20200809, 13:55  #1 
May 2017
ITALY
111011011_{2} 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*((N1)/6)^2+(N1)/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? 
20200809, 17:08  #2 
If I May
"Chris Halsall"
Sep 2002
Barbados
2·59·79 Posts 

20200809, 22:51  #3 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
3^{2}·653 Posts 

20200810, 12:06  #4  
May 2017
ITALY
5^{2}·19 Posts 
Quote:
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*((N1)/6)^2+(N1)/6) mod (6*a+1)^2)2*a*(6*a+1)+(6*a+1)^2 Last fiddled with by Alberico Lepore on 20200810 at 12:09 Reason: ) 

20200810, 13:27  #5 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
3×43×79 Posts 
Thank $DEITY that it is the last one.

20200810, 16:17  #6 
May 2017
ITALY
1DB_{16} 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 
20200810, 17:17  #7 
May 2017
ITALY
5^{2}·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*((N1)/6)^2+((N1)/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 20200810 at 17:19 Reason: Edit: 
20200810, 17:26  #8  
Aug 2006
2×2,969 Posts 
Quote:
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 20200810 at 17:29 

20200810, 17:29  #9 
May 2017
ITALY
111011011_{2} Posts 

20200810, 18:08  #10 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·3·7·109 Posts 

20200810, 18:29  #11 
May 2017
ITALY
5^{2}×19 Posts 

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