what is the remainder when $(17^{3}+19^{3} + 21^{3}+23^{3})$ is divided by 83? what is the remainder when $(17^{3}+19^{3} + 21^{3}+23^{3})$ is divided by 83?
NOTE:$a^{3}+b^{3}=(a+b)(a^2-ab+b^2)$

For every $a$, $(a+1)^3+(a-1)^3=2(a^3+3a)$ and $(a+3)^3+(a-3)^3=2(a^3+3\cdot9a)$ hence $s(a)=(a-3)^3+(a-1)^3+(a+1)^3+(a+3)^3$ is
$$
s(a)=4a^3+2\cdot3\cdot(1+9)a=4a(a^2+15).
$$
Since $83$ is prime, $83$ divides $s(a)$ if and only if $83$ divides $a$ or $83$ divides $a^2+15$. In your post, $a=20$, then $83$ does not divide $a$ but $a^2+15=415=5\cdot83$ hence $83$ divides $s(20)$.
