Any Odd Difference?

$x^3-x=x(x^2-1)=x(x-1)(x+1)=(x-1)x(x+1)$ . Since $x$ is an integer, the expression is the product of three consecutive integers. Within three consecutive integers there must be at least an even integer. Therefore, the product and hence the expression is always even.

$\text{Let }x = 2q + r \text{, r = 0, 1} \\ x^3 - x = x(x^2-1) = x(x+1)(x-1)\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ \large \text{Case 1 . x = 2q }\\ x^3 - x = 2q(2q+1)(2q-1) \\ \implies 2|(x^3-x) \\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \large \text{Case 2. x = 2q + 1} \\ x^3 - x = (2q+1)(2q + 2)(2q) \\ \implies 2|(x^3-x)$

$x^3 - x = x(x^2 - 1)$ .

If $x$ is even then $x(x^2 - 1)$ is even. ( $\because$ Even × any integer = Even).

If $x$ is odd, then $x^2$ is odd. ( $\because$ Odd × Odd = Odd $\forall$ integers). Thus $x^2 - 1$ is even ( $\because$ Odd and Even are consecutive $\forall$ integers). Thus $x(x^2 -1)$ is even.

Therefore, $x^3 - x$ is even $\forall x \in \text{Z}$ .

Trivially, $x^3 \equiv x \pmod{2} \implies x^3 - x \equiv 0 \pmod{2}$ . Hence, $\boxed{True}$ .

(I felt an overwhelming urge to do a one-line proof.)

x3-x=x(x2-1), If x is even, x2-1 is odde and x(x2-1) is even If x is odd, x2-1 is even and x(x2-1) is even