[Solved] Modulo Arithmetic in Modified Geometric Progression

Question

Note that if r is a primitive root modulo p.
Then we can reduce complexity of the sum.

We have to find S = a1*1 + a1*r + a1*r^2 + ... + a1*r^n. Then we write S in the closed form as S = a1*(r^n - 1) / (r - 1).

Now it can be reduced to:

 a1*(r^n - 1) / (r - 1) = S (mod p)
=> a1*r^n = S * (r - 1) + 1 (mod p)

Now take discrete logarithm with base r both sides,

   log(a1*r^n) = log_r(S*(r-1) + 1 (mod p))
   =>log_r(a1) + n*log_r(r) = log_r(S*(r-1) + 1 (mod p))
   =>n*log_r(r) = log_r(S*(r-1) + 1 (mod p)) - log_r(a1) (mod(p-1))
   =>n*1 = log_r(S*(r-1) + 1 (mod (p-1))) - log_r(a1) (mod (p-1))

Note that if a1 is 1 then the last term is 0.

Let S = 6, r = 3, and m = 7, a1 = 1.
Then, we want to solve for n in the following congruence:

     (3^n - 1)/(3 - 1) = 6 (mod 7)
=> 3^n - 1 = (3 - 1) * 6 (mod 7)
=> 3^n = 2 * 6 + 1 (mod 7)
=> 3^n = 6 (mod 7)

Then we take the discrete logarithm of both sides:

     log_3(3^n) = log_3(6) (mod (7-1))
=> n * log_3(3) = log_3(6) (mod 6)
=> n * 1 = 3 (mod 6)
=> n = 3 (mod 6)

So, n = 3.

You can use Baby-step Giant-step algorithm to solve this in O(sqrt(m)).
If you want implementation in code I will provide you.

Accepted Answer