Prove the divisibility

Question

For every positive integer prove that$$n^7 -n$$ is divisible by 7 My turn : Let $$k^7 -k = 7m$$ Now i will try to prove that $$(k+1)^7 -(k+1) = 7p$$ $$(k+1)((k+1)^6 -1)$$ but i stopped here ! What should i do ?

Answer 1

Let us consider the possible remainder when taken $7$ modulo

$$\begin{align}n & \quad 0 \quad 1\quad 2 \quad 3 \quad 4 \quad 5\quad 6 \\ n^2 &\quad 0 \quad 1\quad 4 \quad 2 \quad 2 \quad 4\quad 1 \\ n^6 &\quad 0 \quad 1\quad 1 \quad 1 \quad 1 \quad 1\quad 1 \\ n^7 &\quad 0 \quad 1\quad 2 \quad 3 \quad 4 \quad 5\quad 6 \\ \end{align} $$

And hence , $$\begin{align}n^7 -n &\quad \color{red}{\boxed{0 \quad 0\quad 0 \quad 0 \quad 0 \quad 0\quad 0} } \end{align} $$

It is trivial to observe that $n^7 - n$ leaves $0$ as a remainder when divided by $7$ , and so It is divisible by $7.$