Solve for system of diophantine equations $\cases{x+1=a^2 \cr x^3-x^2+1=b^2}$

I just can found a trivial solution $x=0$. Is there any other ?
 A: Here's an elementary solution.  First note that we may assume $a,b \ge 0$ since negative values yield exactly the same $x$.  It's easy to see that $a=0$ doesn't work and $a=1$ does, so let's assume $a \ge 2$.
The two equations are obviously equivalent to $(a^2-1)^3 - (a^2-1)^2 + 1 = b^2$.  The LHS expands to $a^6 - 4a^4 + 5a^2 - 1 = (a^3-2a)^2 + (a^2-1)$.
Since $a \ge 2$, both $a^3-2a$ and $a^2-1$ are positive.  In general, whenever $n^2+m$ is a square ($n,m$ positive integers) we must have $m \ge 2n+1$ by considering the smallest square after $n^2$.
Therefore, if $(a^3-2a)^2 + (a^2-1)$ is a square, $a^2-1 \ge 2(a^3-2a)$.  But it's easy to check this is impossible for $a \ge 2$, so there are no further solutions.
This simple technique is often very effective for detecting square values of monic even-degree polynomials in general.
A: I assume you are only looking for integral solutions to the system of equations. First consider the equation $b^2=x^3-x^2+1$. This defines an elliptic curve $E$. One can show that $E$ has rank $1$ and trivial torsion subgroup, so $E(\mathbb{Q})$ is generated by one single point, namely $(x,b)=(0,1)$. In other words,
$$E(\mathbb{Q})=\{ nP : n\in\mathbb{Z},\ P=(0,1)\}.$$
By inspection of the first few multiples of $P$ we find the following integral points
$$\pm P = (0,\pm 1),\ \pm 2P = (1,\pm 1),\ \pm 3P = (4,\pm 7)$$
and $4P=(-3/4, 1/8)$ and $5P=(28/9, -125/27)$. Now using the theory of heights one can show that the only integral points on $E$ are $\pm P$, $\pm 2P$, and $\pm 3P$.
Thus, there are only three possible values for $x$ in your system, namely $x=0,1,4$. But $1+1$ and $4+1$ are not perfect squares. So the unique solution to your system (in integers) is $x=0$, and $a=b=1$.
