Legendre Symbol sum

Question

Let $p \equiv 3 \pmod 8$. Show:

$\sum_{a=1}^{p-1} a\Big(\frac{a}{b}\Big)=\sum_{a=1}^{(p-1)/2} (2a-p)\Big(\frac{a}{b}\Big)$

and $\sum_{a=1}^{p-1} a\Big(\frac{a}{b}\Big)=\sum_{a=1}^{(p-1)/2} (p-4a)\Big(\frac{a}{b}\Big)$

I manged to prove the first identity by making use of the fact that p is congruent to 3 (mod 4) but I have to idea for the second one.

Answer 1

Try to reduce it to first identity by changing summation index to $a' = 2a$.