Let $x,y,z$ be 3 coprime integers where $u|x, v|y, w|z$, is $x^3+y^3+z^3\equiv 0 \pmod{(uvw)^3}?$

Question

Let $u,v,w \neq \pm1$ be 3 non-zero integers respective factors of 3 relatively prime integers $x,y,z$. Is the following equivalence possible: $$x^3+y^3+z^3\equiv 0 \pmod{(uvw)^3}?$$ It is obvious if $x^3+y^3+z^3=0$ there are only trivial integral solutions. I am curious about the general case above. Any hints?

Answer 1

You are looking for solutions of the congruence

$$(au)^3+(bv)^3+(cw)^3\equiv0\mod(uvw)^3$$

where $au$, $bv$, and $cw$ are (pairwise) relatively prime and $u,v,w\gt1$. (Note, negative signs can be absorbed into the $a,b,c$.) Let's see if we can construct a simple example, using the smallest available values for $u,v,w$, namely $2,3,5$.

These values for $u,v,w$ mean we are looking for simultaneous solutions of the congruences

$$\begin{align} (3b)^3+(5c)^3&\equiv0\mod8\\ (2a)^3+(5c)^3&\equiv0\mod27\\ (2a)^3+(3b)^3&\equiv0\mod125\\ \end{align}$$

with restrictions to keep $2a,3b,5c$ relatively prime.

The three congruences are satisfied by solutions of the simpler congruences

$$\begin{align} 3b+5c&\equiv0\mod8\\ 2a+5c&\equiv0\mod27\\ 2a+3b&\equiv0\mod125\\ \end{align}$$

The first of these is clearly satisfied if we set $b=c=1$, which can only help keep $2a,3,5$ relative prime. With this choice for $b$ and $c$, a little finagling reduces the other two congruences to

$$\begin{align} a&\equiv11\mod27\\ a&\equiv61\mod125\\ \end{align}$$

The Chinese Remainder Theorem guarantees a solution, and it's clear that the solution is divisible by neither $3$ nor $5$. If you work things out explicitly, you find $a=686$ solves the final two conguences and hence

$$(1372)^3+3^3+5^3\equiv0\mod(2\cdot3\cdot5)^3$$

is a solution of the OP's congruence.