If $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a,b$ or $c$ is divisible by $7$.

Question

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?

Answer 1

By Fermat's theorem $x^6\equiv1\pmod7$ holds for all $x$ coprime to $7$, so $(x^3)^2\equiv1\pmod7$ hence $$(x^3)^2-1=(x^3-1)(x^3+1)\equiv0\pmod7$$ and hence $x^3\equiv\pm1\pmod{7}$.

Answer 2

You want to prove the statement:

$[a^3+b^3+c^3\equiv0\pmod7]\implies[a\equiv0\pmod7]\vee[b\equiv0\pmod7]\vee[c\equiv0\pmod7]$

Instead, prove the equivalent statement:

$[a\not\equiv0\pmod7]\wedge[b\not\equiv0\pmod7]\wedge[c\not\equiv0\pmod7]\implies[a^3+b^3+c^3\not\equiv0\pmod7]$

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Now, observe the following:

• $n\equiv1\pmod7 \implies n^3\equiv1^3\equiv 1\equiv1\pmod7$
• $n\equiv2\pmod7 \implies n^3\equiv2^3\equiv 8\equiv1\pmod7$
• $n\equiv3\pmod7 \implies n^3\equiv3^3\equiv 27\equiv6\pmod7$
• $n\equiv4\pmod7 \implies n^3\equiv4^3\equiv 64\equiv1\pmod7$
• $n\equiv5\pmod7 \implies n^3\equiv5^3\equiv125\equiv6\pmod7$
• $n\equiv6\pmod7 \implies n^3\equiv6^3\equiv216\equiv6\pmod7$

Therefore, $[a\not\equiv0\pmod7]\wedge[b\not\equiv0\pmod7]\wedge[c\not\equiv0\pmod7]\implies$ one of the following:

• $a^3+b^3+c^3\equiv1+1+1\equiv 3\equiv3\not\equiv0\pmod7$
• $a^3+b^3+c^3\equiv1+1+6\equiv 8\equiv1\not\equiv0\pmod7$
• $a^3+b^3+c^3\equiv1+6+1\equiv 8\equiv1\not\equiv0\pmod7$
• $a^3+b^3+c^3\equiv1+6+6\equiv13\equiv6\not\equiv0\pmod7$
• $a^3+b^3+c^3\equiv6+1+1\equiv 8\equiv1\not\equiv0\pmod7$
• $a^3+b^3+c^3\equiv6+1+6\equiv13\equiv6\not\equiv0\pmod7$
• $a^3+b^3+c^3\equiv6+6+1\equiv13\equiv6\not\equiv0\pmod7$
• $a^3+b^3+c^3\equiv6+6+6\equiv18\equiv4\not\equiv0\pmod7$

Answer 3

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 $7$.