Reputation: 29
Stuck with a question for Complexity class: "You are given a,b in Z_N* and want to complete a^-1 and b^-1 mod N. By now we know how to calculate it via using the Euclidean algorithm twice, with a,N and b,N. My question is that how can one do it with a single invocation of the Euclidean Algorithm (Finding the GCD) and 3 multiplications mod N?
Upvotes: 0
Views: 53
Reputation: 32667
We know the following (all congruences mod N)
a * b * (a * b)^-1 = 1
b * (a * b)^-1 = a^-1
a * (a * b)^-1 = b^-1
So if we calculate (a * b)^-1
, the partial inverses a^-1
and b^-1
can be calculated with a multiplication.
Upvotes: 1