How to simplify the multiplication of mod inverses

Question

Given: $(x^k$ $mod$ $l$$)^{-1}$ $*$ $(x^k$ $mod$ $l$$)^{-1}$, how can the two be combined? For example, could the two $x^k$ terms be combined in some way?

Answer 1

This is a case of simple modular arithmetic.
It is true that $(a \mod c) \cdot (b \mod c) = ab \mod c.$

Proof

We can use the division algorithm to write $a,b$ in terms of $c$.

$a = c\cdot q_1 + r_1, q_1,r_1 \in \mathbb{Z}$

$b = c\cdot q_2 + r_2, q_1,r_2 \in \mathbb{Z}$

Clearly, $(a \mod c) = r_1$ and $(b \mod c) = r_2$. So $(a \mod c) \cdot (b \mod c) = r_1r_2$.

Also $(ab \mod c) = ((c^2q_1q_2 + cq_1r_2 + c q_2 r_1+ r_1 r_2) \mod c)$

$\implies (c(cq_1q_2 + q_1r_2+q_2r_1) + r_1r_2) \mod c= r_1r_2$

Now since these two are equal, we can conclude that $(a \mod c) \cdot (b \mod c) = ab \mod c.$

I'll let you continue on from here as you see fit.