Calculate d from n, e, p, q in RSA?

Question

Not sure if this is the correct place to ask a cryptography question, but here goes.

I am trying to work out "d" in RSA, I have worked out p, q, e, n and ø(n);

p = 79, q = 113, e = 2621

n = pq                   ø(n) = (p-1)(q-1)
n = 79 x 113 = 8927      ø(n) = 78 x 112 = 8736

e = 2621
d = ???

I cant seem to find d, I know that d is meant to be a value that.. ed mod ø(n) = 1. Any help will be appreciated

As an example would be e = 17, d = 2753, ø(n) = 3120

17 * 2753 mod 3120 = 1

Answer 1

You are looking for the modular inverse of e (mod n), which can be computed using the extended Euclidean algorithm:

function inverse(x, m)
    a, b, u := 0, m, 1
    while x > 0
        q := b // x # integer division
        x, a, b, u := b % x, u, x, a - q * u
    if b == 1 return a % m
    error "must be coprime"

Thus, in your examples, inverse(17, 3120) = 2753 and inverse(2621, 8736) = 4373. If you don't want to implement the algorithm, you can ask Wolfram|Alpha for the answer.

Answer 2

The algorithm you need is the Extended Euclidean Algorithm. This allows you to compute the coefficients of Bézout's identity which states that for any two non-zero integers a and b, there exist integers x and y such that:

ax + by = gcd(a,b)

This might not seem immediately useful, however we know that e and φ(n) are coprime, gcd(e,φ(n)) = 1. So the algorithm gives us x and y such that:

ex + φ(n)y = gcd(e,φ(n))
           = 1
Re-arrange:
ex = -φ(n)y + 1

This is equivalent to saying ex mod φ(n) = 1, so x = d.

Answer 3

For example you need to get d in the next:
3*d = 1 (mod 9167368)

this is equally:
3*d = 1 + k * 9167368, where k = 1, 2, 3, ...

rewrite it:
d = (1 + k * 9167368)/3

Your d must be the integer with the lowest k.
Let's write the formula:
d = (1 + k * fi)/e

public static int MultiplicativeInverse(int e, int fi) {
    double result;
    int k = 1;
    while (true) {
        result = (1 + (k * fi)) / (double) e;
        if ((Math.Round(result, 5) % 1) == 0) {
            //integer 
            return (int)result;
        } else {
            k++;
        }
    }
} 

let's test this code:

Assert.AreEqual(Helper.MultiplicativeInverse(3, 9167368), 6111579); // passed