m kn n 0k − ∑ =

This condition implies that
log2(n)/2 - 15 < log2(p), log2(q) < log2(n) / 2 + 15
The generation of an RSA modulus of exactly k bits could be done with the following
algorithm:
• Choose a random prime number p in the range ]2k/2-15, 2k/2+15[;
• Choose a random prime number q in the range [2k-1/p, 2k/p[;
• If the condition 0.1 < |log2(p)-log2(q)| < 30 is not satisfied, go back to the first step;
• Let n be the product of p and q.
A more complicated method that avoids the third step altogether but produces differently
distributed primes is:
• Choose a random prime number p in the range [2k/2-9/20, 2k/2+15[;
• Choose a random prime number q in the range]a,b[ where a=max(ceil(2k-1/p)-1, p.2-
30) and b=min(2k/p, p.2-1/10);
• Let n be the product of p and q.
W.Chocianowicz – Kryptologia – semestr zimowy 2008/2009
str.138