# Thread: Congruence and Euler's function

1. ## Congruence and Euler's function

Please help on following:

Show if $m>1$ then $a^m \equiv a^{m-\phi(m)}(mod \ m)$ for all natural numbers $a$.

Basic Euler function is this $a^{\phi(m)} \equiv 1 (mod \ m)$ but how to build it up in order to answer the above question?

Thank you

2. $a^{\phi(m)} \equiv 1\implies a^{-\phi(m)}\equiv 1$. Now multiply both sides by $a^m$. (Here negative exponents means inverses mod $m$).

3. I get it, thank you.
Forgot about inverse.