<s>[INST] If $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a,b$ or $c$ is divisible by $7$. 
Show that if $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a$, $b$ or $c$ is  divisible by $7$.

Here it seems Fermat's theorem  has no use. We could consider many different cases of remainders of $a,b,c$ modulo $7$ but that's tedious.
Any ideas for a much shorter solution?
 [/INST] Suppose that none of $a,b,c$ is divisible by $7$.
Write $a^3+b^3 \equiv -c^3$, square both sides, and use Fermat:
$ a^6 + 2a^3b^3 + b^6 \equiv c^6 $
$ 1+2a^3b^3 + 1\equiv 1 $
$ (ab)^3 \equiv 3$
But $3$ is not a cube mod $7$.
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