RSA Arithmetic

Greatest Common Divisor of a and b:

gcd(a,b) {
  if(b==0){
    return a;
  }
  else{
    return gcd(b,a % b);
   }

The Time Calculation:

We are given integers m and n with m > n >0 . Then m = kn+r with n > r ≥0. In particular m ≥ n+r .

Performing the same calculation with n and r we have n = k'r+r' with r > r' ≥0. In particular n ≥ r+r'

So

       (m+n) ≥ (n+r+n)+r ≥ 2(r+r')+r ≥ 2(r+r')

Renaming m , n , r and r' as m₀ , n₀ , m₁ , n₁

       (m₀+n₀) ≥ 2¹(m₁+n₁)

Repeating :

       (m₀+n₀) ≥ 2^k(m_k+n_k)

Finally if,
       2^k ≥ (m₀+n₀)

We know we can find the gcd with 2k divisions.

An Example:

p=7 and q=13 so pq=91

p-1=6=2¹3¹ and q-1=12=2²3¹

Choose e=5
 
We are looking for d such that de=1 mod((p-1)(q-1))=1 mod(72).

The Steps:

• 72 -14*5 = 2
   
•  5 - 2*2=1
   
•  5 - 2*(72-14*5)=1
   
• (-2)*72 +29*5 = 1
   
• d=29
   

The Test:

3^5*29 = 3¹⁴⁵
 
= 1522586358169246802159262479225089070726226750574991661790882326344643
 
=3 mod(91)