Find all integer solutions: $x^4+x^3+x^2+x=y^2$ Find all integer solutions of the following equation:
$$x^4+x^3+x^2+x=y^2$$
 A: The equation 
$$y^2 = x^4 + x^3 + x^2 + x = x(x+1)(x^2+1)\tag{*1}$$
has four trivial solutions over $\mathbb{Z}\times\mathbb{Z}$:
$$(x,y) = (-1,0), (0, 0),(1,\pm 2)$$
To determine whether there are other non-trivial solutions, let's look at an equivalent problem:
$$y^2 = \begin{cases}
z(z + 1)(z^2 + 1), &\text{ for } z = x \ge 2\\
z(z - 1)(z^2 + 1), &\text{ for } z = -x \ge 2\tag{*2}
\end{cases}$$
Notice 
$$\gcd(z(z\pm 1),z^2+1) = \gcd( \pm z - 1, z^2 + 1) = \gcd(\pm z - 1,2) = 1 \text{ or } 2$$
Any solution of $(*2)$ must have one of the following two factorizations:
$$\begin{cases}
z(z\pm 1) &= p^2,\\
z^2 + 1   &= q^2,\\
y &= \pm pq;\end{cases}
\quad\text{ or }\quad
\begin{cases}
z(z\pm 1) &= 2p^2,\\
z^2 + 1   &= 2q^2,\\
y &= \pm 2pq.\end{cases}\quad\text{ with } \gcd(p,q) = 1.$$
We can rule out the factorization on the left because for $z \ge 2$, $z^2 + 1$ is never a square. For the factorization on the right, we can rewrite the $x$ part as
$$\left\{\begin{array}{ccrl}
(2 z \pm 1 )^2 &-& 8p^2 &= 1\\
z^2 &-& 2q^2 &= -1
\end{array}\right.\tag{*3}$$
Define $A, B : \mathbb{Z} \to \mathbb{Z}$ by
$$A(k) + B(k) \sqrt{2} = (1+\sqrt{2})^k$$
It is known the positive integer solutions of Diophantine equations like those in $(*3)$
has the form
$$\left\{\begin{array}{ccl}
2 z\pm 1 &=& A(2m)\\
2p &=& B(2m)\\
z &=& A(2n-1)\\
q &=& B(2n-1)\\
\end{array}\right.
\quad\quad\text{ for } m , n \in \mathbb{Z}_{+}.
$$
It is clear $2 z \pm 1 \ge z \implies 2m \ge 2n-1$.
Notice $\frac{A(k+1)}{A(k)} \to 1+\sqrt{2} > 2$ as $k \to \infty$.
For any $n$ large enough such that $A(2n) > 2 A(2n-1) + 1$, there are no way to find
a $m$ to satisfy $(*3)$. A brute force computation shows that this
happens as early as $n \ge 2$. 
This leaves us the cases $n = 1$. It is easy to check
for $n = 1$, one can choose $m = 1$ to make $(*3)$ works. However,
$$
m = n = 1 \quad\implies\quad
\left\{\begin{array}{ccl}
2 z\pm 1 &=& A(2) = 3 \\
2p &=& B(2) = 2\\
z &=& A(1)  = 1\\
q &=& B(1)  = 1\\
\end{array}\right.
\quad\iff\quad
(x,y) = (1,\pm 2)
$$
corresponds to a pair of trivial solutions we have covered before. The conclusion is aside from the four trivial solutions, $(*1)$ doesn't have any non-trivial solution.
A: The following shows that the only solutions are
$$(0,0),(-1,0),(1,2),(1,-2)$$
by using a birational transformation to an Elliptic curve.  

1. Birational transformation to an Elliptic curve
Denote
$$C: v^2-u^4-u^3-u^2-u=0$$
It is clear that if $u=0$, then $v=0$. For the other case, $u\neq 0$, we can divide by $u^4$:
$$\dfrac{v^2}{u^4}-1-\dfrac{1}{u}-\dfrac{1}{u^2}-\dfrac{1}{u^3}=0$$
Rearranging:
$$\left(\dfrac{v}{u^2}\right)^2=\left(\dfrac{1}{u}\right)^3+\left(\dfrac{1}{u}\right)^2+\left(\dfrac{1}{u}\right)+1$$
Which is in fact an Elliptic curve:
$$E: Y^2=X^3+X^2+X+1$$
In other words, we have a birational transformation $\phi$:
\begin{align*}
\phi: C &\to E\\
(u,v) &\mapsto \left(\dfrac{1}{u},\dfrac{v}{u^2}\right) = (x,y)
\end{align*}
(We know this type of transformation exists since we can compute genus of $C$ to be 1.)  
If $(u,v)\in \Bbb Z^2$ is an integer solution for $C$, then $\left(\dfrac{1}{u},\dfrac{v}{u^2}\right)$ is a rational point on $E$.  

In other words, it suffices to find all rational points of the form $\left(\dfrac{1}{u},\dfrac{v}{u^2}\right)$ on $E$.  

In particular, notice that $\dfrac{1}{u}$ has numerator $1$ which is very restrictive. We shall make use of this constraint to make our main deduction.  

2. Rational points on $E$
Using Sage, we find that the Mordell-Weil group over $\Bbb Q$ has the structure
$$E(\Bbb Q)\cong \Bbb Z_2\times \Bbb Z$$
Where the order two torsion is $P=(-1,0)$ and the free group generator is $Q=(0,1)$.
In other words all rational points are written as:
$$E(\Bbb Q)=[a]P\oplus [b]Q, \;\;\;\; (a,b)\in [0,1]\times \Bbb Z$$
Clearly $Q$ does not correspond to a solution. On the other hand, $P=(-1,0)$ has $x$-numerator 1. Hence we let $u=-1$:
$$v^2=(-1)^4+(-1)^3+(-1)^2+(-1)=0,$$
and we find a solution in $C$:
$$(u,v)=(-1,0)$$

3. A useful lemma
We shall be using this later:  

If $(X,Y)\in E(\Bbb Q)$, then:
  $$(X,Y)=\left(\dfrac{m}{e^2},\dfrac{n}{e^3}\right)$$
  for some
  $$m,n,e\in \Bbb Z,\;\;\;\;\gcd(m,e)=\gcd(n,e)=1$$
  (This can be shown using $p$-adic valuations.)  


4. Solving the main problem
We have seen that $Q=(0,1)$ does not correspond to a solution. Similarly,
$$[2]Q=\left(\dfrac{-3}{4}, \dfrac{-5}{8}\right)$$
so that $[2]Q$ also does not correspond to a solution.  
Now let $R=P$ or $R=[a]Q, a\geq 2$. We want to show that $R\oplus Q$ cannot correspond to a solution. Then by induction
$$\{P\oplus [a]Q,[a]Q\;|\;a\geq 1\}$$
cannot be a solution and we would have covered $[r]P\oplus [s]Q, (r,s)\in [0,1]\times \Bbb N$.  
Since $R\neq Q$, we can derive a formula for point addition with $Q$:
$$R\oplus Q=\left(\dfrac{2+x-2y}{x^2},\dfrac{4 + 3 x + 2 x^2 + x^3 - 4 y - x y}{x^3}\right)$$
Recall that if indeed some $(u,v)$ maps to a rational point in $E$, then the $x$-coordinate satisfies:
$$x(R\oplus Q)=\dfrac{2+x-2y}{x^2}=\dfrac{1}{u}$$
Now using our lemma in (3):
$$\dfrac{2+x-2y}{x^2}=\dfrac{2+(m/e^2)-2(n/e^3)}{(m/e^2)^2}=\dfrac{1}{u}$$
$$\dfrac{(2e^3+me-2n)e}{m^2}=\dfrac{1}{u}$$
From which we draw the conclusion that
$$e|m^2$$
Since $\gcd(m,e)=1$, this can only happen if $e=\pm 1$. 
Either way, this implies that $(x,y)=(m/e^2,n/e^3)$ are integral points.  
(Perhaps can be solved without needing this information, but quite tedious.)  
Sage tells us that the only integral points are
$$(-1, 0), (0,\pm 1), (1,\pm 2), (7,\pm 20)$$
of which $(0,\pm 1)=\pm Q$ is not possible by construction of $R$. 
Individual checking of the rest tells us that only $(-1,\pm 0),(1,\pm 2)$ results in $u=\pm 1$. We have seen the case $u=-1$ in section (2). For $u=1$::
$$v^2=1^4+1^3+1^2+1=4$$
From which we deduce $v=\pm 2$. Thus far we have the points
$$(u,v)\in \{(0,0),(-1,0),(1,2),(1,-2)\}$$
From the induction, we conclude that there are no other possible candidates corresponding to
\begin{align*}
R &=P\oplus [a]Q, a\geq 1\\
R &= [a]Q, a\geq 1
\end{align*}
It remains to cover the rest of the cases. Since $P$ is order two, $[-1]P=P$. Then immediately
$$P\oplus [a]Q = [-1]P\oplus [a]Q, a\geq 0$$
so that these are also not solutions.  
The inverse of a point does not change the $x$-coordinates:
$$-(x,y)=(x,-y)$$
Therefore if $R$ is not a solution, then $-R$ is also not a solution. This shows that
$$-(P\oplus [a]Q)=[-1]P\oplus [-a]Q=P\oplus [-a]Q$$
$$-([a]Q)=[-a]Q, a\geq 1$$
are not solutions.  
Therefore we have covered all the cases and we are done.  

5. Sage codes for reference

E = EllipticCurve([0,1,0,1,1]);
  tor = E.torsion_points();
  rank = E.rank();
  gens = E.gens();
  integralPoints = E.integral_points();

A: When $|x|>1$ we have 
$$(2x^2+x)^2<4(x+x^2+x^3+x^4)<(2x^2+x+1)^2.$$ 
A: 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:
$$(x,y)=(0,0),(1,2),(1,-2).$$
