Invertible elements $\mathbb{Z} / p^2 \mathbb{Z}$

Question

Let $p$ be a prime. We are curious about the invertible elements of the quotient ring $\mathbb{Z} / p^2 \mathbb{Z}$. What we do know is that according to the Euler totient function there are $\phi (p^2)=p^2-p$ such elements. We want to determine precisely those elements $a*b= 1 \mod p^2$ or equivalently $\gcd(a,p^2)=1$. How would we determine such elements if all we know is that $p$ is a prime. Why is this important?

Answer 1

It is important $p$ is a prime because that way you know $gcd(a,p^2)$ might be only $1,p$ or $p^2$. It is clear the elements in the quotient ring which are divisible by $p$ are not invertible. So $0,p,2p,3p,...,(p-1)p$ are $p$ not invertible elements. So you have only $p^2-p$ elements left in the ring, hence all the other elements must be invertible.

Answer 2

Observe that $\gcd(a,p^2)$ can have one of the three values $1,p,p^2$. Since $a <p^2$, so we can rule out $p^2$. So we just need to discard things which are multiples of $p$. So from all the $p^2$ elements we remove $0,p,2p,\ldots,(p-1)p$. The leftovers are all invertible.