Calculating pow(a,b) mod n

I want to calculate a^bmod n for use in RSA decryption. My code (below) returns incorrect answers. What is wrong with it?

unsigned long int decrypt2(int a,int b,int n)
{
    unsigned long int res = 1;

    for (int i = 0; i < (b / 2); i++)
    {
        res *= ((a * a) % n);
        res %= n;
    }

    if (b % n == 1)
        res *=a;

    res %=n;
    return res;
}

Solution 1:

You can try this C++ code. I've used it with 32 and 64-bit integers. I'm sure I got this from SO.

template <typename T>
T modpow(T base, T exp, T modulus) {
  base %= modulus;
  T result = 1;
  while (exp > 0) {
    if (exp & 1) result = (result * base) % modulus;
    base = (base * base) % modulus;
    exp >>= 1;
  }
  return result;
}

You can find this algorithm and related discussion in the literature on p. 244 of

Schneier, Bruce (1996). Applied Cryptography: Protocols, Algorithms, and Source Code in C, Second Edition (2nd ed.). Wiley. ISBN 978-0-471-11709-4.

Note that the multiplications result * base and base * base are subject to overflow in this simplified version. If the modulus is more than half the width of T (i.e. more than the square root of the maximum T value), then one should use a suitable modular multiplication algorithm instead - see the answers to Ways to do modulo multiplication with primitive types.

Solution 2:

In order to calculate pow(a,b) % n to be used for RSA decryption, the best algorithm I came across is Primality Testing ¹⁾ which is as follows:

 int modulo(int a, int b, int n){
    long long x=1, y=a; 
    while (b > 0) {
        if (b%2 == 1) {
            x = (x*y) % n; // multiplying with base
        }
        y = (y*y) % n; // squaring the base
        b /= 2;
    }
    return x % n;
}

See below reference for more details.

¹⁾Primality Testing : Non-deterministic Algorithms – topcoder

Solution 3:

Usually it's something like this:

while (b)
{
    if (b % 2) { res = (res * a) % n; }

    a = (a * a) % n;
    b /= 2;
}

return res;