Prove that ${n^5 - n}$ is divisible by 5 I need to prove by induction if ${n^5 - n}$ is divisible by 5 and I have no idea how I would do it. 
I am trying to prove it for several hours now, I started with ${n^5 - n} \mod  5 = 0$ but then I realized that I have no idea how to use modulo for transformations.
The next thing I tried was $(5 + n)\frac{n^5 - n}5 = \frac{(n+1)^5 - n+1}5$ but it didn't get me anywhere.
How would you do it?
 A: Let $\displaystyle f(n): n^5-n$ is divisible by $5$ holds true for $n=m$
$\displaystyle\implies 5\mid(m^5-m)$
Now, $\displaystyle f(m+1)-f(m)=(m+1)^5-(m+1)-(m^5-m)=\sum_{1\le r \le 4}\binom5 rm^r$ which is divisible by $5$ 
as each $\displaystyle\binom5r,1\le r\le4$ is divisible by $5$
Finally, establish the base case i.e., $n=1$
Generalization :
More generally prime $p$ divides $\displaystyle\binom pr$ (can you prove this?) for $1\le r\le p$ or more genrally $p\not\mid r$
So, we can prove  $\displaystyle n^p-n$ will be divisible by $p$ (Fermat's Little Theorem)
A: Let $p$ be any prime number. Consider the polynomial $f=X^p-X=X(X^{p-1}-1)\in \mathbb F_p[X].$ This polynomial is the zero function $\mathbb F_p\to\mathbb F_p$ if and only if $n^p-n$ is divisible by $p$ for all $n\in\mathbb Z$. Clearly $f(0)=0$ and $\mathbb F_p^\times$ is a group of order $p-1$ so for all $a\in\mathbb F_p^\times$ we have $a^{p-1}=1$. Now apply this to $p=5$.
A: Hint $\,\ {\rm mod}\ 5\!:\ n\equiv 0,\,\pm1,\,$ or $\, \pm2.\ $  $\ 0^5\!\equiv 0,\,\ (\pm1)^5\!\equiv \pm1,\,\ (\pm2)^5\!\equiv \pm32\equiv \pm2,\ $ so $\ n^5\!\equiv n.$
A: You can always try brute force in case you are stuck otherwise:
$$(n+1)^5=1+5 n+10 n^2+10 n^3+5 n^4+n^5$$
$$(n+1)^5-(n+1)=4 n+10 n^2+10 n^3+5 n^4+n^5$$
$$(n+1)^5-(n+1)-\big(n^5-n\big)=5 n+10 n^2+10 n^3+5 n^4=5(n+2n^2+2n^3+n^4)$$
A: Any natural number can be represented by $5k + q$, where $q = 0,1,2,3,4$. If we prove it for all mentioned $q$, then it will work with every natural.
$n^5 - n = n(n^4 - 1) = n(n - 1)(n + 1)(n^2 + 1)$.
By the nature of this expansion if we plug in $ n = 5k + q$,  where $ q =0,1,4$, we will  get something of the form $5k*f(k)$, which is divisible by $5$.
If $q = 2,3$, then $n(n-1)(n+1)$ will not return a multiple of $5$, so we look at $n^2 + 1$ :
$n^2 + 1 = (5k + q)^2 + 1 = 25k^2 + 10kq + q^2 + 1$
If $q = 2$, then $n^2 + 1 = 25k^2 + 20k + 5$
If $q = 3$, then $n^2 + 1 = 25k^2 + 20k + 10$ 
In both cases, the resulting coefficients are divisible by $5$. The polynomial is divisible by $5$, for all mentioned $q$. Thus the polynomial satisfies the required condition for all naturals.
A: I would like to give you a more simple approach to this question:
=>n^5-n is given to you.
So, on taking n common n(n^4-1) is you got.
so this expression has to be divisible by 5 the multiplication of both n and n^4-1 have to divisible by 5. And, Clearly for each number we have n^4-1 have to be divisible by 5.
Now, calling the number
Any number which has even in unit's place like 2,4,6,8
2^4=6
4^4=6
6^6=6
8^6=6
Any number which have even number in unit's place is leading to in 6 and subtracting 1 we got 5 so each expression is divided by 5
And, similarly for odd numbers 
1^4=1
3^4=1
7^4=1
9^4=1
And, subtracting 1 from we got 0 so it is again divisible by 5.
And, when we have 5 at unit's place n is divisible by 5.
Sorry , for the poor presentable form, but I think this will work
