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How can I prove that:

$ (g^b \bmod{p})^a \bmod{p} = (g^a \bmod{p})^b \bmod{p}$

where p is a prime number, g is a primitive root of p, and a and b are integers.

While I understand that $(g^b)^a = (g^a)^b$ , I cannot figure out how to deal with the mod functions...

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1 Answer 1

You can expand $(g^b \bmod p)$ as $g^b + pK$ by the definition of $\text{mod}$.

Then $(g^b + pK)^a = g^{ab} + \binom{a}{1}g^{(a-1)b}pK + \ldots + (pK)^a$

Since all but the first of the terms in the binomial expansion contain a factor of $p$, $(g^b \bmod p)^a \equiv g^{ab} \bmod p$

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How g^b mod p = g^b + pK? – codeomnitrix Jul 25 '14 at 9:42
@codeomnitrix, $a \equiv b\pmod c$ is a shorthand notation for $\exists k : a - b = ck$ – Peter Taylor Jul 30 '14 at 9:34
Thanks Peter... – codeomnitrix Jul 30 '14 at 15:35

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