Let $\chi$ be a multiplicative non trivial character of $F_p$ and $\rho$ be a character of order 2. Show that $\sum\chi(1-t^2)=J(\chi,\rho).$
[Hint: Evaluate $J(\chi,\rho)$ using the relation $N(x^2=a)=1+\rho(a)$]
I have read Rosen on my own, and I have understood characters and Jacobi sums, but can't do this question. Please give me some hints.
Note that $\rho$ is not just any character of order 2, but the unique character of order $2$, i.e. $\rho(a) = \left( \frac{a}{p}\right)$. Then, as in the hint, we have $N(x^2=a) = 1 + \rho(a)$, because $x^2=a$ either has 0 solutions, or 2 distinct solutions modulo $p$, and so in either case exactly $1 + \left(\frac{a}{p}\right)$ solutions.
So we have $$J(\chi,\rho)=\sum_{a \not\equiv 0,1} \chi(1-a) \rho(a) = \sum_a \chi(1-a) (N(x^2=a)-1) = \sum_a \chi(1-a)N(x^2=a) - \sum_a \chi(1-a).$$
Now since $\chi$ is a non-trivial character, its sum over ${1,2,...,p-1}$ is 1, therefore its sum over $\{2,...,p-1\}$, i.e. the second term above, is equal to $0$.
So now $$J(\chi,\rho)=\sum_{a \not\equiv 0,1} \chi(1-a)N(x^2=a) = \sum_{a \in S} 2\chi(1-a),$$
where $S$ is the set of square residues mod $p$ except 0 or 1, i.e. $S = \{ b^2: 1< b< p\}$. Noting that each square residue $b^2\in S$ occurs twice as $b$ ranges over $\{1<b<p\}$ we can write
$$ \sum_{a\in S} 2\chi(1-a) = \sum_{1 < b < p} \chi(1-b^2).$$