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!

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
$$m^2 \equiv 0 \mod 13\\
m \equiv 0 \mod 13$$
