Why $x^5y\equiv y^5x\pmod {240}$ if $x,y$ have similar parity? Let $x,y$ be any 2 integers of similar parity. Why do we have :$$x^5y\equiv y^5x\pmod {240}$$ 
Any hint anyone?

Factorising we get $xy(x-y)(x+y)(x^2 + y^2) \equiv 0 \mod 2^4\cdot 3\cdot 5$
Because $x, y$ have the same parity then, if $x,y$ even all factors are divisible by 2, else if $x,y$ odd all but $xy$ are divisible by 2 and $(x^2 - y^2)$ is divisible by 8. Therefore $xy(x-y)(x+y)(x^2 + y^2) \equiv 0 \mod 2^4$ 
Using  Fermat's little theorem, we have $x^5 \equiv x \mod 5$ and $y^5 \equiv y \mod 5$ therefore $x^5y\equiv y^5x\pmod {5}$ also $x^3 \equiv x \mod 3$ and $y^3 \equiv y \mod 3$ therefore $x^5 \equiv x \mod 3$ and $y^5 \equiv y \mod 3$
