A cool divisibility problem

Since $m^5 - m \; = \; m(m^4 - 1) \; = \; m(m^2 - 1)(m^2 + 1) \; = \; (m-1)m(m+1)(m^2+1)$ has three consecutive factors $m-1,m,m+1$ , it is clear that $6$ divides $m^5 - m$ for any integer $m$ . Fermat's Little Theorem tells us that $5$ divides $m^5 - m$ for all integers $m$ . Hence we deduce that $30$ divides $m^5 - m$ for all integers $m$ . It follows that $30$ divides $(a^5 + b^5 + c^5) - (a + b + c) \; = \; (a^5 - a) + (b^5 - b) + (c^5 - c)$ for all integers $a,b,c$ . Thus, if $a,b,c$ are integers such that $30$ divides $a+b+c$ , then $30$ divides $a^5 + b^5 + c^5$ as well.

$30 = 2*3*5 \\ \\ a \equiv a^5 \pmod 2 \\ a \equiv a^5 \pmod 3 \\ a \equiv a^5 \pmod 5$ Then, $(a+b+c) \equiv (a^5+b^5+c^5) \pmod {30}$

this is Indonesian MO