Fermat’s Little Theorem states that for a prime $p$ and an integer $a$ not divisible by $p$
$$ a^{p-1} \equiv 1 \hspace{0.05in} (\text{mod } p) $$
Problem (Belorussia, 1965)
Given that $a^{41}+b^{41}+c^{41}$ is divisible by $83$. Is it true that $abc$ is also divisible by $83$?
Solution
Answer: yes, it is true.
Let $x$ be an integer not divisible by $83$. Notice that $83$ is prime and therefore by Fermat’s Little Theorem
$$ x^{82} \equiv 1 \hspace{0.05in} (\text{mod } 83) $$
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This implies that $x^{82} – 1$ is divisible by $83$. From the equality
$$x^{82} – 1 = \left( x^{41} – 1 \right) \left( x^{41} +1 \right)$$
we see that $x^{41}$ should always be congruent to $\pm 1$ modulo $83$. If none of the numbers $a$, $b$, $c$ are divisible by $83$, then the sum $a^{41}+b^{41}+c^{41}$ cannot be equal zero modulo $83$. Therefore at least one of the numbers $a$, $b$, $c$ should be divisible by $83$ and their product as well.
