<s>[INST] Primitive instances where $c^3|(a^3+b^3)$ with $a,b,c\in\mathbb{N}$ It is known that by Fermat's Last Theorem there are no solutions to $a^3+b^3=c^3$ for $a,b,c\in\mathbb{N}$. I wondered about how multiplying the $c^3$ by a constant would change this fact.
Accordingly, I have been looking into instances where $c^3|(a^3+b^3)$ for $a,b\in\mathbb{N}$. In other words, solutions to the Diophantine Equation:
$a^3+b^3=dc^3$ where $a,b,$ and $c$ are pairwise co-prime and $a,b,c>0$
Obviously there are some trivial solutions. If $c=1$, for example, $a$ and $b$ can be any integers, and $d$ can be chosen to simply be $a^3+b^3$. 
By requiring that $a,b,$ and $c$ are pairwise co-prime and that $c\not=1$, we eliminate the trivial solutions, and what remains is of much more interest. 
For $a,b,c,d\le20$, there are 5 solutions:
$4^3+5^3=7*3^3$
$2^3+7^3=13*3^3$
$1^3+8^3=19*3^3$
$3^3+5^3=19*2^3$
$1^3+19^3=20*7^3$
Under $100$ there are $16$ solutions, as found by Mathematica.
My  question about this equation: Has it been studied previously? Are there infinitely many primitive solutions (which it seems like there are)? If so, can they be parametrized?
 [/INST] It's actually very easy to identify an infinite number of solutions.  Suppose that $m^3+1$ is divisible by $3^3=27$, whichnisctrue for $m=8$.  Then
$(m+9)^3+1=(m^3+1)+(27m^2+243m+729)$
is also divisible by $27$.  Having $1$ as one of the cubes guarantees relative primarily for all these solutions, but we can reasonably expect infinitely many relatively prime solutions to arise similarly from $m^3+n^3$ for other values of $n$.
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