# 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.

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!

• The normal way you would solve this problem, if it were true, would be to look at the equation $\pmod 3$ and check the 8 cases of $\{a, b, c\} \subseteq \{1, 2\} \pmod 3$. – DanielV Feb 11 at 10:21
• Lets assume that all integers are strictly bigger than $1,$ then $2\le c\le 5$ and $2\le b\le 18.$ Therefore it is not difficult to compute all possible such triples (even by hand). – Bumblebee Feb 14 at 0:18

## 1 Answer

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