How do you prove that $ n^5$ is congruent to $ n$ mod 10? How do you prove that $n^5 \equiv n\pmod {10}$ Hint given was- Fermat little theorem. Kindly help me out. This is applicable to all positive integers $n$
 A: Here's a direct proof: note that
$$n^5 - n = n(n^4-1) = n(n-1)(n+1)(n^2+1)$$
You need to prove that this number is divisible by $2$ and $5$.


*

*For divisibility by $2$, note that either $n$ or $n+1$ must be even.

*For divisibility by $5$, you should prove that if neither $n$ nor $n \pm 1$ is divisible by $5$ (i.e. if $n \equiv 2\ \text{or}\ 3 \bmod 5$) then $n^2+1$ is divisible by $5$.
A: Hint: In order to use Fermat's Little Theorem, you need the modulus to be prime which $10$ is not. However, $10 = 2\times 5$. Use Fermat's Little Theorem for each prime and see if you can combine these results to arrive at your claim.
A: By Fermat's Little Theorem, $n^5 \equiv n \pmod 5$
Also $n \equiv 0 \ or \ 1 \pmod 2 \implies n^5 \equiv n \pmod 2$
Chinese Remainder Theorem guarantees the presence of a solution $\pmod {10}$ as $(2,5) = 1$
From the second congruence, $n^5 = 2k + n$
Substitute into the first congruence, $2k+n \equiv n \pmod 5 \implies 2k \equiv 0 \pmod 5 \implies k \equiv 0 \pmod 5$
So $k = 5m$
Substitute that into second congruence, $n^5 = 2(5m) + n  = 10m +n$
Therefore, $n^5 \equiv n \pmod {10}$ $\rm{(QED)}$
