number theory - Calculating sum of geometric series (mod m)

Question

I have a series

S = i^(m) + i^(2m) + ...............  + i^(km)  (mod m)   

0 <= i < m, k may be very large (up to 100,000,000),  m <= 300000

I want to find the sum. I cannot apply the Geometric Progression (GP) formula because then result will have denominator and then I will have to find modular inverse which may not exist (if the denominator and m are not coprime).

So I made an alternate algorithm making an assumption that these powers will make a cycle of length much smaller than k (because it is a modular equation and so I would obtain something like 2,7,9,1,2,7,9,1....) and that cycle will repeat in the above series. So instead of iterating from 0 to k, I would just find the sum of numbers in a cycle and then calculate the number of cycles in the above series and multiply them. So I first found i^m (mod m) and then multiplied this number again and again taking modulo at each step until I reached the first element again.

But when I actually coded the algorithm, for some values of i, I got cycles which were of very large size. And hence took a large amount of time before terminating and hence my assumption is incorrect.

So is there any other pattern we can find out? (Basically I don't want to iterate over k.) So please give me an idea of an efficient algorithm to find the sum.

Answer 1

This is the algorithm for a similar problem I encountered

You probably know that one can calculate the power of a number in logarithmic time. You can also do so for calculating the sum of the geometric series. Since it holds that

1 + a + a^2 + ... + a^(2*n+1) = (1 + a) * (1 + (a^2) + (a^2)^2 + ... + (a^2)^n),

you can recursively calculate the geometric series on the right hand to get the result.

This way you do not need division, so you can take the remainder of the sum (and of intermediate results) modulo any number you want.