If $ab \equiv 1 \pmod{p}$, is $a \equiv b \pmod{p}$? If $ab \equiv 1 \pmod{p}$, is $a \equiv b \pmod{p}$?
I can see that $b$ is an inverse of $a$ modulo p. But what property does the inverse of $a$ has but $a\bar{a} \equiv 1 \pmod{p}$?
Thank you
 A: No, $a$ is congruent to its inverse mod $p$ iff $a\equiv \pm 1 \bmod p$.
A: Consider a set $P = \{1, 2, 3, ...., p-1\}$; $|P|=p-1, p > 2$ is prime.
Apart from $1$ and $p-1$, where
$$1^2 ≡ (p-1)^2 ≡ 1  \pmod p,$$ 
consider further two positive integers $m$ and $n$ such that neither are multiples of $p$. Therefore each are congruent modulo $p$ to two of the $p-3$ members of $P$. 
Now, $mn ≡ 1 \pmod p$ implies that $mn - 1 = kp$ for some positive integer $k$.
This we can rewrite as 
$$mn - kp = 1$$
The equation
$rx + sy = k (k, r, s, x, y ∈ ℕ)$
has solutions for $x$ and $y$ $\iff \gcd(x, y)$ divides $k$. In which case there are $\frac{k}{\gcd(x, y)}$ solutions.
Since $\gcd(m,p) = \gcd(n, p) = 1$ (and of course $1|1$), the above equation has a unique solution for the positive integers $m$ and $n$.
Now the only two members of $P$ which satisfy
$$mn ≡ 1 \pmod p ⇒ m ≡ n \pmod p$$
are
$$m, n ≡ 1 \pmod p \text{ and } m, n ≡ p-1 \pmod p$$
It follows that there are $\frac{p-3}{2}$ pairs of integers $m, n$ such that $$mn ≡ 1 \pmod p \text{ where }
m ≢ n \pmod p$$
