 mersenneforum.org Letzte attempt to factor RSA
 Register FAQ Search Today's Posts Mark Forums Read  2020-08-09, 13:55 #1 Alberico Lepore   May 2017 ITALY 1110110112 Posts Letzte attempt to factor RSA Given N in the form N=(6*a+1)*(6*b+1) then if this is solvable solve 2*((N+6*(6*a+1)-1)/6)^2+(N+6*(6*a+1)-1)/6 mod ((6*a+1)^2) =(( (2*((N-1)/6)^2+(N-1)/6) mod (6*a+1)^2)-2*a*(6*a+1)) the factoring is solved Example N=91 solve 2*((91+6*(6*a+1)-1)/6)^2+(91+6*(6*a+1)-1)/6 mod ((6*a+1)^2) =((465 mod (6*a+1)^2)-2*a*(6*a+1)) wolfram solves it https://www.wolframalpha.com/input/?...86*a%2B1%29%29 but how do you solve it?   2020-08-09, 17:08   #2
chalsall
If I May

"Chris Halsall"
Sep 2002

2·59·79 Posts Quote:
 Originally Posted by Alberico Lepore ...but how do you solve it?
I don't.   2020-08-09, 22:51   #3
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

32·653 Posts Quote:
 Originally Posted by Alberico Lepore Example N=91
This is so ridiculous.    2020-08-10, 12:06   #4
Alberico Lepore

May 2017
ITALY

52·19 Posts Quote:
 Originally Posted by Alberico Lepore Given N in the form N=(6*a+1)*(6*b+1) then if this is solvable solve 2*((N+6*(6*a+1)-1)/6)^2+(N+6*(6*a+1)-1)/6 mod ((6*a+1)^2) =(( (2*((N-1)/6)^2+(N-1)/6) mod (6*a+1)^2)-2*a*(6*a+1)) the factoring is solved Example N=91 solve 2*((91+6*(6*a+1)-1)/6)^2+(91+6*(6*a+1)-1)/6 mod ((6*a+1)^2) =((465 mod (6*a+1)^2)-2*a*(6*a+1)) wolfram solves it https://www.wolframalpha.com/input/?...86*a%2B1%29%29 but how do you solve it?
there are special cases Example 247

where the formula is

solve 2*((N+6*(6*a+1)-1)/6)^2+(N+6*(6*a+1)-1)/6 mod ((6*a+1)^2) =( (2*((N-1)/6)^2+(N-1)/6) mod (6*a+1)^2)-2*a*(6*a+1)+(6*a+1)^2

Last fiddled with by Alberico Lepore on 2020-08-10 at 12:09 Reason: )   2020-08-10, 13:27 #5 xilman Bamboozled!   "𒉺𒌌𒇷𒆷𒀭" May 2003 Down not across 3×43×79 Posts Thank $DEITY that it is the last one.   2020-08-10, 16:17 #6 Alberico Lepore May 2017 ITALY 1DB16 Posts maybe it can help in the resolution [2*((91+6*(6*a+1)-1)/6)^2+(91+6*(6*a+1)-1)/6] mod ((6*a+1)^2) =(465 mod ((6*a+1)^2))-2*a*(6*a+1) , [2*((91+6*(6*b+1)-1)/6)^2+(91+6*(6*b+1)-1)/6] mod ((6*b+1)^2) =(465 mod ((6*b+1)^2))-2*b*(6*b+1) , (6*a+1)*(6*b+1)=91   2020-08-10, 17:17 #7 Alberico Lepore May 2017 ITALY 52·19 Posts If it can help you could fix this solve [[(2*((N+6*(6*a+1)-1)/6)^2+(N+6*(6*a+1)-1)/6) mod (N+6*(6*a+1))] -[[2*((N-1)/6)^2+((N-1)/6)]mod N]] mod (6*a+1)=0 Example N=91 solve [[(2*((91+6*(6*a+1)-1)/6)^2+(91+6*(6*a+1)-1)/6) mod (91+6*(6*a+1))] -10] mod (6*a+1)=0 Edit: Special cases must also be taken into account here Last fiddled with by Alberico Lepore on 2020-08-10 at 17:19 Reason: Edit:   2020-08-10, 17:26 #8 CRGreathouse Aug 2006 2×2,969 Posts Quote:  Originally Posted by Alberico Lepore Given N in the form N=(6*a+1)*(6*b+1) then if this is solvable solve 2*((N+6*(6*a+1)-1)/6)^2+(N+6*(6*a+1)-1)/6 mod ((6*a+1)^2) =(( (2*((N-1)/6)^2+(N-1)/6) mod (6*a+1)^2)-2*a*(6*a+1)) the factoring is solved Example N=91 If I ever need to factor a semiprime coprime to 6 with at most 2 decimal digits, I'll know where to turn to -- assuming your method can handle the other eight candidates. Here's an alternate method. Check if the last digit is 5. If so, the number is divisible by 5; otherwise, the number is divisible by 7. Last fiddled with by CRGreathouse on 2020-08-10 at 17:29   2020-08-10, 17:29 #9 Alberico Lepore May 2017 ITALY 1110110112 Posts Quote:  Originally Posted by CRGreathouse If I ever need to factor a semiprime coprime to 6 with at most 2 decimal digits, I'll know where to turn to -- assuming your method can handle the other eight candidates. (Hint: they're all divisible by 5 or 7.) N=(6*a+5)*(6*b+1)*5 N=(6*a+5)*(6*b+5)*25   2020-08-10, 18:08 #10 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 22·3·7·109 Posts Quote:  Originally Posted by xilman Thank$DEITY that it is the last one.
Amen to that!

Alberico Lepore, you promised that this will be your last attempt. Please keep that promise!   2020-08-10, 18:29   #11
Alberico Lepore

May 2017
ITALY

52×19 Posts Quote:
 Originally Posted by Batalov Amen to that! Alberico Lepore, you promised that this will be your last attempt. Please keep that promise!
it was meant in chronological order   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Alberico Lepore Alberico Lepore 23 2020-05-20 09:51 Alberico Lepore Alberico Lepore 9 2020-05-01 09:41 jvang jvang 10 2019-02-03 23:17 jasong Programming 4 2005-11-29 22:24 JHansen Factoring 34 2005-05-27 19:24

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