Does the Diophantine equation $y^2=x^2x^3+x^4$ have solutions other than $x=1,y=1$? Interestingly, the Diophantine equation $y^2=x^2x^3+x^5$ has such solutions: $x=3,y=15$, $x=5,y=55$, $x=56,y=23464$. However I don't know if there are any other solutions. The solution $x=3,y=15$ is related to the Mordell's equations $y^3=x^2+2$ and $y^3=x^2+18$ which have well known sole solutions $x=5,y=3$ and $x=3,y=3$. Indeed it follows from $3^3=5^2+2$ and $3^3=3^2+18$ that $3^5(15)^2(3^33^2)=0$.

2$\begingroup$ The solutions to $y^{2} = x^{2}  x^{3} + x^{5}$ all come from the integral points on $E : y^{2} = x^{3}  x + 1$. Magma indicates that the only solutions are $x = 1$, $x = 0$, $x = 1$, $x = 3$, $x = 5$ and $x = 56$. $\endgroup$ – Jeremy Rouse Jul 23 '14 at 12:18
The only integer solutions $(x,y)$ are $(0,0)$ and $(1,\pm 1)$.
Your equation is $y^2=x^2(1x+x^2)$. This can only have integer solutions if $p(x)=1x+x^2$ is a square or $y=0$. For $x>1$ we have $(x1)^2<p(x)<x^2$, so $p(x)$ is not a square. For $x<0$ we have $x^2<p(x)<(x1)^2$, so $p(x)$ is not a square. The only possibilities are $x=0$ and $x=1$, and they indeed give solutions.

2$\begingroup$ Alternately, for $x\neq 0$, $z:=y/x$ is an integer, and the equation factors as $(2z2x+1)(2z+2x1)=3$. It is easy to finish from here. $\endgroup$ – GH from MO Jul 23 '14 at 12:25

1$\begingroup$ Or $z^{2} = (x+\omega)(x+\omega^{2}),$ where $\omega = e^{\frac{2 \pi i }{3}}$ (and $z$ must be prime to $3$). This leads (if $x+\omega$ is not a unit) to $(x+ \omega) = (a +b\omega)^{2}$ for integers $a$ and $b.$ Hence $1 = 2abb^{2},$ so that either $b= a = 1$ and $x =0$ or $b = a = 1, x =0.$ The possibility $x + \omega = \omega^{2}$ leads to $x=1$. $\endgroup$ – Geoff Robinson Jul 23 '14 at 12:46

$\begingroup$ I think Joonas solution is not complete. For example we have to prove that the system $x^2=a^3,p(x)=a$ does not have integer solutions for some integer $a$. $\endgroup$ – Zurab Silagadze Jul 23 '14 at 12:59

1$\begingroup$ @ZurabSilagadze: Joonas's solution is fine. If $y^2=x^2(1x+x^2)$ holds for $y\neq 0$, then $1x+x^2$ is a rational square and also an integer, hence an integral square. On the other hand, he shows that unless $x=0$ or $x=1$ holds, $1x+x^2$ lies strictly between two neighboring integral squares, so it is not an integral square. I outlined an other approach in my comment, which avoids inequalities. It is based on the fact that $3$ cannot be factored in many ways over the integers. $\endgroup$ – GH from MO Jul 23 '14 at 13:06

2$\begingroup$ @GHfromMO: Thanks, now I see the point. I have accepted Joonas's solution as the correct one. You solution is cute. $\endgroup$ – Zurab Silagadze Jul 24 '14 at 8:46