Question

a Mer sehne Mer Senne pr me otler then3

Answer 1

is not a quadratic residue modulo as there exists no solution for the equation

as

r 3 mod 3

12 13 mod 3 22-43 mod 3

So we consider the case when

p2

Now $\frac{q-1}{2}=\frac{2^p-2}{2}=2^{p-1}-1$

Also by Law of Quadratic Reciporocity

3 (q-1)3-1

Hence

(p-1)(3-1 3

Since is odd so

2p = 22H 142-1

Now

$4\equiv 1\mod 3\\ \implies 4^r.2\equiv 2 \mod 3$

Thus

$q= 4^r.2-1\equiv 1 \mod 3$

Thus q is a quadratic residue modulo 3.

Hence

$(\frac{q}{3})=1\\$

Using (*) we have

$(\frac{q}{3})=-1$

Thus is a quadratic non-residue modulo