@e-sushi Nice one: "seven of the ten most common vulnerabilities seen across federal agency networks at the issuance of this directive would be addressed through complying with the required actions in this directive related to web security" xD

@Lery “Governmental InfoSec” ;)

In case anyone is interested in grabbing a micro-bounty…

@Evinda surely you can ask, whether I will be able to help is a different question though ;)

I haven't understood the proof from the part: The $\frac{1}{2}(p-1)$ elements.. till the end.

Why do we pick i to belong in the interval [0,p-1) ? Do you have an idea? @SEJPM

@Evinda because you want to talk about all squares

and all squares have the form $g^{2i}=(g^i)^2$

and if you were to go bigger than $1/2(p-1)$ you'd repeat

How do we get that the number of the $g^{2i}$s, $0 \leq i<p-1$ is $\frac{1}{2}(p-1)$ ? @SEJPM

@Evinda because $g^k$ generates $p-1$ elements and exactly half of them are captured by introducing the $2$

(note also that you are doing math in $\mathbb Z_{p-1}^*$ with $i$ and $k$ with $p-1$ even

First of all do we know that when we have $g^{2i}$ that this will be a quadratic residue modulo p, and if we consider $g^{2i+1}$ that it will be not? @SEJPM

@Evinda well, $x$ is a quadratic residue if there is some $y$ such that $y^2=x\bmod p$, right?

so if we pick $g^{2i}$ we know it will be a quadratic residue because $(g^{i})^2=g^{2i}$

Yes, and if we consider $g^{2i+1}=(g^{i})^2 g$ this will be not a quadtratic residue, since g cannot be written as $g^{2k}$, for some k, right? @SEJPM

@Evinda 2i+1 is odd and thus you can't find any (integer) k such that 2k=2i+1 modulo an even number (p-1)

I see... And we see that for $i \in [\frac{1}{2}(p-1),p)$ and for $j=i-\frac{p-1}{2} \in [1, \frac{p-1}{2})$, it holds that $g^{2i}=g^{2j}$, so there are in total $\frac{p-1}{2}$ quadratic residues, right? @SEJPM

yes

I see, thank you :) @SEJPM

:)