Show that if $\displaystyle a$ is a quadratic residue $\displaystyle (\mod{p})$ and $\displaystyle ab \equiv 1 (\mod{p})$ then $\displaystyle b$ is a quadratic residue $\displaystyle (\mod{p})$.

I am not sure where to go with this proof, I started with:
$\displaystyle x^2 \equiv a (\mod{p})$ because $\displaystyle a$ is a quadratic residue.
$\displaystyle x^2b \equiv ab (\mod{p})$
$\displaystyle x^2b \equiv 1 (\mod{p})$

Stumped right here.

Then I thought maybe using Legendre's symbols would be better.
$\displaystyle (a/p) = 1$ because $\displaystyle a$ is a quadratic residue

So I have two starting points and no finish line in sight.

Thanks in advance for any help.

But not sure where to go to show that $\displaystyle (b/p) = 1$

You do know the multiplicative property of the Legendre symbol, right? (That $\displaystyle (a/p)(b/p)=(ab/p)$).

Also, $\displaystyle (1/p)=1$.

$\displaystyle (a/p) = 1$
$\displaystyle (a/p)*(b/p) = 1*(b/p)$
$\displaystyle (ab/p) = (b/p)$
$\displaystyle (1/p) = (b/p)$
$\displaystyle 1 = (b/p)$