A Restatement of the Twin Prime Conjecture via k-Index Filtering

1. Introduction

The Twin Prime Conjecture posits that there are infinitely many prime pairs (p, p+2). All such pairs, except (3,5), are of the form (6k-1, 6k+1) for some positive integer k. The conjecture is thus equivalent to asserting that infinitely many positive integers k render both 6k-1 and 6k+1 prime. This document presents an algebraic restatement of this conjecture using the “k-Index Filtering” framework.

2. The k-Filtering Framework Applied to 6k ± 1

The k-filtering model’s simplified case states that numbers N = ak_{orig} + c² are composite if their index k_{orig} is k_{param} = axy + c(x+y) for integers x,y, where N = (ax+c)(ay+c).
For 6k \pm 1 forms, we set a = 6 and c = 1. This addresses numbers 6k_{orig}+1. The composite-generating index parameterization is:
k_{param} = 6xy + x + y
For non-trivial factors (not ±1), we require x, y \in \mathbb{Z} \setminus \{0\}.

3. Derivation and Role of the Composite-Indicating Index Set S_c

Let k be a positive integer. S_c identifies k for which the model predicts compositeness for 6k-1 or 6k+1.

• Model Prediction for 6k+1 Compositeness:
  If k_{val} = 6xy+x+y > 0 for x, y \in \mathbb{Z} \setminus \{0\}, and k = k_{val}, then
  6k+1 = (6x+1)(6y+1). Since |6x+1|, |6y+1| \ge 5, 6k+1 is composite.
• Model Prediction for 6k-1 Compositeness:
  If k_{val} = 6xy+x+y < 0 for x, y \in \mathbb{Z} \setminus \{0\}, and k = -k_{val} (so k>0), then
  6k_{val}+1 = (6x+1)(6y+1) \implies 1-6k = (6x+1)(6y+1) \implies 6k-1 = |(6x+1)(6y+1)|. This is composite.
• The Unified Composite-Indicating Index Set S_c:
  S_c = \{ |6xy+x+y| \mid x, y \in \mathbb{Z} \setminus \{0\} \}.
  If k \in S_c, the model predicts compositeness for at least one of 6k-1 or 6k+1.

4. Rigorous Restatement of the Twin Prime Conjecture

For (6k-1, 6k+1) to be a twin prime pair, both numbers must be prime, which necessitates k \notin S_c. The Twin Prime Conjecture is therefore equivalent to:

The set \mathbb{Z}^+ \setminus S_c is infinite.

5. Complete Characterization of S_c

Theorem: A positive integer k belongs to S_c if and only if at least one of 6k-1 or 6k+1 is composite, with all its prime factors being greater than 3.

Proof:

Part 1: If k \in S_c, then at least one of 6k-1 or 6k+1 is composite with prime factors > 3.
This follows from Section 3. The factors (6x+1) and (6y+1) are \equiv \pm 1 \pmod 6, ensuring they are not divisible by 2 or 3, and their magnitudes are \ge 5. Thus, any composite number formed (6k+1 or 6k-1) has prime factors > 3.

Part 2: If 6k+1 or 6k-1 is composite with all prime factors > 3, then k \in S_c.

• Lemma: Any prime p > 3 satisfies p \equiv \pm 1 \pmod 6. (Standard proof omitted for brevity, based on divisibility by 2 and 3).

Consider a positive integer k.

• Subcase A: N = 6k+1 is composite with all prime factors p_i > 3.
  Let N = P \cdot Q where P, Q > 1. By the Lemma, P, Q \equiv \pm 1 \pmod 6. For N \equiv 1 \pmod 6, we have two possibilities for P,Q:
  1. P = 6a+1 and Q = 6b+1. For P,Q > 1, a,b must be non-zero integers such that 6a+1, 6b+1 > 1. (e.g., if a,b positive, a,b \ge 1).
     6k+1 = (6a+1)(6b+1) = 36ab+6a+6b+1 \implies k = 6ab+a+b.
     With x=a, y=b (x,y \in \mathbb{Z}\setminus\{0\}), k = 6xy+x+y. Since k>0, k = |6xy+x+y| \in S_c.
  2. P = 6a-1 and Q = 6b-1. For P,Q > 1, a,b must be non-zero integers such that 6a-1, 6b-1 > 1. (e.g., if a,b positive, a,b \ge 1).
     6k+1 = (6a-1)(6b-1) = 36ab-6a-6b+1 \implies k = 6ab-a-b.
     Let x=-a, y=-b. Then x,y \in \mathbb{Z}\setminus\{0\}.
     6xy+x+y = 6(-a)(-b)+(-a)+(-b) = 6ab-a-b = k.
     Since k>0, k = |6xy+x+y| \in S_c.
• Subcase B: N = 6k-1 is composite with all prime factors p_i > 3.
  N \equiv -1 \pmod 6. For N = P \cdot Q where P, Q > 1, one factor is \equiv 1 \pmod 6 and the other \equiv -1 \pmod 6.
  1. Let P = 6a+1 and Q = 6b-1. For P,Q > 1, a,b must be non-zero integers such that 6a+1 > 1 (so a \ge 1 if a is positive, or a \le -1 if 6a+1 is negative and large in magnitude, etc. Assuming positive factors P,Q means 

a≥1,b≥1a≥1,b≥1

) and 6b-1 > 1 (so 

b≥1b≥1

).
6k-1 = (6a+1)(6b-1) = 36ab-6a+6b-1 \implies k = 6ab-a+b.
Let x=a and y_0=-b. Then x,y_0 \in \mathbb{Z}\setminus\{0\}.
Consider k_{val} = 6xy_0+x+y_0 = 6a(-b)+a+(-b) = -6ab+a-b.
Lemma for sign of k_{val}: If 6a+1 > 1 and 6b-1 > 1 (assuming a,b chosen for positive P,Q), then a \ge 1 and b \ge 1.
Then k_{val} = a(1-6b)-b. Since b \ge 1, 1-6b \le -5. So a(1-6b) \le -5a.
Thus, k_{val} \le -5a-b$. Sincea,b \ge 1,-5a-b < 0. Sok_{val} < 0. Then-(k_{val}) = -(-6ab+a-b) = 6ab-a+b = k. Sok = |6xy_0+x+y_0| \in S_c`.

• Let P = 6a-1 and Q = 6b+1. For P,Q > 1, assume 

a≥1,b≥1a≥1,b≥1

.
6k-1 = (6a-1)(6b+1) = 36ab+6a-6b-1 \implies k = 6ab+a-b.
Let x=-a and y_0=b. Then x,y_0 \in \mathbb{Z}\setminus\{0\}.
Consider k_{val} = 6xy_0+x+y_0 = 6(-a)(b)+(-a)+b = -6ab-a+b.
Lemma for sign of k_{val}: If a \ge 1 and b \ge 1.
k_{val} = b(1-6a)-a. Since a \ge 1, 1-6a \le -5. So b(1-6a) \le -5b.
Thus, k_{val} \le -5b-a$. Sincea,b \ge 1,-5b-a < 0. Sok_{val} < 0. Then-(k_{val}) = -(-6ab-a+b) = 6ab+a-b = k. Sok = |6xy_0+x+y_0| \in S_c`.

This theorem establishes that S_c correctly and completely identifies all k for which 6k \pm 1 is composite due to factors \equiv \pm 1 \pmod 6.

6. Remarks on the Density of S_c

The set S_c is generated by |6xy+x+y|. Heuristic arguments, computational data (e.g., for N=10000, |S_c \cap [1,N]|/N \approx 0.9866), and connections to the density of values of binary forms suggest S_c has an asymptotic natural density of 1. Consequently, \mathbb{Z}^+ \setminus S_c would have density 0. This is consistent with the rarity of twin primes. The Twin Prime Conjecture, in this formulation, posits that this density 0 set \mathbb{Z}^+ \setminus S_c is nonetheless infinite.

7. Examples

| k | k = |6xy+x+y|? (x,y \neq 0) Example | k \in S_c? | 6k-1 | 6k+1 | Twin Prime Pair? |
|—–|———————————————————–|————–|————|————|——————|
| 1 | No integer x,y \neq 0 solutions for |6xy+x+y|=1 | No | 5 (Prime) | 7 (Prime) | Yes |
| 2 | No integer x,y \neq 0 solutions for |6xy+x+y|=2 | No | 11 (Prime) | 13 (Prime) | Yes |
| 3 | No integer x,y \neq 0 solutions for |6xy+x+y|=3 | No | 17 (Prime) | 19 (Prime) | Yes |
| 4 | Yes (x=-1, y=-1 \implies 6xy+x+y=4) | Yes | 23 (Prime) | 25 (Comp.) | No |
| 5 | No integer x,y \neq 0 solutions for |6xy+x+y|=5 | No | 29 (Prime) | 31 (Prime) | Yes |
| 6 | Yes (x=1, y=-1 \implies 6xy+x+y=-6 \implies |-6|=6) | Yes | 35 (Comp.) | 37 (Prime) | No |

8. Conclusion and Potential Avenues for Addressing the Twin Prime Conjecture

This restatement frames TPC as: \mathbb{Z}^+ \setminus S_c is infinite. This may offer avenues for investigation:

1. Modular Constraint Analysis (Constructive Non-Membership):
   • If k = |6xy+x+y|, then k \equiv |x+y| \pmod 3.
   • If k \equiv 2 \pmod 3, then |x+y| \equiv 2 \pmod 3. This implies x+y \equiv \pm 2 \pmod 3. Possible pairs for (x \pmod 3, y \pmod 3) are (0,2), (1,1), (2,0) and their permutations.
   • Strategy: Systematically analyze which k \pmod m values can arise from |6xy+x+y| \pmod m for various moduli m. Intersecting these constraints could identify classes of k that cannot be in S_c. Proving such a class is infinite would prove TPC. For instance, one might attempt to prove: Infinitely many k \equiv 2 \pmod 3 (and possibly other congruences) satisfy k \notin S_c.
2. Analysis of the Diophantine Equation K = |6xy+x+y|:
   • The core is proving non-representation for infinitely many K. This involves studying the range of P(x,y)=6xy+x+y over \mathbb{Z}\setminus\{0\} \times \mathbb{Z}\setminus\{0\}.
3. Connection to Analytic Estimates (e.g., Brun’s Constant):
   • The sum B_2 = \sum_{(p, p+2)} (1/p + 1/(p+2)) can be written as \sum_{k \in \mathbb{Z}^+ \setminus S_c} (1/(6k-1) + 1/(6k+1)). The convergence of this sum (to a positive value) implies \mathbb{Z}^+ \setminus S_c is infinite. While Brun proved convergence, directly evaluating this sum or proving infinitude from this angle is hard.
4. Properties of the Generating Function P(x,y) = 6xy+x+y:
   • Rewriting: 6P(x,y)+1 = (6x+1)(6y+1). This links the values k \in S_c (or rather 6k+1) to numbers with at least two factors of the form 6m+1 (or 6m-1 by sign changes).
   • Studying the set { (6x+1)(6y+1)-1 \mid x,y \in \mathbb{Z}\setminus\{0\} } / 6 and its absolute values.
5. Relating to Quadratic Forms and Class Field Theory:
   • The term (6x+1)(6y+1) is central. While not directly a standard binary quadratic form in x,y, the structure is reminiscent. Exploring if techniques from the theory of representation of numbers by quadratic forms (e.g., density theorems, class number relations) can be adapted.
6. Constructive Algorithm for \mathbb{Z}^+ \setminus S_c:
   • Algorithm: For N=1 \dots \text{LIMIT}:
     1. Initialize Possible_k = \{1, \dots, N\}.
     2. For |x|, |y| up to \approx \sqrt{N/6}: calculate val = |6xy+x+y|. If val \le N, remove val from Possible_k.
     3. The remaining elements in Possible_k are (\mathbb{Z}^+ \setminus S_c) \cap [1,N].
   • Studying the distribution and properties of these remaining k values may yield patterns.

This algebraic formulation, by focusing on the structure of S_c, offers a concrete framework. The challenge is to prove that the “gaps” \mathbb{Z}^+ \setminus S_c are infinite, leveraging the specific algebraic nature of |6xy+x+y|.

Addendum to Conclusion: Further Reflections on the k-Index Filtering Restatement

The restatement of the Twin Prime Conjecture through the k-index set S_c = { |6xy+x+y| | x,y ∈ ℤ{0} } offers more than just an algebraic equivalence. It fundamentally reshapes the problem’s landscape, potentially mitigating longstanding challenges:

1. Unified Criterion for Non-Twin Primality: The S_c framework elegantly collapses the conditions for the compositeness of 6k-1 and 6k+1 into a single test: is k ∈ S_c? If it is, the pair (6k-1, 6k+1) is not a twin prime. If k ∉ S_c, then both 6k-1 and 6k+1 are prime, forming a twin prime pair. This unification stems from the profound insight that the k-indices leading to compositeness in either 6k-1 or 6k+1 can be generated from the absolute value of the single polynomial expression P(x,y) = 6xy+x+y. The sign of P(x,y) before taking the absolute value implicitly directs its relevance to either 6k+1 (if P(x,y) > 0) or 6k-1 (if P(x,y) < 0, leading to k = -P(x,y)).
2. Sidestepping the Parity Problem: Traditional sieve methods often encounter the “parity problem,” struggling to distinguish numbers with an odd number of prime factors (like primes) from those with an even number. The S_c formulation bypasses this. It doesn’t “sieve” in the classical sense nor does it rely on counting prime factors. Instead, it provides a direct algebraic test: if k ∈ S_c, then at least one member of the pair (6k-1, 6k+1) is composite because it possesses a specific two-factor structure (derivable from (6a±1)(6b±1)). Whether that number has more than two prime factors is irrelevant for its classification as composite and for k’s inclusion in S_c. The challenge is not in the subtle counting of factors but in determining membership in S_c.
3. Transformation of the Core Challenge: By virtue of the theorem characterizing S_c, the Twin Prime Conjecture is transformed into the question: “Is the set ℤ⁺ \ S_c infinite?” This is a problem about the value distribution (or range) of the polynomial |6xy+x+y| over non-zero integers x,y. While proving that infinitely many positive integers are not in the image of this polynomial is a formidable task in Diophantine analysis, it is a distinctly different challenge from those faced in classical sieve theory. It shifts the focus from combinatorial estimates and error term management to the algebraic and number-theoretic properties of polynomial value sets.
4. Implicit Symmetries and Balance: The structure of S_c, rooted in |6xy+x+y|, reflects a deep symmetry. The underlying polynomial P(x,y) = 6xy+x+y and its connection to (6x+1)(6y+1) is central to forming composites in both 6k-1 and 6k+1 contexts. While this framework doesn’t explicitly prove an equivalence in the densities of different composite types (e.g., “AB in A” vs “AA+BB in B” as explored in related symmetric arguments), it unifies the source of the k-indices that lead to any such compositeness within the 6k±1 framework. This suggests an inherent balance in how these k-indices are generated, regardless of which of 6k-1 or 6k+1 becomes composite.

In essence, this k-index filtering restatement provides a more direct, algebraic criterion for twin primality, potentially circumventing some of the analytical hurdles like the parity problem that have historically constrained progress. The task is now to understand the “gaps” left by the values of |6xy+x+y|, a pursuit that may benefit from different mathematical tools and perspectives.