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Old 2020-08-09, 13:55   #1
Alberico Lepore
 
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Default 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?
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Old 2020-08-09, 17:08   #2
chalsall
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Quote:
Originally Posted by Alberico Lepore View Post
...but how do you solve it?
I don't.
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Old 2020-08-09, 22:51   #3
retina
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Quote:
Originally Posted by Alberico Lepore View Post
Example N=91
This is so ridiculous.
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Old 2020-08-10, 12:06   #4
Alberico Lepore
 
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Quote:
Originally Posted by Alberico Lepore View Post
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: )
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Old 2020-08-10, 13:27   #5
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Thank $DEITY that it is the last one.
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Old 2020-08-10, 16:17   #6
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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
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Old 2020-08-10, 17:17   #7
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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:
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Old 2020-08-10, 17:26   #8
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Quote:
Originally Posted by Alberico Lepore View Post
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
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Old 2020-08-10, 17:29   #9
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Quote:
Originally Posted by CRGreathouse View Post
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
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Old 2020-08-10, 18:08   #10
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Quote:
Originally Posted by xilman View Post
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!
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Old 2020-08-10, 18:29   #11
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Quote:
Originally Posted by Batalov View Post
Amen to that!

Alberico Lepore, you promised that this will be your last attempt. Please keep that promise!
it was meant in chronological order
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