Integer solutions of $x^3+y^3=z^2$ Is there any integer solution other than $(x,y,z)=(1,2,3)$ for $x^3+y^3=z^2$?
 A: 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.
A: 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.
A: 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. 
A: 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.
A: There's $(2,2,4)$ lurking here too.
A: I think I have found a rather general solution. Let us choose any co-prime positive integers $u$ and $v$. From $u^3+v^3=(u+v)(u^2-uv+v^2)$, we form $[u(u^3+v^3)]^3+[v(u^3+v^3)]^3=[(u^3+v^3)^2]^2$. In this way, we can find infinitely many integer solutions. However, $(1,2,3)$ is the only co-prime integer solution triplet.
A: If we take $$x=n(n^2-3)$$ $$y=3n^2-1$$ for $n\in\mathbb{N},$  $$x^2+y^2=(n^6-6n^4+9n^2)+(9n^4-6n^2+1)=n^6+3n^4+3n^2+1=(n^2+1)^3$$ then the above equation has infinitely many integer solutions. 
A: 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.
A: You can work in $\mathbb{Z}[\omega],$ where $\omega = e^{\frac{2 \pi i}{3}}.$ 
There are two essential cases to deal with. In both cases, you can easily reduce to considering the case that $x,y,z$ are pairwise coprime ( you can afterwards multiply $x$ and $y$ by $d^{2}$ and $z$ by $d^{3}$ for some integer $d$). The case when $3$ does not divide $z$ is easier.
In that case, you can assume that $x + y = a^{2}$ for some integer $a$ and that $x + \omega y = (b + \omega c)^{2}$ for integers $b$ and $c.$ Then you can see that 
$x^{3} + y^{3} = (a(b^{2}-bc +c^{2}))^{2}.$ Note that the second equation gives
$x = b^{2}-c^{2}$ and $y = 2bc - c^{2}$ since $1 + \omega + \omega^{2} = 0.$ 
Hence we additionally need the integers $a,b,c$ to satisfy $a^{2} + 3c^{2} = (b+c)^{2},$ but this is easy to arrange.
In the (still pairwise coprime case that $3$ divides $z,$ you must have $x + y = 3^{2m-1}a^{2}$ for some integer $a$ coprime to $3$, and $x + \omega y = (1- \omega)(b + \omega c)^{2}$ for some integers $b$ and $c$ with $b+c$ not divisible by $3$.
A: Well you can and draw another formula.  $x^3+y^3=z^2$
$x=(b^2-a^2)(b^2+2ba-2a^2)c^2$
$y=(b^2-a^2)(2b^2-2ab-a^2)c^2$
$z=3(b^2-a^2)^2(a^2-ab+b^2)c^3$
The most interesting thing there is that the formula that led, like should not give mutually simple solutions, but after sokrasheniya on common divisor can be obtained and are relatively prime solutions. This means that the formula itself describes as relatively prime so no. Coprime solutions - there are private solutions.
A: If $n$ and $m$ are relatively prime positive integers,
there are infinitely many positive integral solutions to
$x^n + y^n = z^m$.
Proof: there are positive integers $u$ and $v$
such that $un - vm = -1$, or $un+1 = vm$.
Let $x = pa^u$, $y = q a^u$, and
$z = a^v$.
Then we want
$(p^n+q^n)a^{un} = a^{vm}
= a^{un+1}
$,
or $p^n+q^n = a$.
So, from $n$ and $m$ get $u$ and $v$.
Then choose any $p$ and $q$,
and compute $a = p^n+q^n$.
Then get $x$, $y$, and $z$ as above
($x = pa^u$, $y = q a^u$, and
$z = a^v$).
For this case,
with $n=3$ and $m=2$,
$u = 1$ and $v = 2$,
so choose $p$ and $q$
and let $a = p^3+q^3$
$x = p a$, $y = qa$, and $z = a^2$.
As a check,
$x^3+y^3 = (pa)^3 + (qa)^3
= (p^3+q^3)a^3
= a^4$
and $z^2 = (a^2)^2 = a^4$.
Of course, if you want $x$ and $y$ to be relatively prime,
it is harder.
A: Here is the way to produce ALL integer solutions, not necessarily coprime, and not just some random infinite families.
For any integer $m$, let $D_2(m)$ be the set of integers $w$ such that $w^2$ is a divisor of $m$. For example, $D_2(18)=\{\pm 1,\pm 3\}$.
Then the set of all integer solutions to $x^3+y^3=z^2$ can be described as
$$
(x,y,z)=\left(\frac{u(u^3+v^3)}{w^2}, \, \frac{v(u^3+v^3)}{w^2}, \frac{(u^3+v^3)^2}{w^3} \, \right), \quad u,v \in {\mathbb Z}, \quad w \in D_2(u^3+v^3).
$$
Condition $w \in D_2(u^3+v^3)$ ensures that $x,y,z$ are integers, and a direct substitution proves that they satisfy the equation. Let us prove that, conversely, all integer solutions are covered by this family. Indeed, solution $(0,0,0)$ is covered. Let $(x,y,z)$ be any other solution, let $d$ be the greatest common square-free divisor of $x$ and $y$, and write $x=du$, $y=dv$ for some integers $u,v$. Then $d(u^3+v^3)=(z/d)^2$ is an integer, hence $z/d$ is an integer, say $z'$. Then $d \cdot (u^3+v^3)=(z')^2$ implies the existence of integers $a,b,w$ such that $d=ab^2$ and $u^3+v^3=aw^2$. Because $d$ is square-free, $b^2=1$, hence $d=a=\frac{u^3+v^3}{w^2}$. So, take $u,v$ arbitrary, $w \in D_2(u^3+v^3)$, and then express $x,y,z$. We obtain the stated formulas.
The method works for all equations of the form $F(x,y)=z^2$, where $F$ is a cubic form with integer coefficients.
