Recent content by Lexaila

Thank you, it solves that question! But if I have a different example, such as ( (p+3)/2 /p) = (2/p)(3/p) and when I try to use ##\left(\dfrac{2}{p}\right)=(-1)^{\frac{p^2-1}8}##, ##\left(\dfrac{3}{p}\right)=\left(\dfrac{p}{3}\right)(-1)^{\frac{p-1}2}(-1)^{\frac{3-1}2}## for, for example, p...

I'm still a bit confused, could you please give me an example from the table?

Thank you for your reply! In our table we must also use QRL+- in between (2/p) and (3/p). I understand how to fill the rest of the table with 1 and -1, but not this QRL+- column and how it determines the final result of (p-6/p)...

How do we make the table to show that p-6 is a quadratic residue modulo p if p \equiv 1,5,7,11 (mod 24)?

(p-6/p)=(-1/p)(2/p)(3/p) Make a table, so at the head row you have p(mod24), (-1/p), (2/p), QRL+-, (p/3) and finally (p-6/p), with in the head column below p (mod 24): 1,5,7,11