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Factoring p times q through modulus of p plus b times x and q plus e times x Obtaining b times e
Factoring p times q through modulus of p plus b times x and q plus e times x Obtaining b times e
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机译:通过p的模数乘以p乘以q加b乘以x和q加e乘以x
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#$%^&*AU2014100414A420140605.pdf#####Factoring p * q through modulus of p + b * x and q + e * x: Obtaining b * e Paul C. Bingham Cheffers 19th April 2014 Abstract An algorithm is shown that can factor p * q if two modulus p + b * x and q+e*x are set so that q+e*x ~ q+,t, p+b*x ~ p+,r/p and q+e*x is a factor of p+b*x by factoring of a e*b modular residue when it is an actual product. It is shown that making p+ b*x a multiple of q+e *x is considerably simpler when x = 4 and when p mod 4 = q mod 4, which is the case in most RSA semiprimes. Finally, the case for when p - b * 4 = q + e * 4 is shown, and this case leads to the discovery of b * e where b + e = (p - q)/4. 1
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机译:#$%^&* AU2014100414A420140605.pdf #####通过p + b * x的模数分解p * q和q + e * x:获得b * e保罗·宾厄姆·谢弗2014年4月19日抽象显示了一种算法,如果两个模数p + b * x可以分解p * q和q + e * x设置为q + e * x〜q +,t,p + b * x〜p +,r / p和q + e * x通过将e * b模数残差分解为p + b * x的因子实际产品。证明使p + b * x是q + e * x的倍数当x = 4且p mod 4 = q mod 4时,它要简单得多。大多数RSA半素数都是这种情况。最后,什么时候p-b * 4 = q + e * 4被显示,这种情况导致发现b * e其中b + e =(p-q)/ 4。1个
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