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 February 1st, 2018, 03:52 AM #1 Member   Joined: Jan 2015 From: italy Posts: 39 Thanks: 1 Is this a bug in RSA? Is this a bug in RSA? This method as described is not deterministic but can become so in this way it works only if q = (a ^ 2 + b ^ 2) / 2 admits integer solution an example is RSA100 of the RSA challenge Suppose we want to factor 629 Suppose the ratio q / p < 3 -> q_max = 43 a is odd k is even always (p*(a+k)*a)^2+(629-p*a^2)^2=(629)^2 varying a starting from 1 we see which equations admit integer solutions a=1 NULL a=3 NULL (p*(a+k)*a)^2+(629-p*a^2)^2=(629)^2 , a=5 p=1258/(k^2+10*k +50) p*k^2+10*k*p+50*p=1258 dividing everything by p k^2+10*k-2*q+50=0 since k is even k = 2 * t (2*t)^2+2*10*t-2*q+50=0 then k=2*t q=2*t^2+10*t+25 for t=1 -> k=2 & q=37 What do you think?
 February 3rd, 2018, 11:52 AM #2 Senior Member   Joined: Aug 2012 Posts: 2,139 Thanks: 623 You'll have better luck posting this to a cryptography discussion forum, perhaps https://www.reddit.com/r/crypto/ Thanks from greg1313

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