divisibility problem by muhammad dihya

Let $N = x^n - (x \bmod 5)^n$ . We need to show if $N \bmod 5 =0$ for \integer $x > 4$ . For integer $x > 4$ , we can always express $x$ as $x = 5m + x \bmod 5$ , where $m$ is a positive integer. Then

$\begin{aligned} N & \equiv x^n - (x \bmod 5)^n \text{ (mod 5)} \\ & \equiv (5m+x \bmod 5)^n - (x \bmod 5)^n \text{ (mod 5)} \\ & \equiv (x \bmod 5)^n - (x \bmod 5)^n \text{ (mod 5)} \\ & \equiv \boxed{0} \text{ (mod 5)} \end{aligned}$

True , $N$ is always divisible by 5 for all $x > 4$ and $n \in \mathbb N$ .

$x^n - (x \mod 5)^n \equiv x^n-(x^n \mod 5) \equiv (x^n-x^n) \pmod{5} \equiv 0 \pmod{5}$

Hence the result follows.