Semiprime factorization and custom proof system

Suppose we have N=pq, with p and q are unknown odd primes. We can encode factorization problem in the system of polynomial equations. For instance, p= 1+ \sum_{k=1}^n 2^k x_k, q= 1+ \sum_{k=1}^n 2^k y_k, where n = \lfloor \log_2 N \rfloor is one less the number of bits required to represent N, and x_k, y_k are binary indeterminates. By abuse of notation, we can write the following system of equations (here x, y \in \mathbb{C}^n)

\left\{ \begin{array}{lll} f_1= &x_k^2-x_k & =0 \\ g_1= &y_k^2-y_k &=0  \\ h_{1,1}= &\left( 1+ \sum_{k=1}^n 2^k x_k \right) \left( 1+ \sum_{k=1}^n 2^k y_k \right) -N & = 0  \end{array} \right.

This system has only two solutions for semiprimes – x encoding p, y encoding q, and vice versa. Therefore, the Groebner basis will consist of 2n linear equations encoding line passing through (x,y)= (p,q) and (x,y)=(q,p), and one quadratic equation selecting these two points along this line.

Therefore, we want to represent the linear part of Groebner basis using a linear combination of our equations with polynomial coefficients – c+ \sum_k a_k x_k +b_k y_k = P_1 h+ \sum_k P_{2,k} f_k + P_{3,k} g_k with P_{...} \in \mathbb{C}[x_1,..., x_n, y_1, ..., y_n] , Here, the coefficients in the polynomials P are treated as indeterminates. For example, for n=1 the correspondin system reads

\left\{ \begin{array}{lll} f_k= &x_1^2-x_1 & =0 \\ g_k= &y_1^2-y_1 &=0  \\ h_{1,1}= &\left( 1+ 2 x_1 \right) \left( 1+ 2 y_1 \right) -N & = 0  \end{array} \right.

The linear bases is expressed by

\begin{array}{l} c+ a_1 x_1 +b_1 y_1 = \\ \left( r_{1,1} x_1 +r_{1,2} y_1 +r_{1,3} \right) (x_1^2-x_1)+ \\ \left( r_{2,1} x_1 +r_{2,2} y_1 +r_{2,3} \right) (y_1^2-y_1)+ \\ \left( r_{3,1} x_1 +r_{3,2} y_1 +r_{3,3} \right) (1+2 x_1+2 y_1+4 x_1 y_1-N) \end{array}.

and we have the following system for coefficients of different monomials

\left\{ \begin{array}{rrl}    x_1^3: & r_{1,1}&=0 \\  x_1^2y_1: &r_{1,2} +4r_{3,1} &=0 \\  x_1^2:& r_{1,3} - r_{1,1} +2 r_{3,1} &=0\\  x_1y_1^2:& r_{2,1}+4r_{3,2} &=0\\  x_1y_1:& -r_{1,2}-r_{2,1} +2r_{3,1}+2r_{3,2}+4r_{3,3}&=0\\  x_1:& -a_1 -r_{1,3}-r_{3,1}(N-1)+2r_{3,3} &=0\\  y_1^3:& r_{2,2} &=0\\  y_1^2:&  r_{2,3}-r_{2,2} +2r_{3,2}&=0\\  y_1:& -b_1 -r_{2,3} -r_{3,2} (N-1) +2r_{3,3} &=0\\  1:&  c+ r_{3,3} (N-1) &=0\\   \end{array} \right.

\left\{ \begin{array}{rrl}    x_1^2y_1: &-r_{1,2} &=4r_{3,1} \\  x_1y_1^2:& -r_{2,1}&=4r_{3,2} \\  x_1^2:& -r_{1,3} &=2 r_{3,1}\\  y_1^2:&  -r_{2,3} &=2r_{3,2}\\  x_1y_1:& 2r_{3,3}&=-3(r_{3,1}+r_{3,2})\\  x_1:&  r_{3,1}N+3r_{3,2} &=a_1\\  y_1:&  r_{3,2} N+3r_{3,1} &=b_1\\  1:&    -3(r_{3,1}+r_{3,2})(N-1) &=2c\\   \end{array} \right.

(N r_{3,1}+3r_{3,2}) x_1+   ( N r_{3,2} +3r_{3,1} ) y_1 -   \frac{ 3}{2}(r_{3,1}+r_{3,2})(N-1)= \\  \left( -4r_{3,1} y_1 -2r_{3,1} \right) (x_1^2-x_1)+ \\ \left( -4r_{3,2} x_1 -2r_{3,2} \right) (y_1^2-y_1)+ \\ \left( r_{3,1} x_1 +r_{3,2} y_1 - \frac{ 3}{2}(r_{3,1}+r_{3,2}) \right) (1+2 x_1+2 y_1+4 x_1 y_1-N)

So the left-hand side of the equation can be written as

[r_{3,1}, r_{3,2}] \left[ \begin{array}{ccc}   N &3 & \frac{ 3}{2}(1-N) \\ 3 &N  &\frac{ 3}{2}(1-N)  \end{array} \right]   \left[ \begin{array}{c} x_1\\y_1\\1 \end{array} \right]

What we can see here is that the matrix in the above expression has rank 2, unless N=3, where the rank of the matrix is 1. When N=3 the nullspace of the matrix is

\left[ \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right], \left[ \begin{array}{c} -1 \\ 1 \\ 0 \end{array} \right]

which correspond to two solutions – x_1= 1, y_1= 0 \rightarrow 3= 3*1 , and x_1= 0, y_1= 1 \rightarrow 3= 1*3 . Otherwise the null space is expressed as

\left[ \begin{array}{c} \frac{3(N-1)}{2(N+3)} \\ \frac{3(N-1)}{2(N+3)} \\ 1 \end{array} \right]

So, for N=1 x_1=y_1= 0 \rightarrow 1= 1*1, and for N=9 x_1= 1, y_1= 1 \rightarrow 9= 3*3 .

Overall, despite we started with 9+3 indeterminates there are only 2 are relevant to the problem. The question is now how does it scales with n. Or another question, is it at all possible to find an algorithm that will select only relevant monomials in P_{...}.

Let’s try to extend analysis to the case n=2.
To be continued.

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