Prove that for all integers $a, b, c$ if $a+b^3+c^5=6001$ then at least one of $a,b,c$ is a multiple of three.

Question

Prove that for all integers $a, b, c$ if $a+b^3+c^5=6001$ then at least one of $a,b,c$ is a multiple of three.

Do I start with cases? How should I go about proving this?

Thanks for your help!

Answer 1

It's not true, e.g. $a=5999$, $b=1$, $c=1$.