Integer solutions of $x^3+y^3=z^2$

Is there any integer solution other than $(x,y,z)=(1,2,3)$ for $x^3+y^3=z^2$?

Much as for Pythagorean triples, there are parametrizations for the coprime solutions to such equations (this is generally true for $x^p+y^q=z^r$ whenever the sum of the reciprocals of $p, q$ and $r$ exceeds $1$). One can find these on pages 467 to 470 of Henri Cohen's excellent "Number theory II" (Springer GTM 240) : one such parametrization (there are $3$ in total) is to take $$ x=-3s^4+6t^2s^2+t^4, \; y=3s^4+6t^2s^2-t^4, \; z = 6 s t ( 3 s^4+t^4), $$ where $s$ and $t$ are coprime, of opposite parity, and $3$ does not divide $t$. Of course, this is up to exchange of $x$ and $y$.

As noted earlier, non-coprime solutions are readily found.

Note that if $(x,y,z)$ is a solution, then so are $(y,x,z)$ and $(a^2x,a^2y,a^3z)$ where $a \in \mathbb{Z}$.

Below are some primitive solutions, you can use them and follow the above technique to generate infinite solutions. Hence, we will leave out these duplicates.

$$(1,2,3); (2,2,4); (2,46,312); (7,21,98); (10,65,525); (11,37,228); (14,70,588); (22,26,168)$$ and so on.

You can multiply $x$ and $y$ by $a^2$, and $z$ by $a^3$.

There is the boring $(0,0,0)$. And the almost equally boring $(-1, 1,0)$. And then we have $2^3+2^3=4^2$ and its relatives.

I did a little experiment. I got these.

1 2 3
2 1 3
2 2 4
4 8 24
7 21 98
8 4 24
8 8 32
9 18 81
18 9 81
21 7 98

Some are duplicates. I wrote this program in Python.

for k in range(1,100):
    for j in range(1,100):
        for l in range(1,100):
            if(k**3 + j**3 == l**2):
                print k, j, l

It's crude but you might tinker with it to see what else you churn up.

It may be of help to consider that $x^3 + y^3 = ( x + y ) \cdot ( x^2 - xy + y^2)$, so one could start by looking for values of $x$ and $y$ for which those factors are equal (and then for values where one factor "completes a square" with the other).

ADDENDUM: I had a little time to think more on this during my snowy walk home... We can show that if we set $x = y$ , the two factors are $2x \cdot x^2$, which only gets us the (2, 2, 4) family of solutions [apart from $(0,0,0)$].

In fact, if we set $y = kx$ ($k$ integral), the factors are $[k + 1]x \cdot [k^2 - k + 1]x^2$, so we have $ k + 1 = k^2 - k + 1 \Rightarrow k = 0, 2$ if we require the two factors $( x + y ) $ and $ ( x^2 - xy + y^2)$ to be equal. [Just noticed that $k = 2$ will produce, for instance, the $(1,2,3)$ family.] So almost all the solutions are ones for which one of these factors contains a divisor just once that also appears in the other factor just once. An example is $(4,8,24)$, for which the factors are $(4+8) \cdot (4^2 - 4 \cdot 8 + 8^2) = ( 2^2 \cdot 3 ) \cdot (3 \cdot 4^2)$ . This ought to help narrow the search somewhat.