Can anyone explain how to show that $n^{5} -n ≡0$ mod $30$ for every $ n \in \mathbb{N} $

Question

I first tried to answer this using proof by induction, however my problem got more complicated when I got to the induction step. Is there another way of solving this problem?

Answer 1

Note that $$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1).$$

First, $n(n-1)$ is even because these are two consective numbers.

Second, $(n-1)n(n+1)$ is divisible by $3$ because these are three consective numbers.

Third, one of $n,n-1,n+1,n^2+1$ is divisible by $5$. In mod $5$, if $n\equiv 0$, then it's obvious. If $n\equiv 1$, then $n-1\equiv 0$. If $n\equiv 2\ \text{or}\ 3$, then $n^2+1\equiv 0$. If $n\equiv 4$, then $n+1\equiv 0$.

Hence, $n^5-n$ is divisible by $2\times 3\times 5=30$.

Answer 2

Verify the equation for 0,1,2,...,29. If it is valid for these numbers, the equation is valid for all $n \in \mathbb{N}$.

Answer 3

HINT

By Fermat's little theorem $$n^5\equiv n \pmod 5\\n^3\equiv n\pmod 3\\n^2\equiv n\pmod 2$$

Answer 4

$$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2-4+5)$$

$$=\underbrace{n(n-1)(n+1)(n-2)(n+2)}_{\text{The product of }5 \text{ consecutive integers}}+5\cdot\underbrace{n(n-1)(n+1)}_{\text{The product of }3 \text{ consecutive integers}}$$

See The product of n consecutive integers is divisible by n factorial