I suppose I already suggested that one computes a^{-1} mod b
as a^{b-1} mod b, using a plain old modexp.I realise that this will be asymptotically slower, in this setting O(n^3) vs O(n^2), but it ought have a much lower constant factor. Torbjörn _______________________________________________ gmp-devel mailing list gmp-devel@gmplib.org https://gmplib.org/mailman/listinfo/gmp-devel
