Tag Info

Once you have a generator $g$ of a cyclic group $G$ of order $n$, all the other generators are given by $g^a$ where $\gcd(a,n)=1$. So you pick a random element, test it: $$\forall p|n,\qquad g^{n/p}\neq e,$$ and as soon as you find a generator, you find them all. Since the order of $(\mathbb{Z}/251\mathbb{Z})^*$ is $250=2\cdot 5^3$, any element $a\in ... 1 Yes, there are better approaches. A better approach is to verify that$g^{n-1} \equiv 1 \mod n$(which is always true) and$g^{\frac{n-1}{d}} \not \equiv 1 \mod n$for the factors$d$of$n-1$. This can be optimized a little bit. See this answer by André Nicolas for a bit more there. Once you've found one, you can get all the others "for free." In ... 1 if i understand you right, that is not possible. if you take a=3, b=5, c=7, it is the same thing as if you take a=4,b=4, c=7 that means, if you are able to get c if you know the real$a_1$and$b_1$, you can get the same c if you use "wrong" a,b, lets say$a_2,b_2$such that$a_1+b_1 = a_2 b_2\$