Question regarding an IMO number theory solution

Question

I am reading this Art of Problem Solving solution of an IMO question. I do not understand this statement "There exists no numbers $1<m_1,m_2\leq p-2$ such that $m_1m_2\equiv1\text{ (mod }p^2\text{)}$ as $(p-2)(p-2)<p^2$ and $2^2>1$. Therefore, at most half of the values where $a_0^{p-1}\equiv 1\text{ (mod }p^2\text{)}$ are in range $1<a_0\leq p-2$. " where $p\ge5$ is a prime number. I understand the first sentence, but fail to see the second follows from that. Could someone please give an explicit derivation?

Answer 1

Suppose we have $1 < m_1, m_2 leq p - 2$. Then we have $1 < m_1 m_2 \leq (p - 2)^2$.

Now since $p \geq 2$, we have $(p - 2)^2 = p^2 - 4p + 4 \leq p^2 - 4 < p^2$. Thus, we have $1 < 4 \leq m_1 m_2 \leq (p - 2)^2 < p^2$. That is, $1 < m_1 m_2 < p^2$.

Now suppose $m_1 m_2 \equiv 1 \mod p^2$. Then there exists some integer $j$ with $m_1 m_2 = 1 + j p^2$. Since $m_1 m_2 > 1$, we must have $j > 0$. Then $j \geq 1$. But then we would have $1 + p^2 \geq p^2 > m_1 m_2 = 1 + j p^2 \geq 1 + p^2$. This is a contradiction.

I don’t see how the second half of the statement follows from the first half. However, since $p$ is odd, if $a_0^{p -1} \equiv 1 \mod p^2$ then $(p^2 - a_0)^{p - 1} \equiv 1 \mod p^2$. This is what is really needed to show that at most half the solutions $0 \leq a_0 < p^2$ can satisfy $1 < a_0 \leq p - 2$, since each such solution corresponds to another $b_0 = p^2 - a_0$ such that $p^2 - p + 2 \leq b_0 < p^2 - 1$ and the two ranges don’t overlap.