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?

$$x^5y-xy^5=(x^5-x)y-(y^5-y)x$$
Now $p|(a^p-a)$ by Fermat's little theorem for prime $p$
$a^5-a$ is divisible by $a^3-a$
If $x,y$ are even, we are done.
For odd $a,a=2c+1$(say)
$(2c+1)^2=8\dfrac{c(c+1)}2+1=8d+1$ for some integer $d$
$(2c+1)^4=(8d+1)^2=16(d+4d^2)+1$
$\implies(2c+1)^{4m+1}\equiv2c+1\pmod{16}$
$\implies a^5\equiv a\pmod{16}$ for odd $a$
