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?
 A: 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 $y^5 \equiv y \mod 3$
A: $$x^5y-xy^5=(x^5-x)y-(y^5-y)x$$
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 $a$
A: Thanks to Chinese Reminder Theorem we need to check mod $3$, $5$ and $16$ only.
Using that $a^4 \equiv 1 (mod 5)$ for every invertible $a$ in $ \mathbb Z_5$, and $a^4 \equiv a^2 \cdot a^2 \equiv 1 (mod 3)$ for every invertible $a$ in $ \mathbb Z_3$ we are almost done. Now notice that if $x$, $y$ have the same parity, if they are both even we are done mod $16$. Otherwise $(2d+1)^4 = (4d^2 + 4d +1)^2\equiv 1 + 8(d^2+d) \equiv 1 (mod 16)$.
