Diophantine equation $x^2 + 32x = y^3$ I am trying to find all solutions to the Diophantine equation $x^2 + 32x = y^3$.
I think that the first step is to factorise: $x(x+32)=y^3$.
If $x$ is odd, then $x+32$ is also odd. The common divisors of $x$ and $x+32$ must divide 32, and since all the divisors of 32 are even except 1, then we must have gcd$(x,x+32)=1$. Therefore $x$ and $x+32$ are cubes, and since there are no cubes which are 32 apart, then there are no solutions when $x$ is odd.
I am stuck on what to do when $x$ is even. Is the same approach the best way to do it, i.e. consider all the possible common divisors when $x$ is even? I don't see how to proceed from this point.
 A: Let $z = x + 16$.  Then we have $(z-16)(z+16) = y^3$.
Case A:
If $z-16$ and $z+16$ are coprime, each must be a cube, and then we have two cubes whose difference is $32$. As the difference between cubes increases fairly fast, it is easy to check that there are no such cubes. $[3^3 - 2^3 = 19, 4^3 - 3^3 = 37, \cdots]$.  
Case B:
If $z-16$ and $z+16$ share a factor $k > 1$, as $\gcd(z-16, z+16) = \gcd(32, z + 16),$ we must have $k \mid 32$.  It is easy to see that $k$ must be the highest power of $2$ which divides either of $z-16$ or $z+16$, unless one of them is zero.  
Assume first that neither of $z-16$ or $z+16$ are zero.  Then
$k^2 \mid LHS \implies k^2 \mid y^3 \implies k = 8$, a perfect cube, as there is no other factor of $2$ in the LHS.  Let $u = \dfrac{z}{8}$ in that case, and we have $(u-2)(u+2) = v^3$ as the equation now.  Now $u-2$ and $u+2$ must be coprime and hence cubes themselves. So we have two cubes whose difference is $4$.  Easy to verify such cubes do not exist.
Hence the only case remaining is when either $z-16=0$ or $z+16=0$.  In both options, we have obvious solutions with $y=0$, corresponding to $x = 0$ or $x = -32$. 
A: If $2\nmid x$ then $x$ and $x+32$ need to be both cubes, that is impossible. So there exist $x_1,y_1 \in \mathbb{Z}$ such that $x=2x_1$, $y=2y_1$, and we get $$x_1(x_1+16)=2y_1^3$$
If $2\nmid x_1$ then $2\nmid x_1+16$, so there exists $x_2 \in\mathbb{Z}$ such that $x_1=2x_2$. But then $2\mid \text{gcd}(x_1,x_1+16)$, implying that $4\mid 2y_1^3$, i.e. there exists $y_2 \in \mathbb{Z}$ such that $y_1=2y_2$. Substituing we get $$x_2(x_2+8)=4y_2^3.$$
Going ahead at the same way, we get that $2\mid x_2$, setting $x_2=2x_3$ for some $x_3 \in \mathbb{Z}$ we get $$x_3(x_3+4)=y_2^3.$$ If $2\nmid x_3$ then $2\nmid x_3y_2(x_3+4)$, and in particular $x_3$ and $x_3+4$ have to be both cubes of some integers, and that's again impossible. So by force $2\mid x_3$, implying that $2\mid y_2$, i.e. there exist $x_4,y_4$ such that $x_3=2x_4$, $y_2=2y_4$. Now we get $$x_4(x_4+2)=2y_4^3;$$ as before $2\mid \text{gcd}(x_4,y_4)$, we set $x_4=2x_5, y_4=2y_5$ and we get $$x_5(x_5+1)=4y_5^3$$
To ease finally the notation, set $\alpha=2x_5-1$ and $\beta=2y_5$: the above equation is exactly equivalent to $$\alpha^2-1=2\beta^3, \text{ with }\alpha,\beta \in \mathbb{Z}.$$
Since $2\nmid \alpha$ then $2=\text{gcd}(\alpha+1,\alpha-1)$. So, we have exactly two cases.
Case 1.
$\alpha+1=2a^3, \alpha-1=b^3$ for some integers $a,b$. $b$ has to be even, so that $$\left(2\frac{a}{b}\right)^3+\left(-\frac{2}{b}\right)^3=4.$$
Case 2.
$\alpha+1=c^3, \alpha-1=2d^3$ for some integers $c,d$. $c$ has to be even, so that $$\left(2\frac{d}{c}\right)^3+\left(\frac{2}{c}\right)^3=4.$$
[Thanks to Yimin for the correction: we have to rule out cases $b=0$ and $c=0$ before dividing all the remaining part, from which we obtain exactly the two trivial solution $x=0$ and $x=-32$]
In both cases we should get a rational solution $(X,Y) \in \mathbb{Q}^2$ to the equation $$X^3+Y^3=4.(*)$$
Now, both Euler (see Algebra, 2, 1770, Art.247) and Dirichlet (see Werke, II, Anhang, 352-3) proved by descent that (*) has no rational solutions.
A: I will leave it to others to find an elementary proof, but I just wanted to say that this equation defines an elliptic curve $E:x^2+32x=y^3$ (usually the so-called Weierstrass equation would be written $y^2+32y=x^3$, but that doesn't matter). The Mordell-Weil theorem states that the set of rational points is a finitely generated abelian group, so
$$E(\mathbb{Q})\cong T_E \times \mathbb{Z}^{R_E},$$
where $T_E$ is a finite subgroup, and $R_E\geq 0$ is called the rank of the elliptic curve. There are methods to calculate $T_E$ and $R_E$ (such as the Nagell-Lutz theorem, or the method of $2$-descent), and these methods are implemented in software such as Sage and Magma. In this case:
$$T_E=\{\mathcal{O},(0,0),(-32,0)\}\cong \mathbb{Z}/3\mathbb{Z},$$
where $\mathcal{O}$ is the point at infinity with projective coordinates $[1,0,0]$, and $R_E=0$. So there are no other rational or integral points other than $(0,0)$ and $(-32,0)$.
I calculated the torsion and rank using the online Magma calculator:

E:=EllipticCurve([0,0,32,0,0]);
Rank(E);
TorsionSubgroup(E);

