20190821, 13:15  #1 
May 2017
ITALY
2^{3}·3^{2}·7 Posts 
Potential Coppersmith attack on RSA
The sum S of consecutive odd cubes from 1 to n = 2 * h + 1 is
[(h+1)^2]*(2*h^2+4*h+1)=2*h^4+8*h^3+11*h^2+6*h+1 [see attachment] That said One approach to RSA factorization could be this If a multiple K * N of the number to be factored N is sum of cubes then 2*h^4+8*h^3+11*h^2+6*h+1=N*K then using Coppersmith is found h in polynomial times once found h GCD((h+1),N)=p it can be done? 
20190821, 14:28  #2 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2×3×1,811 Posts 

20190821, 17:17  #3 
May 2017
ITALY
1F8_{16} Posts 

20190821, 17:58  #4 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2A72_{16} Posts 

20190821, 18:12  #5 
May 2017
ITALY
111111000_{2} Posts 

20190821, 20:40  #6 
"Curtis"
Feb 2005
Riverside, CA
2·2,477 Posts 

20190822, 07:14  #7 
May 2017
ITALY
2^{3}×3^{2}×7 Posts 

20190822, 11:53  #8 
May 2017
ITALY
1F8_{16} Posts 
Then the two new questions are:
Would any of you know how to transform 2*h^4+8*h^3+11*h^2+6*h+1=N*K into a polynomial that satisfies the conditions of Coppersmith? Do you know other methods to find h? 
20190822, 15:18  #9 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3^{2}·1,061 Posts 

20190823, 02:57  #10  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
14170_{8} Posts 
Quote:


20190824, 13:37  #11 
May 2017
ITALY
2^{3}×3^{2}×7 Posts 
I managed to bring any number into the polynomial
81 * X = a1 * n ^ 3 + a2 * n ^ 2  a3 * n + a4 Coppersmith can be used by multiplying the whole polynomial by a prime number P so that n0 <(81 * P) ^ 1/3? or (81 * P) * X =  (a1 * P) * n ^ 3 + (a2 * P) * n ^ 2  (a3 * P) * n + (a4 * P) would you kindly tell me if in this case you can use Coppersmith? 
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