Given $p = nq$, $n \not\equiv q$, and $\forall m: m ≡ m^{ed} \bmod n$, how can we show that $ed \equiv 1 \bmod \lambda(n)$? My idea was to show it using the $k$ exponent, but I got stuck. So:
From Fermat's Little Theorem we know, that $$ m^p = m \bmod p $$ $ m^{\lambda(n)} = 1 \bmod n$ (Carmichael theorem)
$$\forall m, m^k \equiv 1 \bmod n$$ so I add $k$: $$m^{\lambda(n) \cdot k} \equiv 1 \bmod n$$
In other words: $$m^{kλ(n)+1} \equiv m \bmod n$$ $$ed = k\lambda(n)+1$$ so: $$m^{ed} = m \bmod n$$ How then do I show that $ed \equiv 1 \bmod \lambda(n)$?