Polig-helman algorithm - if $g$ is a prim root mod $p$ does this mean that in the subproblems that all $A_i$ are prim roots mod $p$?

116 Views Asked by At

I am using the Pohlig-Hellman algorithm to solve a Discrete Log Problem to find $x$ in

$$g^x = h \pmod{p}$$

Where $g, h$ are positive integers and p is prime.

Now presume g is a primitive root mod p, and p is $\beta-smooth$ (p-1 is easily factorised into product of small prime powers).

$N = p - 1 = \prod_{i=1}^{n} q_i^{e_i}$

We can summarize the necessary algorithm calculations in a handy table as

$$\begin{array}{|c|c|c|c|c|} \hline \large q & \large e & \large g^{(p-1)/q^e} mod p & \large h^{(p-1)/q^e} mod p & \mbox{Solve}~ \large \left(g^{(p-1)/q^e} \right)^x = ~ \large h^{(p-1)/q^e}~ \mbox{for} ~ \large x \\ \hline q_1 & e_1 & A_1 & C_1 & \mbox{...}\\ \hline q_2 & e_2 & A_2 & C_2 & \mbox{...}\\ \hline \end{array}$$ ...

Questions:

Does $g$ being a primitive root mod p imply that all $A_i$ are primitive roots mod $p$? Why? If there is such $g$ and $A_i$ s.t. $A_i$ is not a primitive root mod $p$, give an example of this.

1

There are 1 best solutions below

4
On BEST ANSWER

If I understand what's going on in the table, $(A_i)^{q_i^{e_i}}=g^{p-1}=1$, so $A_i$ is not a primitive root modulo $p$.