generating prime p and q, with alpha of element of order q I'm working on VSSS (Feldman) but I cannot understand the statement as follows:



I'm not good in reading mathematical statements (but working on it). Can anyone explain this in simple mathematics. Example will be appreciated. Thank you.

Edit:  

This is what I got so far.
Given the following values: n=4, k=3, s=12, p=5, q=3, g=4
wherein n is no. of shares, k is threshold, s is the secret, p and q are prime, g is the generator  

Using the polynomial function f(x) = 12 + 3x +2x^2.
then, f(2) = 68 mod q = 68 mod 3 = 2   

Computing on commitment:
c1 = g^12 mod p = 4^12 mod 5 = 1
c2 = g^3 mod p = 4^3 mod 5 = 4
c3 = g^2 mod p = 4^2 mod 5 = 1
wherein, cx are commitments and 12,3 &amp; 2 are coef
Verifying:  

g^f(2) mod p = 4^2 mod 5 = 1  

c1^x^0 mod p = 1^2^0 mod 5 = 1
c2^x^1 mod p = 4^2^1 mod 5 = 4
c3^x^3 mod p = 1^2^2 mod 5 = 1
where x is the value of x in f(x) which is 2.

1+4+1 &lt;> 1 ????

What's wrong??? Am I doing it right?
There are generated the primes $p$ and $q$ such that $q | (p-1)$


What that means is that we search for two prime numbers $p$ and $q$, with the constraint that $q$ is a divisor of $p-1$.  That is, $kq = p-1$ for some integer $k$.

In practice, this is easy to do; we first pick a prime $q$ of an appropriate size, and then search for the value $k$ values of the appropriate size until we find a value $p = kq + 1$ which is also prime.


  and $\alpha \in \mathbb{Z}_p^*$ ...


$\mathbb{Z}_p^*$ is the multiplicative group modulo $p$; its elements have an obvious identification to the between 1 and p that are relatively prime to $p$ (and since $p$ is prime, that means the values $[1, p-1]$), and


  an element of order $q$.


What this means is that the smallest positive value $\lambda$ for which $\alpha^\lambda \equiv 1 \pmod p$ is $\lambda = q$.

It turns out to be a lot easier to find such a value $\alpha$ than you'd expect; you pick an arbitrary value $\beta \in [2, p-2]$ and set $\alpha = \beta ^ {(p-1)/q} \pmod p$.  As long as $\beta \ne 1$, then $\alpha$ will have an order precisely $q$.


We know that $\alpha^q = (\beta ^ {(p-1)/q})^q = \beta ^ {p-1} = 1$ by Fermat's Little Theorem (and because $p$ is prime)
We know that $\alpha^r \ne 1$ for any $0 &lt; r &lt; q$, because if $\alpha^r = 1$, then we could show that $\alpha^{\gcd(q, r)} = 1$, and as $q$ is prime, $\gcd(q, r) = 1$, and so we'd have $\alpha^1 = 1$, which we had verified is not the case