$p$ an odd prime and $a,b\in\Bbb Z$ such that $ab\equiv 1\mod p$. Show that legendre symbols $({a\over p}) = ({b\over p})$

Question

Need to prove $ab\equiv 1 \mod p \implies a\equiv b\mod p$

I know that the first statement implies that $b$ is the inverse of $a$ modulo $p$ but can't see how that would prove the second statement.

I also know that from the first statement: $({ab\over p}) = ({1\over p}) = ({a\over p})\cdot( {b\over p})$ but again, stuck with what to do next...

Answer 1

The Legendre symbol is multiplicative and takes values in $\{0, \pm 1\}$, hence neither of $\displaystyle\left({a\over p}\right),\left({b\over p}\right)$ is $0$ (since they multiply to $1$). And if exactly one of them is $-1$ the product is $-1$, so they must be equal.