Find all natural numbers $n$ such that $n^{17}-n$ is divisible by 10 I've encountered this math problem and I don't know how to solve it. What math region,field(however you call it. English is not my first language) is used here ?

Find all natural numbers for which ($n^{17} - n $  ) is divisible by
  10.

Any tips or advice you can give me and if you could share some websites where I can learn more about this.
 A: You can write $$n^{17}-n=(n-1)n(n+1)(n^2+1)(n^4+1)(n^8+1).$$ To satisfy the condition it is enough that one of these factors be divisible by 10. For example, any $n$ that is a multiple of $10$ satisfies the condition. 
In fact any natural number larger than $1$ satisfies the condition. The first three factors are consecutive numbers. If none of those numbers is a multiple of $5$ then (as long as $n>1$) it must be that $n=5k+2$ or $n=5k+3$ for some integer $k\geq 0$. In the first case $$n^2+1=(5k+2)^2+1=5 (5k^2+4k+1),$$ while in the second case $$n^2+1=(5k+3)^2+1=5 (5k^2+6k+2).$$
A: Hint:
$1)$ $n^{17}-n=n(n^4-1)(n^4+1)(n^8+1)$
$2)$ $\phi(10)=4$
A: A systematic approach: just check the cases of $2$ and of $5$ respectively.
It is clear that $n^{17}-n$ is even. Also we see
$$n^{17}-n=(n^4)^{4}\times n-n\equiv 0\pmod5$$
since if $n$ is not divisible by $5,$ then $n^4-1$ is divisible by $5.$  
Hope this helps.
A: You can check all the remnants by $mod \ 10$ simply taking 17th power (there are 10 of them - $1,2, \ldots , 9$) or you can check whether the polynomial mentioned above takes its value divided by 10.
A: I don't have time to develop this, but it seems that all positive numbers satisfy your problem.
In Mathematica:
Reduce[Mod[n^17 - n, 10] == 0, n, Integers]

We have output:
C[1] \[Element] Integers && 
(n == 10 C[1] || n == 1 + 10 C[1] || n == 2 + 10 C[1] ||
n == 3 + 10 C[1] || n == 4 + 10 C[1] || n == 5 + 10 C[1] || 
n == 6 + 10 C[1] || n == 7 + 10 C[1] || n == 8 + 10 C[1] || 
n == 9 + 10 C[1])

