# I. INTRODUCTION


# Yonatan Zilpa

Let s 1 , s 2 , and S be any integers such that S = s 1 s 2 , then (s 2 -s 1 ) 2 + 4S is a perfect square. Indeed (s 2 -s 1 ) 2 + 4S = (s 2 + s 1 ) 2 .

Let M be a semiprime number and let p, q be its prime factors, where q > p. Let d = q -p and let n and x be any integers, such that n divides M -x, then
M -x n -n 2 + 4(M -x) = M -x n + n (1.1) is a positive integer. Thus, if M -x n -n 2 -4x is a non-negative perfect square, then M -x n -n 2 -4x = d 2 . (1.2) Equation (1.2) implies that M -x n -n = 4x + d 2 .
Hence, x must contain a factor t such that
x t -t = d.
The number x must be of the form: where j is an integer. Let k be a positive integer less than p, then substituting
(d + j)jx with k(d + k) in equation (1.2) yields M -(d + k)k n -n 2 -4(d + k)k = d 2 (1.3)
Solving equation (1.3) for d we get the following two solutions 
d 0 = M -k 2 +2kn-n 2 k-n = M -(k-n) 2 k-n d 1 = M -k 2 -2kn-n 2 k+n = M -(M -(d+k)k n -n 2 -4k(d + k) = d 2 M -(d+(k+1))(k+1) n+1 -(n + 1) 2 -4(k + 1)(d + (k + 1)) = d 2 M -(d+(k+2))(k+2) n+2 -(n + 2) 2 -4(k + 2)(d + (k + 2)) = d 2(1.6)
System (1.6) has three equations with three variables n, k, d, however this system is dependent. We may overcome this problem by trying other functions. Let t : Z ? Z be any function, replace n with t(n) and k with u in equation (1.3). Equality (1.5) implies that u -t(n) = p (or t(n) -u = p) and k -n = p (or n -k = p), which gives us a system of equations
u -t(n) = p k -n = p from which we deduce u -k -t(n) + n = 0 or equivalently u = k + t(n) -n.
We get the following equality:
? ? M -d + k + t(n) -n k + t(n) -n t(n) -t(n) ? ? 2 + -4 k + t(n) -n d + k + t(n) -n = d 2 (1.7)
Based on equation (1.7) we can deduce a new system of three equations with three variables k, n and, d. We may find three functions t 1 , t 2 , t 3 : Z ? Z and replace t(n) with t 3 (n) to get the third equation, t(n) with t 2 (n) to get the second equation, and finally t(n) with t 1 (n) to get the first equation. The key here is to select the functions t 1 , t 2 , and t 3 in such a way that our system has a unique solution, where |n -k| ? = 1. When moving d 2 to the left side of equality (1.7) and multiplying it with t 2 (n), the left side of this equality becomes:
?(t, n, k, d) := M -d + k + t(n) -n k + t(n) -n -t 2 (n) 2 -4t 2 (n) k + t(n) -n d + k + t(n) -n -t 2 (n)d 2
(2.1) If t is a polynomial function in R with integral coefficients, then ? can be viewed as a polynomial function from R 3 to R. In this case we also denote the function ?(t, n, k, d) with ? t (x, y, z). We thus get a system of polynomial equations:
? t1 (x, y, z) = 0 ? t2 (x, y, z) = 0 ? t3 (x, y, z) = 0 (2.2)
The problem with d 0 is that the variant of system (1.7) is infinite, any integer n, k such that |n -k| = 1 satisfying this system. However, applying solution d 1 in equality (1.4) and requiring that n, k be positive integers implies that k + n = p.

(3.1)

Replacing n with t(n) and k with u in equation ( 1.3) we get the following system
u + t(n) = p k + n = p(3.2)
from which we deduce u + t(n) -k -n = 0 or equivalently u = n + k -t(n). Now we can replace k with n + k -t(n) and n with t(n) and d with d 1 in equation
(1.3) to get ? ? M -d+ n+k-t(n) n+k-t(n) t(n) -t(n) ? ? 2 -4 n + k -t(n) d + n + k -t(n) = d 2 (3.3)
Since t(n) relies on the second equality of (1.4) and since t(n) differs from n, the first solution in (1.4) won't solve equality (3.3). Hence, by replacing t(n) with polynomial t 1 (n) with positive coefficients we get two independent polynomials.


# Let us denote
t (n, k, d) := M -(d+n+k-t(n))(n+k-t(n)) t(n) -t(n) 2 -4 d + (n + k -t(n)) n + k -t(n)) -d 2 then equality (3.3) becomes t (n, k, d) = 0. (3.4)
If we set t(n) = n, then equation (3.4) is equivalent to (1.3). However, if polynomial t(n) differs from n, then solution d 0 is lost. Hence, for any polynomial t 2 (n) with positive integers that differs from n, polynomials n and t2(n) are independent.

We can repeatedly use the result u = n + k -t(n), obtained from system (3.2), to get the following system of three polynomial equations with three variables:
? ? ? ? ? t1 (n, k, d) = 0 ? t2 (n, k, d) = t1 (t 2 (n), n + k -t(n), d) = 0 t1 (t 3 (n), n + k -t(n), d) + ? t2 (t 3 (n), n + k -t(n), d) = 0. (3.5)
If polynomials t 1 , t 2 , and t 3 differ in pairs and having non-negative integers and if none of these polynomial is zero, then none of the polynomial in system (3.5) depends on the other.

The RSA cryptosystem [4] as well as all public key cryptography implementations rely on the complexity of semiprime factorization. Mathematical attacks based on known relations, such as Pythagorean primes [3] or the use of a polynomial of third degree order [6] have been recently proposed for potential methods for factoring semiprimes numbers. When it comes to factoring large semiprime numbers, well known existing algorithms may consume too much memory and running time. Other algorithms, such as the firefly algorithm [5], may address some of these issues [2].

In this article, we attempt to attack the problem of semiprime factorization by using relationships between M and two different numbers, that are less than M . Using only quadratic relationships, we have constructed a wide variety of systems of three polynomial equations with three variables. A solution of one of one system may lead to a complete factorization of the semiprime number M .

I would like to express my sincere gratitude to Professor Shai Haran, who provided invaluable feedback on the manuscript. His insightful comments and suggestions greatly improved the clarity and rigor of the research findings. I am grateful for his time and expertise, which helped me to refine my research questions and the methodology used to address them. Without his guidance, this paper would not have been possible. ? ?
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(1.4)k+n) 2k+nSince d is a positive integer. The first equality of equation (1.4) implies that|k -n| = 1 or|n -k| = p(1.5)
		
		

			

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