<s>[INST] Mathematical proof that $m^4+8$ is not a cube of an integer if $13$ does not divide $m$ (Table proof provided) My question is how to express my solution to $m^4+8$ is not a cube of an integer if $13$ does not divide $m$ as a mathematical proof.
I understand that to prove by contradiction, the result is to be a cube of an integer:
\begin{align}
 m^4 + 8 & = n^3 \\
 m^4 &= n^3 - 8 \\
 \therefore m^4 &\equiv n^3-8\pmod{13}\\
  \\
\end{align}
That equivalence is however impossible, because of the following table:
We see that $m^4 \equiv n^3-8\pmod{13}$ can only occur when m $\equiv 0 \pmod{13}$ and n = 2, 5 or 6.
This shows that since $13$ can not divide $m$, $m^4+8$ is not a cube of an integer. I am now sure how to express this mathematically.
Any help would be appreciated!
 [/INST] Your method is perfectly fine for this. Just for fun, I'm giving an alternative method in which you can make do with only a table of squares mod 13.
n     0  1  2  3  4  5  6  7  8  9 10 11 12
n^2   0  1  4  9  3 12 10 10 12  3  9  4  1

$$m^4+8 \equiv n^3 \mod 13\\
(m^4+8)^4 \equiv n^{12} \mod 13$$
From Fermat's little theorem we know that $n^{12}\equiv0,1 \mod 13$, so we then get:
$$(m^4+8)^4 \equiv 0,1 \mod 13\\
(m^4+8)^2 \equiv 0,1,12 \mod 13\\
m^4+8 \equiv 0,1,5,8,12 \mod 13\\
m^4 \equiv 0,4,5,6,10 \mod 13$$
Note that $5$ and $6$ are not squares, so they are certainly not fourth powers.
$$m^4 \equiv 0,4,10 \mod 13\\
m^2 \equiv 0,2,6,7,11 \mod 13$$
Note that the only one of these that is a square is $0$ so we are just left with
$$m^2 \equiv 0 \mod 13\\
m \equiv 0 \mod 13$$
</s>