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How To Factor P times Q From One Square Root Of -1
How To Factor P times Q From One Square Root Of -1
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机译:如何从-1的平方根分解P乘Q
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#$%^&*AU2016100469A420160526.pdf#####How To Factor P * Q From One Square Root Of -1 Paul C. Bingham Cheffers 17 December 2015 1 Abstract Abstract The paper shows two methods of factoring two primes, P * Q, with knowledge of one of the Eg mod p * q. and some further operations described in the paper. The first method shows that, for 1 mod 4 semiprime modula that 2 * -12 mod p * q ±(VI + 1)2 mod p * q, and that taking the modular square root of this sum (or the natural square root if 2 * V12 mod p * q is a natural square) will lead to an equation that easily factors either p or q from p * q. A Mathematica procedure is then shown, with output, that shows that out of 245,000 possible p * q modula that 26 have 2 * 12 mod p * q as a natural square. The second method shows that a modular number, X mod p * q, expressed in modular complex notation, a + b * El mod p * q = X has a modular complex conjugate a - b * Eg mod p * q = X, and that the four answers of each of two modular quadratic equations, (x+n)2 mod p * q = X, and (x+n)2 mod p * q Xcc can be combined together to solve for P and Q in the field of P * Q. 1
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机译:#$%^&* AU2016100469A420160526.pdf #####如何从一平方根分解P * Q-1保罗·宾厄姆·谢弗2015年12月17日1摘要抽象本文展示了两种分解两个素数P * Q的方法,其中Eg mod p * q之一的知识。以及其他一些操作本文所述。第一种方法表明,对于1个模4半素模,2 * -12 mod p * q±(VI + 1)2 mod p * q,此和的模平方根(或自然平方根,如果2 *V12 mod p * q是自然平方)将导致一个方程容易地从p * q分解p或q。然后显示一个Mathematica过程,并显示输出在245,000个可能的p * q模中有26个具有2 * 12 modp * q为自然平方。第二种方法表明,模数X mod p * q用模块化复数符号表示,a + b * El mod p * q = X有一个模复共轭a-b * Eg mod p * q = X,并且两个模块化二次方程式的每一个的四个答案,(x + n)2 mod p * q = X,并且(x + n)2 mod p * q Xcc可以组合一起解决P * Q领域中的P和Q。1个
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