2^k-1=\pm2^1-1\Rightarrow x=1,-3\Rightarrow x=1.$ ($x^2$ has to be a square.) Also, this leads $x^2=\pm2^l-1=\pm2^l-1\Rightarrow x^2=1,-3\Rightarrow x=1.$
So, this case leads $(x,y)=(1,\pm 2).$
(2) The case that $x+1$ and $x^2+1$ are coprime.
In this case, we know any two of $x,x+1,x^2+1$ are coprime. So, this leads that each of $x, x+1,x^2+1$ has to be a square number of an integer. However, it is impossible that $x$ and $x+1$ are both square numbers except $x=0$. However, we suppose that $x\not=0$ at the top. Hence, this case has no solution.
Now we reach a conclusion that the solutions of the given equation is only the followings:
$$(x,y)=(0,0),(1,2),(1,-2).$$
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