I am trying to show the following: for $p$ a prime and $p\nmid b$, $$\sum_{a=1}^{p-1}\Bigg(\frac{a(a+b)}{p}\Bigg)=-1$$
My attempt: inspired by other questions on Stack Exchange that are related to the evaluation of the sum $\sum_{a=1}^{p-1}\Bigg(\frac{(a(a+1)}{p}\Bigg)$, I use the fact that each $a$, $1\leq a \leq p-1$ has a unique multiplicative inverse $(mod p)$, so $$\sum_{a=1}^{p-1}\Bigg(\frac{a(a+b)}{p}\Bigg)=\sum_{a=1}^{p-1}\Bigg(\frac{(a+b)/a}{p}\Bigg)=\sum_{a=1}^{p-1}\Bigg(\frac{1+ba^{-1}}{p}\Bigg)$$ but summing over the inverses is the same as summing over just the $a$'s, so this is just $$\sum_{a=1}^{p-1}\Bigg(\frac{a(a+b)}{p}\Bigg)=\sum_{a=1}^{p-1}\Bigg(\frac{1+ba}{p}\Bigg)$$ From here I am unable to proceed. I know I probably at some point have to use the fact that $\sum_{a=1}^{p-1}\Bigg(\frac{a}{p}\Bigg)=0$. I tried re-indexing and other stuff but to no success. Any help is appreciated.
You have a sum $$S=\sum_{a=1}^{p-1}\left(\frac{1+ab}p\right).$$ This is $$-\left(\frac1p\right)+\sum_{a=0}^{p-1}\left(\frac{1+ab}p\right).$$ As $a$ runs through the numbers from $0$ to $p-1$, $1+ab$ runs through all the numbers from $0$ to $p-1$ modulo $p$, so $$S=-1+\sum_{c=0}^{p-1}\left(\frac cp\right)=-1+0=0.$$