gpt pro 5.4 attempt on Math-5369

Problem #11 -- Round 5: many integers have no squarefree shift by a small power of $2$

1) ROUND-5 OBJECTIVE

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

Round 4 showed that there are infinitely many odd integers $n$ for which every exponent \[ 0\le e\le c\,\frac{\log n}{\log\log n} \] is bad (that is, $n-2^e$ is not squarefree), but only in a one-point existence form. The most promising next step is to strengthen that obstruction from existence to abundance. The goal of this round is therefore to prove that for every fixed $c<\tfrac12$, there are many odd integers $n\le X$ whose first possible good exponent exceeds \[ c\,\frac{\log X}{\log\log X}. \] This rules out an entire class of proof strategies: any argument that only inspects exponents up to $(\tfrac12-\varepsilon)\log n/\log\log n$ cannot possibly settle the problem.

2) Round-4 FOUNDATION USED

I use the following Round 4 results without reproving them:

1. Definition of $e_{\min}(n)$: \[ e_{\min}(n):=\min\{e\ge 0: n-2^e\ge 1\text{ and }n-2^e\text{ is squarefree}\}, \] with $e_{\min}(n)=+\infty$ if no such exponent exists.
2. Lemma 11.8 (corrected finite-range CRT killing): for every finite exponent set $E\subset\mathbb Z_{\ge 0}$ there are infinitely many odd integers $n>2^{\max E}$ such that $n-2^e$ is not squarefree for every $e\in E$.
3. Theorem 11.10 (primorial-scale finite-range killing): using the first $K+1$ odd primes gives an odd integer $n$ with $2^K<n<4((K+2)!)^4$ and $n-2^e$ non-squarefree for all $0\le e\le K$.

3) NEW INSIGHT / TOOL (ROUND-5)

The new ingredient is that the Round 4 CRT construction does not merely produce one integer for each $K$; it produces a full arithmetic progression of such integers. Estimating the modulus of that progression sharply enough shows:

• for every fixed $0<c<\tfrac12$, the set of odd integers $n\le X$ with \[ e_{\min}(n)>c\,\frac{\log X}{\log\log X} \] has size at least $X^{1-2c-o(1)}$;
• equivalently, \[ \max_{\substack{n\le X\\ n\text{ odd}}} e_{\min}(n)\ge \left(\frac12-o(1)\right)\frac{\log X}{\log\log X}. \]

Thus Round 5 upgrades the Round 4 infinite-sequence obstruction to a quantitative counting theorem.

4) ATTACK PLAN (ROUND-5)

The exact gap left after Round 4 is that the obstruction was only existential: it showed that some odd integers force large good exponents, but it did not measure how often this happens.

The plan is:

1. strengthen the Round 4 CRT argument into a progression lemma: for each $K$, one residue class modulo a modulus $M_K$ forces every shift $n-2^e$ with $0\le e\le K$ to be non-squarefree;
2. bound $M_K$ from above by \[ \log M_K\le 2K\log K+O(K\log\log K); \]
3. count how many terms of that residue class lie below $X$ when \[ K=\left\lfloor c\frac{\log X}{\log\log X}\right\rfloor, \qquad 0<c<\frac12. \] This produces a lower bound of order $X^{1-2c-o(1)}$.

5) WORK (ROUND-5)

Throughout this round, all unadorned logarithms are natural logarithms.

5.1. A progression version of the finite-range CRT obstruction

Let $p_m$ denote the $m$-th prime, so $p_1=2$, $p_2=3$, $p_3=5$, and so on. For $K\ge 0$, define \[ M_K:=2\prod_{m=2}^{K+2} p_m^2. \]

Lemma 11.12 (CRT progression certificate). For every integer $K\ge 0$, there exists a residue class $a_K\pmod{M_K}$ such that for every integer $n\equiv a_K\pmod{M_K}$ one has \[ p_{e+2}^2\mid (n-2^e)\qquad (0\le e\le K). \] Consequently, every such $n$ with $n>2^K$ satisfies \[ e_{\min}(n)\ge K+1. \]

Proof. Consider the congruence system \[ n\equiv 1\pmod 2, \qquad n\equiv 2^e\pmod{p_{e+2}^2} \qquad (0\le e\le K). \] The moduli $2,p_2^2,p_3^2,\dots,p_{K+2}^2$ are pairwise coprime, so by the Chinese remainder theorem there is a unique residue class $a_K\pmod{M_K}$ satisfying all congruences.

If $n\equiv a_K\pmod{M_K}$, then in particular $n\equiv 2^e\pmod{p_{e+2}^2}$ for each $0\le e\le K$, hence \[ p_{e+2}^2\mid (n-2^e) \qquad (0\le e\le K). \] Therefore each $n-2^e$ is not squarefree. If moreover $n>2^K$, then $n-2^e\ge 1$ for every $0\le e\le K$, so all exponents $e\le K$ are genuinely inadmissible as witnesses, and hence $e_{\min}(n)\ge K+1$. $\square$

5.2. Size of the progression modulus

Lemma 11.13 (upper bound for $M_K$). As $K\to\infty$, \[ \log M_K\le 2K\log K+O(K\log\log K). \] Also, \[ M_K>2^K \] for every $K\ge 0$.

Proof. The lower bound is immediate: \[ M_K=2\prod_{m=2}^{K+2}p_m^2\ge 2\cdot 3^{2(K+1)}>2^K. \]

For the upper bound, use the explicit Rosser--Schoenfeld estimate (already used in Round 4): for all sufficiently large $m$, \[ p_m<m(\log m+\log\log m). \] Hence for all sufficiently large $m$, \[ p_m\le 3m\log m, \] so \[ \log p_m\le \log m+\log\log m+O(1). \] Therefore \[ \log M_K =\log 2+2\sum_{m=2}^{K+2}\log p_m \le C+2\sum_{m=2}^{K+2}\log m+2\sum_{m=2}^{K+2}\log\log m, \] for some absolute constant $C$.

Now \[ \sum_{m=2}^{K+2}\log m=\log((K+2)!)=K\log K+O(K) \] by Stirling's formula, while trivially \[ \sum_{m=2}^{K+2}\log\log m\le (K+1)\log\log(K+2)=O(K\log\log K). \] Combining these estimates gives \[ \log M_K\le 2K\log K+O(K\log\log K), \] as claimed. $\square$

5.3. A counting theorem for large least good exponent

For $0<c<\tfrac12$ and $X\ge 3$, define \[ \mathcal B_c(X):=\left\{n\le X:\ n\text{ odd and }e_{\min}(n)>c\frac{\log X}{\log\log X}\right\}. \]

Theorem 11.14 (many odd integers have no good small exponent). Fix $0<c<\tfrac12$. Then \[ |\mathcal B_c(X)|\ge X^{1-2c-o(1)} \qquad (X\to\infty). \] Equivalently: for every fixed $0<c<\tfrac12$, there are at least $X^{1-2c-o(1)}$ odd integers $n\le X$ such that every exponent \[ 0\le e\le c\frac{\log X}{\log\log X} \] is bad.

Proof. Fix $0<c<\tfrac12$ and let \[ K:=\left\lfloor c\frac{\log X}{\log\log X}\right\rfloor. \] Apply Lemma 11.12 to this $K$, obtaining a residue class $a_K\pmod{M_K}$ with the property that every $n\equiv a_K\pmod{M_K}$ satisfies \[ p_{e+2}^2\mid (n-2^e) \qquad (0\le e\le K). \] Let $b_K$ be the smallest positive integer congruent to $a_K\pmod{M_K}$ and exceeding $2^K$. Because $0<a_K<M_K$ and $M_K>2^K$ by Lemma 11.13, we have \[ b_K<2M_K. \] Then every integer of the form \[ n=b_K+tM_K\qquad (t\ge 0) \] is odd, satisfies $n>2^K$, and obeys $e_{\min}(n)\ge K+1$ by Lemma 11.12. Hence all such $n\le X$ belong to $\mathcal B_c(X)$.

The number of such integers is at least \[ \left\lfloor\frac{X-b_K}{M_K}\right\rfloor+1\ge \frac{X}{M_K}-2. \] So it remains to bound $M_K$.

By Lemma 11.13, \[ \log M_K\le 2K\log K+O(K\log\log K). \] Since \[ K=c\frac{\log X}{\log\log X}+O(1), \] we have \[ K\log K=c\log X+O\!\left(\frac{\log X\,\log\log\log X}{\log\log X}\right), \] and similarly $K\log\log K=o(\log X)$. Therefore \[ \log M_K\le (2c+o(1))\log X, \] that is, \[ M_K\le X^{2c+o(1)}. \] Substituting into the counting bound yields \[ |\mathcal B_c(X)|\ge X^{1-2c-o(1)}, \] which proves the theorem. $\square$

5.4. Consequences

Corollary 11.14.1 (maximal order lower bound). As $X\to\infty$, \[ \max_{\substack{n\le X\\ n\text{ odd}}} e_{\min}(n) \ge \left(\frac12-o(1)\right)\frac{\log X}{\log\log X}. \]

Proof. Let $\varepsilon>0$ be fixed. Apply Theorem 11.14 with $c=\tfrac12-\varepsilon$. Then for all sufficiently large $X$, \[ |\mathcal B_{1/2-\varepsilon}(X)|\ge X^{2\varepsilon-o(1)}>0, \] so there exists an odd integer $n\le X$ with \[ e_{\min}(n)>\left(\frac12-\varepsilon\right)\frac{\log X}{\log\log X}. \] Taking the maximum over odd $n\le X$ gives \[ \max_{\substack{n\le X\\ n\text{ odd}}} e_{\min}(n) \ge \left(\frac12-\varepsilon\right)\frac{\log X}{\log\log X} \] for all sufficiently large $X$. Since $\varepsilon>0$ is arbitrary, the stated $(\tfrac12-o(1))$ lower bound follows. $\square$

Corollary 11.14.2 (proof strategies ruled out). Fix any $\varepsilon>0$. No proof strategy that, for each odd $n$, only inspects exponents \[ e\le \left(\frac12-\varepsilon\right)\frac{\log n}{\log\log n} \] can prove the Erdős conjecture.

Proof. By Corollary 11.14.1, for arbitrarily large $X$ there is an odd $n\le X$ with \[ e_{\min}(n)\ge \left(\frac12-\frac\varepsilon2\right)\frac{\log X}{\log\log X}. \] Since the function $t\mapsto \log t/\log\log t$ is increasing for all sufficiently large $t$, we have \[ \frac{\log n}{\log\log n}\le \frac{\log X}{\log\log X}, \] so, for large $X$, \[ e_{\min}(n)>\left(\frac12-\varepsilon\right)\frac{\log n}{\log\log n}. \] Thus every exponent in that range is bad for this $n$, and any method confined to inspecting only those exponents must fail on infinitely many inputs. $\square$

6) ADVERSARIAL VERIFICATION

Use of Round 4. The new progression lemma is a genuine strengthening of the Round 4 CRT obstruction: Round 4 produced one integer for each $K$, whereas Lemma 11.12 identifies an entire arithmetic progression of such integers.

Admissibility of exponents. The original problem only cares about exponents $e$ with $n-2^e\ge 1$. This is why the proof explicitly introduces $b_K>2^K$. Without that step, the CRT congruences alone would not guarantee relevance to the representation problem.

Counting bound. The estimate \[ \left\lfloor\frac{X-b_K}{M_K}\right\rfloor+1\ge \frac{X}{M_K}-2 \] is valid because $b_K<2M_K$. This is exactly where the bound $M_K>2^K$ is used.

Asymptotic regime. The theorem requires $c<\tfrac12$. This is not an artefact of a sloppy estimate; it comes from the exponent $2c$ in the bound $M_K\le X^{2c+o(1)}$. The present one-prime-per-exponent CRT method does not by itself cross the natural barrier $c=\tfrac12$.

Monotonicity in Corollary 11.14.2. The function $t\mapsto \log t/\log\log t$ is increasing for all sufficiently large $t$, so replacing $X$ by a smaller $n\le X$ is legitimate there.

No conflict with Round 2's Wieferich obstruction. The present construction still uses one distinct prime square per exponent. Therefore it does not contradict, bypass, or weaken Round 2's structural results about reusing a non-Wieferich prime across multiple exponents. Rather, it isolates what can already be forced without any reuse at all.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 5 strictly strengthens Round 4 in a mathematically substantive way:

1. it upgrades the finite-range CRT obstruction from an existence theorem to an arithmetic-progression theorem (Lemma 11.12);
2. it proves a quantitative counting statement: for every fixed $0<c<\tfrac12$, \[ \#\left\{n\le X:\ n\text{ odd and }e_{\min}(n)>c\frac{\log X}{\log\log X}\right\}\ge X^{1-2c-o(1)}; \]
3. it yields the sharp obstruction, within this proof paradigm, that \[ \max_{n\le X\atop n\text{ odd}} e_{\min}(n) \ge \left(\frac12-o(1)\right)\frac{\log X}{\log\log X}. \]

Thus any eventual proof of Problem #11 must cope not merely with isolated integers, but with many integers whose first good exponent is forced above $(\tfrac12-o(1))\log n/\log\log n$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 78%

9) REFERENCES

• J. B. Rosser and L. Schoenfeld, Approximate formulas for some functions of prime numbers, Illinois J. Math. 6 $1962$, 64--94. Used for the explicit upper bound $p_m<m(\log m+\log\log m)$.
• Stirling's formula for $\log((K+2)!)=K\log K+O(K)$.