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!

  • 1
    $\begingroup$ 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 $\{a, b, c\} \subseteq \{1, 2\} \pmod 3$. $\endgroup$ – DanielV Feb 11 at 10:21
  • $\begingroup$ 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). $\endgroup$ – Bumblebee Feb 14 at 0:18

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


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