This question is one of the unsolved problems in my Encryption and Cipher Systems subject:
For prime $p>2$ and $0<a<p\,;$
$ a^{(p-1)/2}(mod \,p)= \begin{cases} 1;\ \ \ \ if \ \ \ \, a\in R_{2}\\ p-1;\ \ \ \ \text{otherwise}\\ \end{cases} $
where $R_{2}$ denotes the set of quadratic residues modulo $p$ that is, there exists $x$ such that : $x^{2} \equiv a \, mod \,p$
I really don't understand what it does mean by the term “Numerically “ also i don’t know how and where to start to write a proof of such problem, So i haven’t tried anything. Any help will be appreciated and thanks a lot
Well, let's make an abstract algebric approach. Let A the group quotient with prime $p$. And let $a$ be element of A/{0,1} As U(p) is generator of A, we know: $$a^{p-1}=1=e$$ where e is element neutre multiplicatif. with $p-1$ is the smallest number which satistifies that equation. Which also means: for all integer $a>1$ and $p$ prime we have:
$(1)$ $$a^{p-1}=1\left(mod\:p\right)$$
if $a∈R_2$ (1) holds. For second case let's assume that there is $a$ integer $m<p$ such that:
$(2)$ $$a^{^{\frac{\left(p-1\right)}{2}}}=m\left(mod\:p\right)$$
Square both sides: $$a^{\left(p-1\right)}=m^2\left(mod\:p\right)$$ By (1) we have $m^2=1$ so $m=1$ or $m=-1$ By (1) $m$ cannot be $1$ because $p-1$ is smallest number which satisfies equation. If $m=1$ we get (2) satisfies same equation with $\frac{\left(p-1\right)}{2}$.
Finally we get $m=-1$