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If you can select the distinct secret primes $p$ and $q$ such that $(p-1)/2$ and $(q-1)/2$ are also prime, then it becomes easy. For a random value $r$, $g = r^2$ will have order precisely $(p-1)(q-1)/4 \approx n/4$ unless $r, r-1$ or $r+1$ happens to not be relatively prime to $n$ (which, if you select $r$ randomly in the range $[2, n-2]$, happens with ...

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