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20210503, 10:34  #1 
May 2017
ITALY
2·3^{2}·29 Posts 
Is it possible to use The Jubjub bird and The frumious Bandersnatch method in this polynomial?
Given the bivariate polynomial
p(m,n) = 675 * m * n + 297 * m + 25 * n + 11 The bivariate Coppersmith method can be used to find m0 and n0 such that (675 * m0 * n0 + 297 * m0 + 25 * n0 + 11) mod (1763) = 0 where is it m0 = 3 ; n0 = 3 Unfortunately I cannot understand the hypotheses Kindly someone could only explain the hypotheses to me ************************************************************************************************ OTHER INFORMATION ************************************************************************************ You can choose some coefficients and their size order of this type of polynomial in O (16). really is O(8) (N3)/8q*(pA)/8[4(A7)*(A5)/8]=A*(q+A48)/8 > N=p*q if we choose A such that pA mod 8 = 0 we can write it this way (N3)/8Q[4(A7)*(A5)/8]=A*X so there are 4 chances to find A and they are 8*h+1 ; 8*h+3 ; 8*h+5 ; 8*h+7 (N3)/8p*(qB)/8[4(B7)*(A5)/8]=B*(p+B48)/8 > N=p*q if we choose B such that qB mod 8 = 0 we can write it this way (N3)/8P[4(B7)*(B5)/8]=B*Y so there are 4 chances to find B and they are 8*k+1 ; 8*k+3 ; 8*k+5 ; 8*k+7 f(N,A,B) is O(16) Example 1763=41*43 220q*(p25)/8[4(257)*(255)/8]=25*(q+2548)/8 220p*(q27)/8[4(277)*(275)/8]=27*(p+B48)/8 220Q[4(257)*(255)/8]=25*X 220P[4(277)*(275)/8]=27*X Q=25*a+11 ; X=10a P=27*b+1 ; X=10b 220Q[4(177)*(175)/8]=17*X Q=17*c+10 ; X=13c ; 9a=13c > c=a+4 220P[4(197)*(195)/8]=19*X P=19*d+9 ; X=12d ; 9b=12d >d=b+3 p=(19*(b+3)+9)(27*b+1)=658*b q=(17*(a+4)+10)(25*a+11)=678*a (658*b)*(678*a)=1763 , (10a)(10b)=((678*a)(658*b)2)/8 (658*b)*(27*1763)/(8*(27*b+1)+1763)=1763 TRUE So I thought about using the bivariate Coppersmith method Q=25*a+11 P=27*b+1 p(m,n)=p(b,a)=P*Q=(27*b+1)*(25*a+11)=Z*1763 
20210503, 18:39  #2 
May 2017
ITALY
2·3^{2}·29 Posts 
N=p*q
with p+q=8*x+4 Last fiddled with by Alberico Lepore on 20210503 at 19:41 
20210504, 09:32  #3 
May 2017
ITALY
2×3^{2}×29 Posts 
At least see if I understand:
in my case given two polynomials for example (25*a+11)*(27*b+1) mod 1763 ==0 and (17*(a+4)+10)*(19*(b+3)+9) mod 1763 == 0 the first question is can they be both modulo 1763? a<IntegerPart[sqrt(1763/2)] b<2*IntegerPart[sqrt(1763/2)] if this method can be applied, it can be solved in polynomial time Did I get it right ? 
20210504, 21:25  #4 
May 2017
ITALY
2×3^{2}×29 Posts 
(658 * b) * (678 * a) 1763 = 0
> (8*a*b65*a67*b+324)=0 which is irreducible pag 3 http://www.cryptouni.lu/jscoron/pub.../bivariate.pdf 
20210505, 08:51  #5 
May 2017
ITALY
1012_{8} Posts 
can we use the Coppersmith method multivariate in this case?
(8*a*b65*a67*b+5) mod 319 = 0 (25*a+11)*(27*b+1) mod 1763 = 0 
20210506, 21:26  #6  
May 2017
ITALY
2·3^{2}·29 Posts 
Quote:
X*Y < W^(2/3δ) ; W = max{a, bX, cY, dXY } we can easily choose  d  XY of the order size N ^ 2 and X * Y of the order size of N. that is the problem? 

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