Proof of the theorem

Question

I have been unsuccessful in trying to find a proof of this theorem the name of which, if existing, I don't know.

For any prime number $p$ and natural number $A$, where $p$ doesn't divide $A$, the remainder of $A^{p - 1}$ when divided by $p$ is $1$.

Any suggestions would be helpful. Thanks!

Answer 1

Not sure it would have helped you find it, but mathematicians writing in English about Fermat's "little" theorem generally prefer to use $a$ rather than $A$.

Also, you're missing the condition that $\gcd(a, p) = 1$. For example, if $a = 14$ and $p = 7$, we then have $14^6 = 7529536 \equiv 0$, not $1 \pmod 7$. If instead we try $a = 12$, which is coprime to $7$, we have $12^6 = 2985984 \equiv 1 \pmod 7$, just like the theorem says.

So this would be a preferable way of stating the theorem:

For any prime number $p$ and number $a$ coprime to $p$, the remainder of $a^{p - 1}$ when divided by $p$ is $1$.

One more thing: Fermat's "little" theorem is more generally useful than his "last" theorem, so some just say "Fermat's theorem" to refer to the "little" theorem.