<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] Let $m=n-1$
$n^4 = m^4+4m^3+6m^2+4m^1+1$
Since $m$ is even, $m=2x$ for some $x\in\mathbb{N}$
$\implies n^4=2^4x^4+4\times2^3x^3+6\times2^2x^2+4\times2x+1$
$16x^4, 32x^3$ are both $0($mod $16)$, leaving $6\times2^2x^2+4\times2x+1$.
If $x$ is odd, $3x+1$ must be even, therefore $8x(3x+1)$ is $0($ mod $16)$.
If $x$ is even, $8x$ must be $0($ mod $16)$.
$\implies n^4=0($ mod $16)$
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