sum of legendre symbol of quadratic non-residue power

Question

What I have done is that

$$\left(\dfrac 1 p\right)+\left(\dfrac a p\right)+\left(\dfrac {a^2} p\right)+\ldots +\left(\dfrac {a^{d-1}} p\right)$$

in this, $\left(\dfrac 1 p\right)=1$, $\left(\dfrac a p\right)=-1$ by assumption. but from $2$ to $d-1$, what can I do?

I think that it will use property of the number of quadratic residue which is $\dfrac {p-1} 2$ and non residue is equal but, in this case, we consider 1~d-1 so I don't know how to apply

Answer 1

Remember that $\left({\cdot\over\cdot}\right)$ is a homomorphism from $\Bbb Z/p\Bbb Z^\times$ to the group $\{\pm 1\}$. But then the kernel has size ${p-1\over 2}$ by Lagrange's theorem. Then the other half of the elements map to $-1$. So you have exactly ${p-1\over 2}$ $+1$s and the same number of $-1$s, so they all cancel out in the sum. In your case, you are looking at a non-square generated subgroup, so you get the same division, in particular you know the kernel intersects $\langle a\rangle$ in an index $2$ subgroup, so the same thing applies to subgroups generated by non-squares.

Answer 2

Hint:
Do you know that $\left(\dfrac{ab}{p}\right)=\left(\dfrac{a}{p}\right)\left(\dfrac{b}{p}\right)$ for $a,b$ prime to $p?$ Also, can you show that $d$ is even?