m kn n 0k − ∑ =
m kn n 0k − ∑ =
m kn n 0k − ∑ =
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This condition implies that<br />
log2(n)/2 - 15 < log2(p), log2(q) < log2(n) / 2 + 15<br />
The generation of an RSA modulus of exactly k bits could be done with the following<br />
algorithm:<br />
• Choose a random prime number p in the range ]2k/2-15, 2k/2+15[;<br />
• Choose a random prime number q in the range [2k-1/p, 2k/p[;<br />
• If the condition 0.1 < |log2(p)-log2(q)| < 30 is not satisfied, go back to the first step;<br />
• Let n be the product of p and q.<br />
A more complicated method that avoids the third step altogether but produces differently<br />
distributed primes is:<br />
• Choose a random prime number p in the range [2k/2-9/20, 2k/2+15[;<br />
• Choose a random prime number q in the range]a,b[ where a=max(ceil(2k-1/p)-1, p.2-<br />
30) and b=min(2k/p, p.2-1/10);<br />
• Let n be the product of p and q.<br />
W.Chocianowicz – Kryptologia – semestr zimowy 2008/2009<br />
str.138