TITLE: The equation $a^3 + b^3 = c^2$ has solution $(a, b, c) = (2,2,4)$.
QUESTION [5 upvotes]: 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

REPLY [3 votes]: 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.