Want to show that, if $P$ is the original plaintext block and $(\gamma^a)'$ is the inverse of $\gamma^a$ modulo $p$, then $$(\gamma^a)'\delta \equiv P \pmod p$$
So, we have:
- $\gamma = \alpha^k \pmod p$
- $\delta = P(\alpha^a)^k mod p$
- $C = (\gamma, \delta)$
Plugging the values above to $(\gamma^a)'\delta$ $$(\gamma^a)'\delta = ((\alpha^k)^a)' P(\alpha^a)^k \equiv P \pmod p$$ $$((\alpha^k)^a)' P(\alpha^a)^k \equiv ((\alpha^k)^a)' (\alpha^a)^k \equiv 1 \pmod p$$
Not sure if I'm doing this correctly...
Hint: We want to show that if $(\gamma^a)'$ is the inverse of $\gamma^a$ modulo $p$, then $(\gamma^a)'\delta \equiv P \pmod p$.
We have:
$\gamma = \alpha^k \pmod p$
$\delta = P(\alpha^a)^k \pmod p$
$C = (\gamma, \delta)$
So, we need to find: $(\gamma^a)'\delta$
Plug in the above values and what do end up with?
$$(\gamma^a)'\delta = ((\alpha^k)^a)' P(\alpha^a)^k \equiv \alpha^{-ak} \cdot \alpha^{ak} P \equiv P \pmod p $$
What allows me to swap and cancel those values?