Talk:Equivalence of Definitions of Legendre Symbol

Definition 2

Could we possibly write Definition 2 as :

$\left({\dfrac a p}\right) \equiv a^{\frac{(p - 1)} 2} \mod p$, with $\left({\dfrac a p}\right) \in \set {-1,0,1}$

This is from a quick Wikipedia lookup. --Anghel (talk) 22:36, 16 October 2022 (UTC)

What do you mean? $a^{\frac{(p - 1)} 2} \mod p$ is never $-1$ according to Definition of modulo operation. --Usagiop (talk) 23:11, 16 October 2022 (UTC)

As usual, it would greatly help if you put the mistake template on the origin of where the actual mistake is. --prime mover (talk) 05:01, 17 October 2022 (UTC)

The page is now open for an equivalence proof to be written. --prime mover (talk) 05:29, 17 October 2022 (UTC)

'$\equiv$' is the equivalence sign, so it means $\left({\dfrac a p}\right)$ and $a^{\frac{(p - 1)} 2}$ are equivalent modulo $p$. Maybe it is better written like this:

$\left({\dfrac a p}\right) \equiv a^{\frac{(p - 1)} 2} \pmod p$, with $\left({\dfrac a p}\right) \in \set {-1,0,1}$

Sorry, you were right. --Usagiop (talk) 15:51, 17 October 2022 (UTC)

Anyway, it looks like the problem has been solved. --Anghel (talk) 15:24, 17 October 2022 (UTC)