Given that $a,b,c$ are there distinct positive integers, prove that among $a^5b-ab^5,b^5c-bc^5,c^5a-ca^5$, there is at least one that is divisible by $8$.
I have no clue what to do and cannot even create cases as these are three random whole numbers. What do I do? Please help me guys!!! Even a hint is graciously accepted!!!
On
The three numbers are cyclic permutations of $ab(a^4-b^4)$. Therefore, at least one of the products must be formed from either $2$ even numbers or $2$ odd numbers.
If $a$ and $b$ are both even, then so is $(a^4-b^4)$, so $ab(a^4-b^4)$ is the product of three even numbers, which must be divisible by $8$.
If $a$ and $b$ are both odd, then $a^4 \equiv b^4 \pmod{8}$ (because any odd number is a square root of $1$ (mod $8$)), so $8|(a^4-b^4)$ and therefore $8|ab(a^4-b^4).$
Note that $x\equiv 0$ (mod $16$) if $x$ is even and $x\equiv 1$ (mod $16$) if $x$ is odd.
Among $a$, $b$ and $c$, at least two are of the same parity. If $a$ and $b$ have the same parity, then $a^4-b^4$ is divisible by $16$.