Number Theory proof involving Legendre symbol Suppose $a,b$ are integers s.t. $ab\equiv 1\pmod{p}$. Show $(\frac{a^2+a}{p})=(\frac{b+1}{p})$.
Not really sure on how to approach this proof.
If I take $a^2+a\equiv 0\pmod{p}$
$a\equiv -a^2\pmod{p}$
$b^2a\equiv -b^2a^2\pmod{p}$
$b\equiv -1\pmod{p}$ so $b+1\equiv 0\pmod{p}$
Not sure how to use this or if I'm able to use this to prove the goal.
OR I can say since $a^2+a\in Q_p$ 
$s^2\equiv a^2+a\pmod{p}$ for some $s\in U_p$ then multiply by $b^2$

Since $(b^2|p) = (b|p)^2 = 1$,  
$$\left(\frac{a^2 + a}{p}\right) = \left(\frac{a^2 + a}{p}\right)\left(\frac{b^2}{p}\right)= \left(\frac{a^2b^2 + ab^2}{p}\right) =\cdots$$
Complete the argument using the condition $ab \equiv 1\pmod{p}$.
