<s>[INST] Prove that for all odd integer $n, n^{4}=1\pmod {16}$ My answer: $n=2k+1$ 
$n^{4}=(2k+1)^{4}$=$16k^{4}+32k^{3}+8k^{2}+24k+1$.
I do not know how to conclude; really needed help here.
 [/INST] $$n^4-1=(n^2+1)(n+1)(n-1)$$
If $n$ is odd, then all the factors are even, and one of the factors $(n+1)$ or $(n-1)$ is divisible by $4$. Hence $2^4|n^4-1$, or in other words $n^4\equiv1\pmod{16}$.
This can be generalized to see that $n^{2^k}\equiv1\pmod{2^{k+2}}$ for odd $n$:
$$n^{2^k}-1=\left(\left(n^{2^{k-1}}+1\right)\left(n^{2^{k-2}}+1\right)\cdots\left(n+1\right)\right)\left(n-1\right)$$
Since $x$ is odd, all the $k+1$ factors in the above are even, and either $(n+1)$ or $(n-1$) is divisible by $4$, so $2^{k+2}|n^{2^k}-1$. 
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