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Jan 3, 2015 at 17:41 comment added poncho @usry: no, what I wrote works just fine; just because you compute $a^j$ and $B \cdot a^{-im}$ modulo $p$ doesn't mess anything up.
Jan 3, 2015 at 15:21 comment added usry Thank you.But B is again in mod p I think same issue still exists there.
Jan 3, 2015 at 15:16 history answered poncho CC BY-SA 3.0