Proving $n^{97}\equiv n\text{ mod }4501770$

Question

How do we show $$n^{97}\equiv n\text{ mod }4501770$$ for all integer $n$? First of all, I thought I could use Fermat's little theorem or Euler's theorem, but I'm not sure if they are applicable here.

Answer 1

As $\displaystyle4501770 =2\cdot3\cdot5\cdot7\cdot13\cdot17\cdot97$

Using Fermat's Little Theorem prime $p$ always divides $(n^p-n)=n(n^{p-1}-1)$

For example, $\displaystyle p=17, n^{97}-n=n(n^{96}-1)=n\{(n^{16})^6-1^6\}$ which is a multiple of $n(n^{16}-1)$ which is divisible by $17$ for all integer $n$

Similarly for the rest of the factors (which are also prime)

Now as $n^{97}-n$ is divisible by $2,3,5,7,13,17,97;$

$n^{97}-n$ will be divisible by their LCM which is their product here(right? why?)

Regarding how $96$ is identified, what is the LCM$(2-1,3-1,5-1,7-1,13-1,17-1,97-1)$

See also : Carmichael Function and calculate $\lambda(4501770)$