If $g \in \Bbb Z_{n^2}^{*}$ and $x_1,x_2 \in \Bbb Z_n$ then help me in proving the following implication.
$g^{n \lambda(n)}\equiv 1 \mod{n^2} \implies g^{(x_1-x_2)\lambda(n)} \equiv 1 \mod{n^2}$
where $\lambda(n)$ is carmichael function.
I know how to prove the left side of the above implication but donno about r.h.s.
You can see it in page #5
In the text there are additional conditions/restrictions on the $x_i,\;$ i.e. you have $r_1, r_2 \in \mathbb{Z}_n^{*}\;$ with $$g^{x_1} r_1^n \equiv g^{x_2}r_2^n \pmod {n^2}.$$ Multiply both sides with $g^{-x_2},\;$ which exists because $g$ is invertible $$g^{x_1-x_2} r_1^n \equiv r_2^n \pmod {n^2}$$ take powers with ${\lambda(n)}$ $$\left(g^{x_1-x_2} r_1^n\right)^{\lambda(n)} \equiv r_2^{n\lambda(n)} \pmod {n^2}$$ $$g^{(x_1-x_2)\lambda(n)} r_1^{n\lambda(n)} \equiv r_2^{n\lambda(n)} \pmod {n^2}$$ and now use Lemma 14 to get rid of the $r_i\;$ terms $$g^{(x_1-x_2)\lambda(n)} \equiv 1 \pmod {n^2}$$
