Solve for integers $x, y$ and $z$: $x^2 + y^2 = z^3.$ 
Solve for integers $x, y$ and $z$:
$x^2 + y^2 = z^3.$

I tried manipulating by adding and subtracting $2xy$ , but it didn't give me any other information, except the fact that $z^3 - 2xy$ and $z^3+2xy$ are perfect squares.
This doesn't give us much information to work on. I don't know if my steps are correct, I do not know how to approach this problem.
Any help would be appreciated.
 A: Let $C$, $D$, $S$ and $T$ be integers, and define
\begin{eqnarray*}
x&=&ab^3X=(C^2+D^2)(CS^3-3DS^2T-3CST^2+DT^3),\\
y&=&ab^3Y=(C^2+D^2)(DS^3+3CS^2T-3DST^2-CT^3),\\
z&=&ab^2Z=(C^2+D^2)(S^2+T^2).
\end{eqnarray*}
Then a routine verification shows that $x^2+y^2=z^3$. I will show that every solution is of this form. Moreover, if we require $S$ and $T$ to be coprime and nonnegative, every solution will have precisely one such representation, making this a proper parametrization.
Let $x$, $y$ and $z$ be integers such that
$$x^2+y^2=z^3.$$
First note that $x$ and $y$ are not both odd, as otherwise we get a contradiction by reducing mod $8$.
Let $d:=\gcd(x,y)$ and let $a$ and $b$ be integers such that $d=ab^3$ and $a$ is cube-free. Then $d^2=a^2b^6$ divides $z^3$ and hence $a$ divides $z$. Writing $x=au$, $y=av$ and $z=aw$ we see that
$$a^3w^3=z^3=x^2+y^2=(ab^3u)^2+(ab^3v)^2=a^2b^6(u^2+v^2),$$
from which it follows that $b^2$ divides $w$ because $a$ is cube-free. So writing $x=ab^3X$, $y=ab^3Y$, $z=ab^2Z$ shows that
$$X^2+Y^2=aZ^3,$$
where $X$ and $Y$ are coprime. Factoring in $\Bbb{Z}[i]$ then shows that
$$aZ^3=(X+Yi)(X-Yi),$$
where $X$ and $Y$ are coprime and not both odd, so the two factors are coprime. Then
$$X+Yi=(A+Bi)(U+Vi)^3,$$
for some integers $A$, $B$, $U$ and $V$ such that $\gcd(A,B)=\gcd(U,V)=1$ and $A^2+B^2=a$ and $U^2+V^2=Z$. Then
\begin{eqnarray*}
X&=&AU^3-3BU^2V-3AUV^2+BV^3,\\
Y&=&BU^3+3AU^2V-3BUV^2-AV^3,
\end{eqnarray*}
and hence for $C=bA$ and $D=bB$ we find that
\begin{eqnarray*}
x&=&ab^3X=(C^2+D^2)(CS^3-3DS^2T-3CST^2+DT^3),\\
y&=&ab^3Y=(C^2+D^2)(DS^3+3CS^2T-3DST^2-CT^3),\\
z&=&ab^2Z=(C^2+D^2)(S^2+T^2).
\end{eqnarray*}

In particular, parametrizations given in the other answers and comments correspond to $(C,D,S,T)=$
$$(1,0,a,b),\qquad(k,k,1,0),\qquad(1,k,1,0),\qquad(a,b,1,0).$$
A: Factoring over Gaussian integers $$(x+iy)(x-iy)=z^3$$ so it is sufficient (but not necessary) for $x+iy$ to be a cube. That is,
$$\begin{align}x+iy&=(a+bi)^3\\
&=(a^3-3ab^2)+\left(3a^2b-b^3\right)i
\end{align}$$
So choose any $a,b$ and set $x=a^3-3ab^2$, $y=3a^2b-b^3$. For example $a=5$, $b=7$ gives $x=-610$, $y=182$. And indeed, $(-610)^2+182^2=74^3$.

This recovers some solution families mentioned in the comments. For example with $a=-k, b=k$, it recovers $x=2k^3,y=2k^3,z=2k^2$.
However it does not produce all solutions. For example in the family $x=1+k^2,y=k+k^3$, there is the the solution $x=5,y=10$. But the family in this answer will not yield that as a solution.
A: First I search for trivial solutions:  $(0,0,0)$ works.  Likewise $(0,1,1)$ and $(1,0,1)$.
$z$ has to be non-negative, as its cube is a sum of squares.  Likewise if a negative value of $x$ or $y$ worked, so would a positive value.  So we can stick to positive $(x,y,z)$ since we already found the solutions with $0$'s.
So, next up would be $z=2$, so $x^2 + y^2 = 8$, which has a pretty immediate solution $(2,2,2)$.
Going to $z = 3$, we would need two squares that add up to $27$, and none work.
We could continue taking values of $z$ one at a time, or we can pick a constraint and see if we can find a set of solutions.  Suppose $x = 1$. Are there any squares that are one less than a cube?
Comparing $1,4,9,16,25,36,49,64,81,100,121,144,169,196,225...$ with
$1,8,27,64,125,216,343,512,729,1000,1331,1728...$. I don't see any such option, but I do notice that $121 + 4 = 125$, so there is a solution $(2,11,5)$.
If we consider even vs odd possibilities:  if $x$ and $y$ are both even, so must $z$ be.  However we can't factor all the $2$'s out.  If $(2a, 2b, 2c)$ is a solution, $(a, b, c)$ is not.
It looks like others are finding better patterns in the comments so I'll leave it there.
