Score:0

Congruence in the Schnorr identification scheme

gb flag
Jan

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 $(\alpha ^a)^{-1} \bmod p$?

Score:1
us flag

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

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