Show that if $\displaystyle p$ is prime and $\displaystyle m \equiv n(mod p-1)$, then for any integer

$\displaystyle a$:

$\displaystyle a^{m} \equiv a^{n} (mod p)$

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- Oct 18th 2012, 09:05 AMmaximus101Number theory question 3-3
Show that if $\displaystyle p$ is prime and $\displaystyle m \equiv n(mod p-1)$, then for any integer

$\displaystyle a$:

$\displaystyle a^{m} \equiv a^{n} (mod p)$ - Oct 26th 2012, 02:55 AMSalahuddin559Re: Number theory question 3-3
Use fermat's little theorem. Simply put, if m == n (mod (p - 1)), m - n = k(p - 1), for some integer k. Now, use both sides as powers of a, some number a. a^(k(p - 1)) can be written as (a^(p - 1))^k, where p is a prime.

Salahuddin

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