Why is $a^{5} \equiv a\pmod 5$ for any positive integer? Why is $a^{5} \equiv a\pmod 5$ for any positive integer?
I feel like it should be obvious, but I just can't see it. Any help appreciated. 
Edit: without Fermat's theorem. 
 A: OK, without using Fermat's Little Theorem (a far more general and elegant result), here's another easy workaround.
Any integer $a$ can be exactly one of $0, 1, 2, -2, -1 \pmod 5$.
Take the fifth powers of each of those and see them reduce back to the original residue in each case.
A: One way to prove this is to prove it by induction. If $n^5\equiv n\pmod 5$ show that $(n+1)^5\equiv n+1\pmod 5$.
Note that $$(n+1)^5=n^5+5n^4+10n^3+10n^2+5n+1\equiv n^5+1\equiv n+1\pmod5$$
The general theorem people have mentioned in comments, Fermat's Little Theorem, states that if $p$ is prime and $a$ is any number: $a^p\equiv a\pmod p$.
A: Consider $$a(a-1)(a-2)(a-3)(a-4)=a(a^4-10a^3+35a^2-50a+24).$$ Taken mod $5$ this becomes
$$a(a^4-1).$$and so $a^5-a \equiv 0 \mod 5$, or $a^5 \equiv a \mod 5$ as required.
A: Note $\ a(a^4\!-1\!)\, =\, a(a^2-1)\overbrace{(a^2+1)}^{\Large \color{#0a0}{a^2-4}\,+\,\color{#c00}5} = \!\!\!\underbrace{a(a^2-1)(\color{#0a0}{a^2-4})}_{\large\color{blue}{ (a-2)(a-1)a(a+1)(a+2)}}\!\!\!\! + \color{#c00}5\,a(a^2-1)$
$\color{#c00}5\,$ divides both summands, the first because $\,\color{#c00}5\,$ divides one of $\,\rm\color{#c00}5\,\  \color{blue}{\rm  consecutive\ integers}$.
A: As others said, it's Fermat's little theorem.  One way to verify it in this particular case is to note that 
$$a^5 - a = a (a - 1)(a+1)(a^2 + 1)$$
If $a \equiv 0, 1$ or $4 \mod 5$, one of the first three factors is $0 \mod 5$.
The other two possibilities are $a \equiv 2$ or $3 \mod 5$, in which case 
$a^2 + 1 \equiv 4 + 1 \equiv 0 \mod 5$.  So in each case, $a^5 \equiv a \mod 5$.
A: A more general solution is the following:
Consider the multiplicative group $\mathbb{Z}_{p}^{\ast}$ of non-zero elements $\mod p$.  This group has $p-1$ elements.  Then for any element in the group, there is a cyclic subgroup $\langle g\rangle$.  The order of this group is the order of $g$ which divides $p-1$, say the order is $k$ and $ak=p-1$.  This means that $g^{p-1}=g^{ak}=(g^{k})^{a}\equiv 1^{a} \equiv 1 \mod p$.  Multiplying again by $g$ you have that $g^{p}\equiv g \mod p$.  That works for all the non-zero elements.  It is obvious for the 0.
