Let $n ∈ ℤ⁺$ with $n > 1$. Prove: If $a ≡ 1$ (mod $n²$), then $a^n ≡ 1$ (mod $n³$)

Question

I don't know where to start or how to prove this at all! Can somebody please explain to me the process of what to do?

Answer 1

Since $a\equiv 1\pmod{n^2}$, then $a=kn^2+1$ for some integer $k$. By the binomial theorem,

$$ (kn^2+1)^n=k^nn^{2n}+\cdots+n(kn^2)+1=k^nn^{2n}+\cdots+kn^3+1$$

Where every term except the last is divisible by $n^3$. The conclusion is now clear.