The equation $a^3 + b^3 = c^2$ has solution $(a, b, c) = (2,2,4)$. The equation $a^3 + b^3 = c^2$ has solution $(a, b, c)  = (2,2,4).$ Find 3 more solutions in positige integers.  [hint.  look for solutions of the form $(a,b,c) = (xz,  yz,  z^2)$
Attempt:
So I tried to use the hint in relation to the triple that they gave that worked.   So I observed that the triple that worked $(2,2,4)$ could also be written as $$2(1,1,2)$$  So i attempted to generalize it to $$n(1,1,2)$$ I thought this may work because I am using the "origin" of where the triple $(2,2,4)$ came from.  This did not work unfortunately.  So I am asking what else could I comsider to devise some system to find these triples? Because I highly doubt they're asking me to just shoot in the dark and pick random numbers
 A: If you substitute $ (x z, y z, z^2) $ for $ (a,b,c) $ then you're looking for solutions to
$ x^3 z^3 + y^3 z^3  = z^4 $
Assuming $ z \neq 0 $, divide both sides by $ z^3 $
$ x^3 + y^3 = z $
Now choose anything for $ x $ and $ y $ and this gives you a $ z $ that satisfies the original equation.  $ (2, 2, 4) $ corresponds to $ x = 1 $, $ y = 1 $, $ z = 1^3 + 1^3 = 2 $.
A: This is not what your exercise is expecting. 
For the purpose to have a reference around, following is what we know about the equation:
$$x^3 + y^3 = z^2$$
According to $\S 14.3.1$ of Henri's Cohen's Number Theory, Vol II,
Analytic and Modern Tools,
the integer solutions of above equation, subject to $\gcd(x,y) = 1$, can be grouped into three families. These families are disjoint, in the sense that any solution of the equation belongs to one and only one family.
The $s$ and $t$ below denote coprime integers satisfying corresponding congruences modulo $2$ and $3$. The three families/parametrizations, up to exchange of $x$ and $y$, are:
$$\begin{align}
1. & \begin{cases}
x &= s(s+2t)(s^2-2ts+4t^2)\\
y &= -4t(s-t)(s^2+ts+t^2)\\
z &= \pm(s^2-2ts-2t^2)(s^4+2ts^3+6t^2s^2-4t^3s + 4t^4)
\end{cases}
& s \text{ is odd },\quad s \not\equiv t \pmod 3\\
\\
2. & \begin{cases}
x &= s^4-4ts^3-6t^2s^2-4t^3s + t^4\\
y &= 2(s^4+2ts^3+2t^3s+t^4)\\
z &= 3(s-t)(s+t)(s^4+2s^3t+6s^2t^2+2st^3+t^4)
\end{cases}
& s \not\equiv t \pmod 2,\quad s \not\equiv t \pmod 3\\
\\
3. & \begin{cases}
x &= -3s^4 + 6t^2s^2 + t^4\\
y &= 3s^4+6t^2s^2-t^4\\
z &= 6st(3s^4+t^4)
\end{cases}
&
s \not\equiv t \pmod 2,\quad 3 \not| t
\end{align}
$$
For more details and derivation, please consult Cohen's book mentioned above.
A: Here is an
infinite number of solutions:
For any positive integer $c$,
$x = (1+c^{3})$,
$y 
=c(1+c^{3})$,
and
$z = (1+c^{3})^2
$.
Check:
$x^3+y^3
=(1+c^3)^3+(c(1+c^3))^3
=(1+c^3)^3(1+c^3)
=(1+c^3)^4
$
and
$z^2
=((1+c^{3})^2)^2
=(1+c^{3})^4
$.
This is gotten from the following,
which is a re-creation
of some algebra of mine
from many years ago:
To find solutions to
$x^n+y^n = z^m$
where
$(n, m) = 1$.
Since $(n, m) = 1$,
there are $a$ and $b$
such that
$am-bn = 1$
or
$am=bn + 1$.
If
$x = u^b$,
$y = v^b$,
$z = w^a$,
then
$u^{bn}+v^{bn}
=w^{am}
=w^{bn+1}
$.
Let
$v = cu$.
Then
$u^{bn}+v^{bn}
=u^{bn}+(cu)^{bn}
=u^{bn}+c^{bn}u^{bn}
=u^{bn}(1+c^{bn})
$.
If
$w = 1+c^{bn}$,
we want
$w^{bn+1}
=w\,w^{bn}
=wu^{bn}
$
or
$u = w$.
Therefore,
a solution is
$x = (1+c^{bn})^b$,
$y = cu
=c(1+c^{bn})^b$,
and
$z = (1+c^{bn})^a
$.
If
$n=3$ and $m=2$,
we want
$2a-3b = 1$.
This is satisfied by
$a=2, b=1$.
We get
$x = (1+c^{3})$,
$y 
=c(1+c^{3})$,
and
$z = (1+c^{3})^2
$.
A: Just because sometimes you can ignore the hint (ha ha), one can also obtain a family of solutions by taking an existing solution, and multiplying both $x$ and $y$ by $k^2$, and multiplying $z$ by $k^3$ for some $k \in \mathbb{N}, k \geq 2$.
For instance, if we take $k = 2$, then the above solution is transformed as
$$
(2, 2, 4) \to (8, 8, 32)
$$
and note that
$$
8^3+8^3 = 512 + 512 = 1024 = 32^2
$$
