Proof on Legendre symbol

Question

Given Legendre symbol $(\frac {y}{p})$. Prove that $(\frac {y}{p})=(\frac{y^{-1}}{p})$

Help me prove this also. $\sum_{y=0}^{p-1} (\frac {y}{p}) = \sum_{y=0}^{p-1} (\frac {y+d}{p})$

Answer 1

Note that if $x$ is a square mod $p$ then $x^{-1}$ is a square, as $(a^2)^{-1} = (a^{-1})^2$. Since $(a ^ {-1}) ^ {-1} = a$, we conclude that $x$ is a square mod $p$ if and only if $x^{-1}$ is a square mod $p$. This answers your first question.

For your second question, note that both sums have exactly the same summands, but with different orders.

Answer 2

Hint:

1. Use that $\left (\frac{a}{p}\right)\left (\frac{b}{p}\right)=\left (\frac{ab}{p}\right)$. What can you multiply $y^{-1}$ with to get $y$? Also what is $\left (\frac{a^2}{p}\right)$?
2. Is assume that $y\in\mathbb{F}_p$, what happens with $y+d$ as $d$ runs over all the values in $\mathbb{F}_p$?