0
$\begingroup$

I have been looking at the Cryptography: Theory and Practice book by Stinson and Paterson and when I came to the Schnorr identification scheme, I read the sentence that goes something like this:

Observe that $v$ can be computed as $(\alpha ^a)^{-1} \bmod p$, or (more efficiently) as $\alpha ^{q-a}\bmod p$.

In this context $\alpha$ is an element having prime order $q$ in the group $\mathbb{Z}_p^*$ (where $p$ is prime and $q\mid p-1$), $a$ is a private key ($0\leq a\leq q-1$), and $v$ is a public key, constructed as $v=\alpha ^{-a} \bmod p$.

My question is, how can we get $\alpha ^{q-a}\bmod p$ from $(\alpha ^a)^{-1} \bmod p$?

$\endgroup$

1 Answer 1

1
$\begingroup$

Note that in any group, the exponent is computed modulo the order of the group. Thus $\alpha^{-a} = \alpha^{-a \bmod q} = \alpha^{q-a}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.