首页> 外国专利> 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

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

摘要

#$%^&*AU2014100414A420140605.pdf#####Factoring p * q through modulus of p + b * x and q + e * x: Obtaining b * e Paul C. Bingham Cheffers 19th April 2014 Abstract An algorithm is shown that can factor p * q if two modulus p + b * x and q+e*x are set so that q+e*x ~ q+,t, p+b*x ~ p+,r/p and q+e*x is a factor of p+b*x by factoring of a e*b modular residue when it is an actual product. It is shown that making p+ b*x a multiple of q+e *x is considerably simpler when x = 4 and when p mod 4 = q mod 4, which is the case in most RSA semiprimes. Finally, the case for when p - b * 4 = q + e * 4 is shown, and this case leads to the discovery of b * e where b + e = (p - q)/4. 1
机译:#$%^&* 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

著录项

  • 公开/公告号AU2014100414A4

    专利类型

  • 公开/公告日2014-06-05

    原文格式PDF

  • 申请/专利权人 PAUL CHEFFERS;

    申请/专利号AU20140100414

  • 发明设计人 CHEFFERS PAUL CLIFTON BINGHAM;

    申请日2014-04-22

  • 分类号G06F17/10;

  • 国家 AU

  • 入库时间 2022-08-21 15:52:27

相似文献

  • 专利
  • 外文文献
  • 中文文献
获取专利

客服邮箱:kefu@zhangqiaokeyan.com

京公网安备:11010802029741号 ICP备案号:京ICP备15016152号-6 六维联合信息科技 (北京) 有限公司©版权所有
  • 客服微信

  • 服务号