Show that if $a^h ≡ 1\mod p$ then $ a^{ph} ≡ 1 \ \mod p^2$.

Question

I don't know how to proceed. I know that regardless of what h is, it divides the order of a modulo $p$. I also know that the order of a divides $\phi(p) \ \text{mod} \ p$, where $\phi$ is Euler's totient function. Hence, by the transitive property, $h$ divides $\phi(p)\ \text{mod} \ p$. But I don't know where to go from here...or if I'm even going in the right direction.

h is not necessarily the order of $a$, right? (Because we don't know if it's the least $h$ such that $a^h$ is congruent to $1$ modulo $p$).

Help please.

Answer 1

Hint: We have $a^h=1+kp$ for some $k$. Take the $p$-th power, using the Binomial Theorem.

Answer 2

Note that $$ a^h = 1+kp $$ so $$ a^{ph} = (1+kp)^p = 1 + kp^2 + \sum_{j=2}^p {p\choose j}k^jp^j $$ and $p^2$ divides every term other than the first one.