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Cryptanalysis of RSA Factorization - Library(ISI Kolkata) - Indian ...

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Chapter 2: Mathematical Preliminaries 34<br />

• R(r 1 ,r 2 ) ≠ 0 if r 1 ,r 2 are algebraically independent, i.e., gcd(r 1 ,r 2 ) = 1, and<br />

• R(r 1 ,r 2 )(x (0)<br />

1 ,...,x (0)<br />

t−1) = 0.<br />

Using the method <strong>of</strong> resultants recursively, we can find a polynomial in x 1<br />

(with root x (0)<br />

1 ) in t −1 iterations. We can then find x (0)<br />

1 easily using some root<br />

finding algorithm. Furthermore, using back-substitution t−1 times, we can find<br />

x (0)<br />

2 ,...,x (0)<br />

t as well, one in each step <strong>of</strong> back-substitution.<br />

Similarly one can use the technique <strong>of</strong> Gröbner Basis [30] to find the roots.<br />

The Gröbner Basis G = {g 1 ,g 2 ,...,g l } is a set <strong>of</strong> polynomials such that<br />

g 1 (x (0)<br />

1 ,...,x (0)<br />

t ) = g 2 (x (0)<br />

1 ,...,x (0)<br />

t ) = ··· = g l (x (0)<br />

1 ,...,x (0)<br />

t ) = 0.<br />

Each polynomial g i can be computed with respect to some ordering that eliminates<br />

the variables. So it is easy to extract the desired root. However, the elimination<br />

<strong>of</strong> variables fails if the variety V(I) <strong>of</strong> the ideal I generated by r 1 ,r 2 ,...,r t is not<br />

zero-dimensional. An interested reader may refer to [30] for more details regarding<br />

Gröbner Basis.<br />

Occasionally, one can collect the roots by examining special structure in the<br />

polynomials as well. For example, if we get a polynomial f(x,y) = ax − by + b<br />

where a,b are constants, then (b,a+1) is quite clearly a root <strong>of</strong> f(x,y).<br />

From our discussion so far, we can conclude that in case <strong>of</strong> solving multivariate<br />

modular polynomials, one needs the following assumption before attempting a<br />

lattice based technique.<br />

Assumption 1: The common root (x (0)<br />

1 ,...,x (0)<br />

t ) can be efficiently collected from<br />

the polynomials r 1 ,...,r t using the method <strong>of</strong> resultants, Gröbner Basis technique<br />

or exploiting the structure <strong>of</strong> the polynomials.<br />

Once we assume the above mentioned statement, (2.1) reduces the condition for<br />

solving a modular polynomial using LLL to the following:<br />

2 ω(ω−1)<br />

4(ω+1−t) det(L)<br />

1<br />

ω+1−t <<br />

N m<br />

√ ω<br />

.<br />

If we treat the number <strong>of</strong> variables t, and the dimension <strong>of</strong> lattice L as constants,<br />

then we can simply state the condition as det(L) < N mω . This technique <strong>of</strong> finding

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