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Math-5369-A6 · revision r1 · on Math-5369

gpt pro 5.2 attempt on Math-5369, version 3

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Contents
  1. Round 3: finite-range CRT obstructions and a lower bound on the least "good" exponent
  2. 1) ROUND-3 OBJECTIVE
  3. 2) Round-2 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-3)
  5. 4) ATTACK PLAN (ROUND-3)
  6. 5) WORK (ROUND-3)
  7. 5.1. A CRT construction killing any prescribed finite set of exponents
  8. 5.2. Quantifying the obstruction: a lower bound on the least possible good exponent
  9. 6) ADVERSARIAL VERIFICATION
  10. 7) FINAL
  11. 8) COMPLETION ESTIMATE (MANDATORY)
  12. 9) REFERENCES

Round 3: finite-range CRT obstructions and a lower bound on the least "good" exponent

1) ROUND-3 OBJECTIVE

Path pursued: (C) obstruction / correction of proof-strategy.

Round 2 reduced any genuine counterexample to a covering problem on the exponent interval $E_L=\{0,1,\dots,L\}$ by residue classes modulo orders $\operatorname{ord}_{p^2}(2)$ (Corollary 11.3.1), and isolated a Wieferich-type obstruction to reusing non-Wieferich primes (Corollary 11.4.2). However, the global quantifier gap \[ \forall\ \text{odd }n\ge 3\ \exists e\ge 0:\ n-2^e\ \text{squarefree} \] remains open.

In this round I prove a sharp obstruction to several natural proof strategies: for any fixed finite set of exponents $E$, one can force all shifts $n-2^e$ ($e\in E$) to be divisible by prime squares via CRT. In particular, no approach that searches only a bounded set of exponents (independent of $n$) can ever prove the conjecture.

Moreover, using a mild prime-existence input (Bertrand's postulate), I quantify this obstruction and show that along an explicit infinite sequence of $n$, the least exponent $e$ for which $n-2^e$ could possibly be squarefree must satisfy \[ e \ \gtrsim\ \sqrt{\log_2 n}. \] This is strictly stronger than any result in Rounds 1--2, and it rules out an entire class of "small-$e$" strategies.

2) Round-2 FOUNDATION USED

I use the following Round 2 results as black boxes:
  1. Lemma 11.3: For fixed odd prime $p$, the set $\mathcal E_p(n)=\{e\ge 0: p^2\mid(n-2^e)\}$ is either empty or a single residue class modulo $\operatorname{ord}_{p^2}(2)$.
  2. Corollary 11.3.1: A counterexample $n$ forces a covering of $E_L$ by such residue classes.
  3. Lemma 11.5: Any counterexample can be reduced modulo the product of its certifying prime squares.
(While the new constructions below do not logically require Lemma 11.3, they are best interpreted as producing explicit exponent-coverings of the type isolated in Corollary 11.3.1.)

3) NEW INSIGHT / TOOL (ROUND-3)

New content introduced here:
  • Theorem 11.6 (finite-range CRT killing): for any finite set of exponents $E$, there are infinitely many odd $n$ for which every $n-2^e$ ($e\in E$) is divisible by an odd prime square.
  • Theorem 11.7 (quantitative obstruction): choosing $E=\{0,1,\dots,K\}$ and selecting primes using Bertrand's postulate yields an explicit upper bound on the constructed $n$ in terms of $K$, implying the lower bound $e_{\min}(n)\gtrsim\sqrt{\log n}$ along an infinite sequence.

4) ATTACK PLAN (ROUND-3)

Gap remaining after Round 2. We still lack a method to guarantee, for each odd $n$, some exponent $e\le \lfloor\log_2(n-1)\rfloor$ such that $n-2^e$ is squarefree.

Round 3 goal. Instead of attempting to close the universal quantifier directly, we:

  1. prove that CRT can force arbitrarily long finite blocks of exponents to be "bad" (all shifts divisible by squares),
  2. thereby rule out strategies that only test finitely many exponents independent of $n$,
  3. quantify the size of such CRT constructions to obtain a lower bound on how large the smallest "good" exponent must sometimes be.
This overcomes a key ambiguity from Round 2: the exponent-covering condition is  easy  to satisfy on a fixed finite set of exponents, so any genuine resolution must exploit the self-referential dependence $L\approx\log_2 n$.

5) WORK (ROUND-3)

5.1. A CRT construction killing any prescribed finite set of exponents

Theorem (Finite-range CRT killing): Let $E\subset\mathbb Z_{\ge 0}$ be finite. Then there exist infinitely many odd integers $n$ such that for every $e\in E$ the integer $n-2^e$ is divisible by the square of an odd prime (hence is not squarefree).

In particular, for every $K\ge 0$ there exist infinitely many odd $n$ such that \[ \forall e\in\{0,1,\dots,K\}:\quad n-2^e\ \text{is not squarefree}. \]

Proof: Choose pairwise distinct odd primes $\{p_e: e\in E\}$. Consider the system of congruences \begin{equation}\label{wsref-0-65-71-3a-43-52-54-73-79-73-74-65-6d} n\equiv 1\pmod 2,\qquad n\equiv 2^e\pmod{p_e^2}\ \ (e\in E). \end{equation} The moduli $2$ and $\{p_e^2: e\in E\}$ are pairwise coprime, so by the Chinese remainder theorem there exists an integer solution $n_0$ to (eq:CRTsystem). Any such $n_0$ is odd (by $n\equiv 1\pmod2$) and satisfies \[ p_e^2\mid (n_0-2^e)\qquad(e\in E), \] so every $n_0-2^e$ is divisible by an odd prime square and therefore is not squarefree.

All solutions are congruent modulo \[ M:=2\prod_{e\in E} p_e^2. \] Hence $n=n_0+tM$ is also a solution for every integer $t\ge 0$, producing infinitely many odd integers $n$ with the stated property. ∎

Corollary (Obstruction to finite-$e$ proof strategies): Fix $K\ge 0$. There exist infinitely many odd integers $n$ such that no representation \[ n=s+2^e\qquad(s\ \text{squarefree}) \] is possible with $e\le K$. Equivalently, any proof of the conjecture must necessarily allow the witnessing exponent $e$ to grow (in general) with $n$.
Proof: Apply Theorem Theorem (Finite-range CRT killing) with $E=\{0,1,\dots,K\}$. ∎

5.2. Quantifying the obstruction: a lower bound on the least possible good exponent

To turn Theorem Theorem (Finite-range CRT killing) into a statement relating $K$ and the size of the constructed $n$, we need a mild way to choose the primes $p_e$ with explicit size control.

Lemma (Bertrand's postulate (external input)): For every integer $m>1$ there exists a prime $p$ with \[ m<p<2m. \]
Theorem (Quantitative finite-range killing): For every integer $K\ge 0$ there exists an odd integer $n$ such that:
  1. $n-2^e$ is not squarefree for every $e\in\{0,1,\dots,K\}$;
  2. $n<2^{K^2+5K+6}$.
Consequently, along an infinite sequence of odd integers $n$ one has the lower bound \[ e_{\min}(n)\ \ge\ \sqrt{\log_2 n}\ -\ 2, \] where $e_{\min}(n)$ denotes the least exponent $e\ge 0$ for which $n-2^e$ could be squarefree (i.e. the least $e$ such that $n-2^e$ is squarefree, if such an $e$ exists; otherwise interpret $e_{\min}(n)=+\infty$).
Proof: Step 1: choose controlled primes. For each $e\in\{0,1,\dots,K\}$, apply Lemma Lemma (Bertrand's postulate (external input)) with $m=2^{e+1}$ to choose a prime $p_e$ satisfying \begin{equation}\label{wsref-0-65-71-3a-70-65-49-6e-74-65-72-76-61-6c} 2^{e+1}<p_e<2^{e+2}. \end{equation} These primes are automatically odd and are pairwise distinct because the intervals $(2^{e+1},2^{e+2})$ are disjoint as $e$ varies.

Step 2: CRT system. Apply Theorem Theorem (Finite-range CRT killing) with $E=\{0,1,\dots,K\}$ and the primes $p_e$ chosen above. Let $M=2\prod_{e=0}^K p_e^2$ and let $n_0$ be the least positive solution to the congruences \[ n\equiv 1\pmod 2,\qquad n\equiv 2^e\pmod{p_e^2}\ (0\le e\le K). \] Then $n_0<M$ and $n_0-2^e$ is divisible by $p_e^2$ for all $0\le e\le K$.

To ensure $n$ is large enough to make the subsequent inequality meaningful (and still preserve the congruences), set \[ n:=n_0+M. \] Then $M\le n<2M$ and still $p_e^2\mid(n-2^e)$ for all $0\le e\le K$. This proves $1$.

Step 3: bound the size of $n$. From (eq:peInterval) we have $p_e<2^{e+2}$ and hence $\log_2 p_e<e+2$. Therefore \[ \log_2 M = 1+2\sum_{e=0}^K \log_2 p_e < 1+2\sum_{e=0}^K (e+2) =1+2\Big(\frac{K(K+1)}{2}+2(K+1)\Big) =K^2+5K+5. \] Since $n<2M$, we obtain \[ \log_2 n<\log_2(2M)=1+\log_2 M< K^2+5K+6, \] which is equivalent to $n<2^{K^2+5K+6}$ and proves $2$.

Step 4: deduce the lower bound on $e_{\min}(n)$. By construction, all exponents $e\le K$ are "bad" (the shifts are not squarefree). Hence $e_{\min}(n)\ge K+1$ (or $e_{\min}(n)=+\infty$). From $\log_2 n< K^2+5K+6=(K+\tfrac52)^2-\tfrac14$, we have $K+\tfrac52>\sqrt{\log_2 n+\tfrac14}$ and thus \[ K+1>\sqrt{\log_2 n+\tfrac14}-\tfrac32\ge \sqrt{\log_2 n}-2. \] Therefore $e_{\min}(n)\ge \sqrt{\log_2 n}-2$ along the constructed sequence. ∎

Remark (Interpretation in the language of Round 2 coverings): Theorem Theorem (Finite-range CRT killing) produces, for any finite exponent set $E$, a "certificate" set of primes $\{p_e: e\in E\}$ such that $E$ is covered by the residue classes corresponding to the congruences $n\equiv 2^e\pmod{p_e^2}$. This is the simplest possible exponent covering: each prime square is used to kill exactly one specified exponent.

The genuine conjecture only fails if one can kill the moving interval $E_L$ where $L=\lfloor\log_2(n-1)\rfloor$. Theorem Theorem (Quantitative finite-range killing) shows that killing an initial interval $\{0,\dots,K\}$ is feasible even when $K$ grows like $\sqrt{\log n}$, hence any proof must exploit structure beyond such "one-prime-per-exponent" CRT constructions.

6) ADVERSARIAL VERIFICATION

CRT coprimality. In (eq:CRTsystem), the moduli are $2$ and $p_e^2$ for distinct odd primes $p_e$, hence pairwise coprime. CRT applies without further hypotheses.

Parity. Because the nontrivial moduli are odd, the congruences $n\equiv 2^e\pmod{p_e^2}$ do not determine parity. Adding $n\equiv 1\pmod 2$ forces $n$ odd and remains compatible.

Squarefree definition. If an odd prime square $p_e^2$ divides $n-2^e$, then $n-2^e$ is not squarefree by definition. No other hidden assumption is used.

Edge cases ($K=0,1$). For $K=0$, one can take $p_0=3$ and obtain solutions $n\equiv 1\pmod{18}$, so $n-1$ is divisible by $9$ (hence not squarefree); this matches the theorem. For $K=1$, choosing $p_0=3,p_1=5$ yields an explicit CRT system with solutions, forcing both $n-1$ and $n-2$ to be divisible by odd prime squares.

Use of Bertrand. Lemma Lemma (Bertrand's postulate (external input)) is the only external input. The disjoint-interval choice ensures distinct primes and yields the claimed size bound. If Bertrand were replaced by a weaker prime-existence bound, the qualitative Theorem Theorem (Finite-range CRT killing) would remain valid (it needs only infinitely many distinct odd primes), but the quantitative inequality in Theorem Theorem (Quantitative finite-range killing) would weaken.

Interaction with Round 2 Wieferich obstruction. The present constructions do not reuse primes across multiple exponents, so Corollary 11.4.2 is not invoked. This is consistent: Theorem Theorem (Finite-range CRT killing) deliberately chooses the least efficient covering (one prime per exponent). Any attempt to make a true counterexample must cover $E_L$ with far fewer primes, hence must reuse primes, where the Wieferich obstruction becomes relevant.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

We still do not have a proof (or disproof) of the global statement for all odd $n\ge 3$. However, Round 3 proves two new theorems that substantially sharpen the obstruction landscape:

  • For any fixed finite exponent set $E$, CRT produces infinitely many odd $n$ for which all $n-2^e$ ($e\in E$) are divisible by odd prime squares (Theorem Theorem (Finite-range CRT killing)). Hence any proof strategy confined to finitely many exponents cannot succeed.
  • Quantitatively, there exist infinitely many odd $n$ for which every $e\le \sqrt{\log_2 n}-2$ is "bad" (Theorem Theorem (Quantitative finite-range killing)). Thus the least possible exponent witnessing the conjectured representation must sometimes be at least on the order of $\sqrt{\log n}$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 62%

9) REFERENCES