gpt pro 5.2 attempt on Math-5369, version 2

Round 2: covering-system reduction and a Wieferich-type obstruction

1) ROUND-2 OBJECTIVE

Path pursued: (C) obstruction track (with computational extension).

Round 1 proved only a fixed-$e$ positive-density result (Lemma 11.2), leaving the global quantifier gap \[ \forall\,\text{odd }n\ge 3\ \exists e\ge 0:\ n-2^e\text{ is squarefree}. \] In Round 2 I focus on narrowing the space of possible counterexamples by extracting necessary algebraic structure that any counterexample must satisfy. Concretely, I reduce counterexamples to a covering system in exponent space governed by the multiplicative orders $\operatorname{ord}_{p^2}(2)$, and I prove a Wieferich-type obstruction to reusing a non-Wieferich prime square across multiple exponents.

I also extend the Round 1 computation (which reached $10^6$) to a substantially larger range ($10^8$) as empirical support.

2) ROUND-1 FOUNDATION USED

I rely on the following Round 1 components (without reproving them):

1. Lemma 11.1: $Q(N)=\frac{6}{\pi^2}N+O(\sqrt N)$ for the count of squarefree integers $\le N$.
2. Lemma 11.2: for each fixed $e\ge 0$, a positive proportion of odd $n$ admit $n=s+2^e$ with $s$ squarefree, with explicit constants.
3. Round 1 computation: verified that every odd $3\le n\le 10^6$ has at least one representation.

3) NEW INSIGHT / TOOL (ROUND-2)

New ingredients introduced in Round 2:

• A counterexample--to--covering-system reduction: for each odd prime $p$, the exponents $e$ for which $p^2\mid (n-2^e)$ form either the empty set or a single congruence class modulo $\operatorname{ord}_{p^2}(2)$ (Lemma 11.3). Hence a counterexample forces a covering of $\{0,1,\dots,L\}$ by such classes.
• An order-lifting dichotomy (Lemma 11.4) giving $\operatorname{ord}_{p^2}(2)\in\{\operatorname{ord}_p(2),\,p\operatorname{ord}_p(2)\}$, and the corollary that a non-Wieferich prime square cannot divide two shifts $n-2^{e_1},n-2^{e_2}$ unless $p\mid(e_1-e_2)$.
• A computational extension: verified (empirically) that every odd $3\le n\le 10^8$ has a representation.

4) ATTACK PLAN (ROUND-2)

Gap after Round 1. The only missing piece for a full proof is the universal statement that every odd $n\ge 3$ has some exponent $e$ with $n-2^e$ squarefree.

Round 2 plan. Instead of attempting uniform squarefree distribution in the sparse set $\{n-2^e\}$, I push the disproof/obstruction line:

1. Derive a structural characterization of counterexamples in terms of exponent coverings by arithmetic progressions with moduli $\operatorname{ord}_{p^2}(2)$.
2. Prove that reusing a prime square $p^2$ to cover two distinct exponents forces a strong congruence constraint on the exponent difference; in particular, if $p$ is not Wieferich base 2 then any two such exponents differ by a multiple of $p$.
3. Use this to rule out an entire class of "efficient" covering-congruence constructions: coverings that attempt to use a small number of large non-Wieferich primes to control many exponents.
4. Extend computation to further restrict where a minimal counterexample could begin.

5) WORK (ROUND-2)

5.1. Setup and notation

Fix an odd integer $n\ge 3$. Define \[ L:=L(n):=\big\lfloor \log_2(n-1)\big\rfloor, \qquad E_L:=\{0,1,\dots,L\}. \] Then $2^e\le n-1$ for all $e\in E_L$, so $n-2^e\ge 1$ for all such $e$.

Call $n$ a counterexample if \[ \forall e\in E_L:\quad n-2^e\ \text{is not squarefree}. \] Equivalently, for each $e\in E_L$ there exists a prime $p$ with $p^2\mid(n-2^e)$.

5.2. Exponents forced by a fixed prime square

Lemma 11.3 (exponent class for a fixed prime square).

Let $p$ be an odd prime and let \[ \mathcal{E}_p(n):=\{e\in\mathbb{Z}_{\ge 0}:\ p^2\mid (n-2^e)\}. \] Then $\mathcal{E}_p(n)$ is either empty, or else it is a single residue class modulo the multiplicative order \[ \operatorname{ord}_{p^2}(2):=\min\{t\ge 1: 2^t\equiv 1\pmod{p^2}\}. \] More precisely: if $e_0\in\mathcal{E}_p(n)$ then \[ \mathcal{E}_p(n)=\{e\ge 0:\ e\equiv e_0\pmod{\operatorname{ord}_{p^2}(2)}\}. \]

Proof.

Assume $\mathcal{E}_p(n)\ne\varnothing$ and fix $e_0\in\mathcal{E}_p(n)$, so $n\equiv 2^{e_0}\pmod{p^2}$. Let $e\ge 0$. Then \[ p^2\mid (n-2^e) \ \Longleftrightarrow\ 2^e\equiv n\equiv 2^{e_0}\pmod{p^2} \ \Longleftrightarrow\ 2^{e-e_0}\equiv 1\pmod{p^2}. \] Since $p$ is odd, $2$ is invertible modulo $p^2$ and the last congruence holds if and only if $\operatorname{ord}_{p^2}(2)\mid (e-e_0)$. This is exactly the claimed residue-class description. $\square$

Corollary 11.3.1 (counterexample implies an exponent covering).

If $n$ is a counterexample, then there exists a finite set of odd primes $\mathcal{P}$ and integers $a_p\ge 0$ such that \begin{align*} &\text{(i)}\quad n\equiv 2^{a_p}\pmod{p^2}\ \text{for each }p\in\mathcal{P},\\ &\text{(ii)}\quad E_L\subseteq \bigcup_{p\in\mathcal{P}}\{e\ge 0:\ e\equiv a_p\pmod{\operatorname{ord}_{p^2}(2)}\}. \end{align*} In words: the exponent interval $E_L$ is covered by residue classes modulo the orders $\operatorname{ord}_{p^2}(2)$ coming from primes whose squares divide at least one shift $n-2^e$.

Proof.

For each $e\in E_L$, choose a prime $p(e)$ with $p(e)^2\mid(n-2^e)$ and let $\mathcal{P}$ be the set of primes that occur. For each $p\in\mathcal{P}$ choose $a_p\in\mathcal{E}_p(n)$ (e.g. the smallest such exponent). Then (i) holds by definition, and (ii) follows from Lemma 11.3 because each $e\in E_L$ lies in $\mathcal{E}_{p(e)}(n)$, which is the residue class of $a_{p(e)}$ modulo $\operatorname{ord}_{p(e)^2}(2)$. $\square$

5.3. Order lifting and the Wieferich obstruction

Lemma 11.4 (order modulo $p^2$ is $r$ or $pr$).

Let $p$ be an odd prime and set $r:=\operatorname{ord}_p(2)$. Then \[ \operatorname{ord}_{p^2}(2)\in\{r,\ pr\}. \] More precisely:

• if $2^r\equiv 1\pmod{p^2}$, then $\operatorname{ord}_{p^2}(2)=r$;
• if $2^r\not\equiv 1\pmod{p^2}$, then $\operatorname{ord}_{p^2}(2)=pr$.

Proof.

Since $r=\operatorname{ord}_p(2)$, we have $2^r\equiv 1\pmod p$. Write \[ 2^r = 1 + pt \] for some integer $t$.

If $p\mid t$ then $2^r\equiv 1\pmod{p^2}$ and therefore the order of $2$ modulo $p^2$ divides $r$. On the other hand, reducing mod $p$ shows the order modulo $p^2$ must be a multiple of $r$. Hence it is exactly $r$.

If $p\nmid t$ then $2^r\not\equiv 1\pmod{p^2}$, so the order modulo $p^2$ is strictly larger than $r$ but still a multiple of $r$. Now consider \[ 2^{pr} = (2^r)^p = (1+pt)^p. \] By the binomial theorem, \[ (1+pt)^p = 1 + p(pt) + \binom{p}{2}(pt)^2+\cdots, \] and every term beyond the first two is divisible by $p^3$ (since $(pt)^2$ already contributes $p^2$ and $\binom{p}{2}$ contributes another factor $p$). In particular, \[ (1+pt)^p \equiv 1\pmod{p^2}, \] so $2^{pr}\equiv 1\pmod{p^2}$ and therefore $\operatorname{ord}_{p^2}(2)$ divides $pr$. Since it is a multiple of $r$ and strictly larger than $r$, the only possibility is $pr$. $\square$

Corollary 11.4.1 (Wieferich characterization via $\operatorname{ord}_{p^2}(2)$).

An odd prime $p$ is Wieferich base $2$ (i.e. $2^{p-1}\equiv 1\pmod{p^2}$) if and only if \[ p\nmid \operatorname{ord}_{p^2}(2). \] Equivalently: $p$ is Wieferich base $2$ if and only if $\operatorname{ord}_{p^2}(2)=\operatorname{ord}_p(2)$.

Proof.

If $p\nmid \operatorname{ord}_{p^2}(2)$, then $\operatorname{ord}_{p^2}(2)$ divides $\varphi(p^2)=p(p-1)$ but not $p$, hence must divide $p-1$. Thus $2^{p-1}\equiv 1\pmod{p^2}$.

Conversely, if $2^{p-1}\equiv 1\pmod{p^2}$ then the order $\operatorname{ord}_{p^2}(2)$ divides $p-1$, so it is not divisible by $p$. Finally, the equivalence with $\operatorname{ord}_{p^2}(2)=\operatorname{ord}_p(2)$ follows immediately from Lemma 11.4. $\square$

Corollary 11.4.2 (obstruction to reusing a non-Wieferich prime square).

Let $p$ be an odd prime that is not Wieferich base $2$. If $p^2\mid (n-2^{e_1})$ and $p^2\mid (n-2^{e_2})$ with $e_1\ne e_2$, then \[ p\mid (e_1-e_2). \]

Proof.

By Lemma 11.3, $e_1\equiv e_2\pmod{\operatorname{ord}_{p^2}(2)}$. By Corollary 11.4.1, since $p$ is not Wieferich we have $p\mid \operatorname{ord}_{p^2}(2)$. Hence $p\mid (e_1-e_2)$. $\square$

Corollary 11.4.3 (quantitative reuse bound in $E_L$).

If $p$ is not Wieferich base $2$, then \[ \big|\mathcal{E}_p(n)\cap E_L\big|\le 1+\Big\lfloor\frac{L}{p}\Big\rfloor. \] In particular, if $p>L$ and $p$ is not Wieferich, then $p^2$ can divide at most one of the shifts $n-2^e$ with $e\in E_L$.

Proof.

By Lemma 11.3, $\mathcal{E}_p(n)$ is a residue class modulo $m:=\operatorname{ord}_{p^2}(2)$. By Corollary 11.4.1, $p\mid m$. Thus any two distinct elements of $\mathcal{E}_p(n)$ differ by at least $p$. An interval of length $L$ contains at most $1+\lfloor L/p\rfloor$ integers spaced by at least $p$. $\square$

5.4. CRT reduction for counterexamples

Lemma 11.5 (reduction modulo the square-modulus of a certificate).

Assume $n$ is a counterexample. Let \[ \mathcal{P}(n):=\{\text{odd primes }p:\ \exists e\in E_L\text{ with }p^2\mid(n-2^e)\}, \qquad M:=\prod_{p\in\mathcal{P}(n)} p^2. \] Let $n_0$ be the least positive residue congruent to $n$ modulo $M$. Then $n_0$ is odd and satisfies: \[ \forall e\in E_{L_0}:\ \exists p\in\mathcal{P}(n)\text{ such that }p^2\mid(n_0-2^e), \qquad L_0:=\big\lfloor \log_2(n_0-1)\big\rfloor. \] In particular, $n_0$ is itself a counterexample (possibly with a smaller exponent range). Consequently, if any counterexample exists, there is one with \[ n < \prod_{p\in\mathcal{P}(n)} p^2. \]

Proof.

Since $M$ is odd, reducing modulo $M$ preserves parity, so $n_0$ is odd. Fix $e\in E_{L_0}$. Then $2^e\le n_0-1\le n-1$, hence $e\in E_L$. Because $n$ is a counterexample, there exists some odd prime $p\in\mathcal{P}(n)$ with $p^2\mid (n-2^e)$. But $n_0\equiv n\pmod{p^2}$ (since $p^2\mid M$), so \[ n_0-2^e\equiv n-2^e\equiv 0\pmod{p^2}, \] which proves the displayed property for $n_0$. Finally, choosing $n_0$ as the least positive residue implies $n_0<M$. If $n$ was already $<M$ this gives no improvement; otherwise it strictly reduces the size of the counterexample. $\square$

5.5. Computational extension (empirical)

Empirical only. I extended the Round 1 brute-force verification from $10^6$ to $10^8$.

Method.

For a given bound $N$:

1. Compute an array $\mathrm{sqfree}[m]$ for $0\le m\le N$ by marking all multiples of $p^2$ (for primes $p\le \sqrt N$) as non-squarefree.
2. For odd $n=2i+1$, mark $n$ as representable if any of the following hold:
   • $e=0$: $\mathrm{sqfree}[n-1]=\mathrm{sqfree}[2i]$;
   • $e\ge 1$: $\mathrm{sqfree}[n-2^e]$.
   This is implemented efficiently on odd indices by shifting boolean slices by $2^{e-1}$.

Result.

For $N=10^8$ the only odd failure is $n=1$; in particular every odd $3\le n\le 10^8$ admits $n=s+2^e$ with $s$ squarefree.

Additional empirical data (Wieferich primes up to $2\cdot 10^5$).

A direct check of primes $p\le 200{,}000$ shows that the only primes satisfying $2^{p-1}\equiv 1\pmod{p^2}$ in this range are \[ 1093\ \text{and}\ 3511. \] (These are exactly the primes $p$ in this range for which $p\nmid\operatorname{ord}_{p^2}(2)$, consistent with Corollary 11.4.1.)

6) ADVERSARIAL VERIFICATION

Quantifier hygiene. Throughout I use $L=\lfloor\log_2(n-1)\rfloor$ so that $n-2^e\ge 1$ for all $e\in E_L$; this avoids the common pitfall of allowing $2^e=n$ which would force $s=0$.

Prime $p=2$. Lemma 11.3 is stated only for odd primes because it uses invertibility of $2$ modulo $p^2$. This omission is harmless here: for $e\ge 2$ the shift $n-2^e$ is odd, hence never divisible by $4$; and for $e\in\{0,1\}$ the squarefree constraint can be checked directly (it is an even/odd parity issue already emphasized in Round 1).

Lemma 11.3 robustness. The step $2^e\equiv 2^{e_0}\pmod{p^2}\iff 2^{e-e_0}\equiv 1\pmod{p^2}$ is valid because $\gcd(2,p^2)=1$ for odd $p$. No hidden assumption on $e-e_0$ being positive is needed: if $e<e_0$ we interpret $2^{e-e_0}$ modulo $p^2$ via inverses, equivalently applying the argument to $|e-e_0|$.

Lemma 11.4 check. The only delicate point is showing that in the $p\nmid t$ case, the order cannot be a proper divisor of $pr$ other than $r$. This is settled because any order modulo $p^2$ must be a multiple of $r$ (by reduction mod $p$), and $p$ is prime, so the only multiples of $r$ dividing $pr$ are $r$ and $pr$.

Lemma 11.5 check. The reduction $n\mapsto n_0$ only asserts counterexample behavior up to $L_0=\lfloor\log_2(n_0-1)\rfloor$, not up to the original $L$. This is important: if $n_0\ll n$ then there is no requirement to certify large exponents $e$ for $n_0$. The proof correctly restricts to $e\le L_0$ and uses $L_0\le L$.

Computation sanity. The sieve marks $m$ as non-squarefree iff it is divisible by $p^2$ for some prime $p\le \sqrt N$; this is exact. The representability marking is purely logical OR of exact squarefree tests and hence cannot create false positives. As a cross-check, the code recovers the Round 1 result (no failures for odd $3\le n\le 10^6$) as a special case.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 does not close the universal quantifier gap. However it adds genuine structural constraints on any hypothetical counterexample:

• A counterexample forces a covering of the exponent interval $E_L$ by residue classes modulo $\operatorname{ord}_{p^2}(2)$ (Corollary 11.3.1).
• Reusing a non-Wieferich prime square across two exponents forces their difference to be divisible by $p$ (Corollary 11.4.2), giving the explicit reuse bound $|\mathcal{E}_p(n)\cap E_L|\le 1+\lfloor L/p\rfloor$ (Corollary 11.4.3).
• Any counterexample can be reduced modulo the product of its certifying prime squares without losing the counterexample property for its (possibly smaller) exponent range (Lemma 11.5).

Empirically, the statement holds for all odd $3\le n\le 10^8$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 55%

9) REFERENCES

No external theorems beyond standard facts about multiplicative order and elementary binomial expansions were invoked. All computations were performed directly (NumPy/SymPy).