How to show that $2730\mid n^{13}-n\;\;\forall n\in\mathbb{N}$

Solution 1:

HINT:

$$n^{13} \equiv n^5 \cdot n^5 \cdot n^3 \equiv n \cdot n \cdot n^3 \equiv n^5 \equiv n \pmod 5$$

$$n^{13} \equiv n^6 \cdot n^7 \equiv n \pmod 7$$

Also you've missed $3$ as prime factor. But that should be easy.

Solution 2:

Like user99680,

Using Fermat's Little Theorem $p|(n^p-n)$ where $n$ is any integer and $p$ is any prime

$\displaystyle n^{13}-n=n(n^{12}-1)=n\left((n^6)^2-1\right)=n(n^6-1)(n^6+1)=(n^7-n)(n^6+1)$ which is divisible by $\displaystyle n^7-n$ which is always divisible by $7$ for all integer $n$

Similarly, $\displaystyle n^{13}-n=n(n^{12}-1)=n\left((n^4)^3-1\right)$ $\displaystyle=n(n^4-1)(n^8+n^4+1)=(n^5-n)(n^8+n^4+1)$ which is divisible by $\displaystyle n^5-n$ which is always divisible by $5$ for all integer $n$