gpt pro 5.4 attempt on Math-641, version 2

Erdős Problem #9 (Round 4): correcting the Round-3 certificate and extracting an explicit infinite family

1) ROUND-4 OBJECTIVE

Path chosen: (C) obstruction/correction.

Round 3 claimed a gap-free certification of one explicit integer $N_0\in A$, but that certification contains a concrete numerical flaw: several listed primes do not have the claimed multiplicative orders, and the displayed decimal value of $N_0$ is not compatible with the modulus condition $N_0\equiv 0\pmod{M_1}$.

The most promising direction is therefore to repair the numerical data, identify the minimal corrected statement, and then strengthen it. The corrected statement is stronger than Round 3: using the valid Round 2 order table, one obtains an explicit infinite family \[ \{N_\nu\}_{\nu\ge 11}\subseteq A, \] and this family satisfies the sharp size estimate \[ \#\{\nu\ge 11: N_\nu\le X\}=\log_2\log_2 X+O(1). \] Hence the corrected CRT construction yields an explicit $\gg \log\log X$ lower bound inside $A$, but it is intrinsically too sparse to prove positive upper density.

2) Round-3 FOUNDATION USED

I use the following parts of the previous rounds.

1. Round 2 / Round 3 reduction lemma. If an odd integer $n$ is neither of the form $p+2^a$ with $a\ge 1$ nor of the form $p+2^a+2^b$ with distinct $a,b\ge 1$, then $n\in A$.
2. Round 3 Fermat-divisor lemma. If $a>b\ge 1$ and $r=v_2(a-b)$, then \[ 2^{2^r}+1\mid 2^a+2^b. \]
3. Round 3 Crocker-type lemma. If \[ N=w\prod_{j=0}^{n-1}B_j\le 2^{2^n}-1,\qquad w\equiv 1\pmod{16}, \] and each $B_j>1$ divides $2^{2^j}+1$, then $N\neq p+2^a+2^b$ for all primes $p$ and all distinct $a,b\ge 1$.
4. Round 2 corrected numerical table. The exact-order table from Round 2 is the valid one; the Round 3 replacement table is not.

I do not use the erroneous numerical assertions from Round 3.

3) NEW INSIGHT / TOOL (ROUND-4)

The new ingredients are:

1. Error isolation. I prove that the Round 3 numerical certificate is invalid as written.
2. Corrected exact-order data. I revert to the Round 2 prime choices, which do satisfy the required order conditions.
3. Family construction. Instead of a single $N_0$, I define for every $\nu\ge 11$ a least positive CRT solution $N_\nu$.
4. Sparse-family theorem. I prove \[ \#\{\nu\ge 11:N_\nu\le X\}=\log_2\log_2 X+O(1), \] which both gives an explicit infinite family in $A$ and rigorously shows that this fixed CRT method cannot by itself establish $\overline d(A)>0$.

4) ATTACK PLAN (ROUND-4)

The exact gaps left after Round 3 are now:

1. The Round 3 order table must be checked entry-by-entry.
2. The Round 3 displayed decimal $N_0$ must be checked against the defining congruences.
3. After repairing the data, the construction should be upgraded from a single exceptional integer to a uniform family.
4. Finally, one should quantify how many integers this repaired construction actually produces up to $X$.

The plan is:

1. Prove that the Round 3 numerical table is wrong.
2. Replace it by the valid Round 2 table.
3. For each $\nu\ge 11$, define $M_\nu=(2^{2^\nu}-1)/G$ with fixed $G=(2^{1024}+1)/45592577$, and define $N_\nu$ by the same corrected congruence conditions.
4. Prove $N_\nu\in A$ for every $\nu\ge 11$.
5. Prove the counting law $\#\{\nu\ge 11:N_\nu\le X\}=\log_2\log_2 X+O(1)$.

5) WORK (ROUND-4)

5.1. The Round-3 numerical certificate is incorrect

Let $M_1=(2^{2048}-1)/G$ with $G=(2^{1024}+1)/45592577$ as in Round 3.

Proposition 5.1. The numerical table displayed in Round 3 is incorrect, and the displayed Round 3 decimal value of $N_0$ is not the CRT integer defined there.

Proof. A deterministic exact computation gives the following counterchecks:

• $48691=23\cdot 29\cdot 73$, so it is not prime; in particular it cannot serve as a prime with $\operatorname{ord}_{48691}(2)=45$.
• $\operatorname{ord}_{811}(2)=270\neq 48$.
• $\operatorname{ord}_{87211}(2)=54\neq 72$.
• $\operatorname{ord}_{233}(2)=29\neq 72$.
• $\operatorname{ord}_{5419}(2)=42\neq 90$.
• $\operatorname{ord}_{2521}(2)=1260\neq 120$.
• $\operatorname{ord}_{1153}(2)=288\neq 144$.
• $\operatorname{ord}_{6700417}(2)=64\neq 144$.
• $\operatorname{ord}_{1303}(2)=651\neq 180$.
• $\operatorname{ord}_{118831}(2)=11883\neq 180$.
• $\operatorname{ord}_{19231}(2)=9615\neq 360$.

Thus at least $11$ of the $28$ claimed order constraints fail.

Moreover, the decimal integer printed in Round 3 has only $202$ decimal digits, while $M_1$ has $316$ decimal digits. Hence that printed integer is strictly between $0$ and $M_1$ and therefore cannot satisfy \[ N_0\equiv 0\pmod{M_1}. \] So the Round 3 numerical certificate must be discarded. $\square$

5.2. Corrected exact-order data

We retain the same overlapping congruence system $(a_i,m_i)$ as in Rounds 2--3: \[ \begin{aligned} &(0,3),(0,5),(1,9),(1,10),(8,12),(8,15),(4,18),(7,20),(5,24),(29,30),\\ &(2,36),(14,36),(17,40),(34,45),(43,45),(13,48),(37,48),(16,60),(19,60),\\ &(26,72),(62,72),(52,90),(37,120),(49,144),(121,144),(103,180),(106,180),(229,360). \end{aligned} \] Its least common multiple is $720$, and every positive integer exponent lies in at least one of these residue classes modulo its corresponding modulus.

The corrected prime choices are exactly the Round 2 ones: \[ \begin{array}{rcl@{\qquad}rcl} (0,3)&\mapsto&7,&(34,45)&\mapsto&631,\\ (0,5)&\mapsto&31,&(43,45)&\mapsto&23311,\\ (1,9)&\mapsto&73,&(13,48)&\mapsto&97,\\ (1,10)&\mapsto&11,&(37,48)&\mapsto&673,\\ (8,12)&\mapsto&13,&(16,60)&\mapsto&61,\\ (8,15)&\mapsto&151,&(19,60)&\mapsto&1321,\\ (4,18)&\mapsto&19,&(26,72)&\mapsto&433,\\ (7,20)&\mapsto&41,&(62,72)&\mapsto&38737,\\ (5,24)&\mapsto&241,&(52,90)&\mapsto&18837001,\\ (29,30)&\mapsto&331,&(37,120)&\mapsto&4562284561,\\ (2,36)&\mapsto&37,&(49,144)&\mapsto&577,\\ (14,36)&\mapsto&109,&(121,144)&\mapsto&487824887233,\\ (17,40)&\mapsto&61681,&(103,180)&\mapsto&181,\\ &&&(106,180)&\mapsto&54001,\\ &&&(229,360)&\mapsto&168692292721. \end{array} \] Let these primes be denoted $p_1,\dots,p_{28}$ in the displayed order. Also let \[ p_{29}=8191=2^{13}-1,\qquad q=45592577,\qquad G=\frac{2^{1024}+1}{q}. \] We fix \[ c\equiv 1\pmod{8191}. \]

Finite certificate used below. A deterministic computation verifies:

1. each listed $p_i$ is prime and $\operatorname{ord}_{p_i}(2)=m_i$;
2. every residue class modulo $720$ is covered by at least one $(a_i,m_i)$;
3. $1\not\equiv p_i+2^e\pmod{8191}$ for all $1\le i\le 28$ and $0\le e\le 12$;
4. with \[ L:=16\cdot 8191\cdot\prod_{i=1}^{28}p_i, \] one has \[ L<G. \]

5.3. The corrected family $N_\nu$

For every integer $\nu\ge 11$, define \[ M_\nu:=\frac{2^{2^\nu}-1}{G}. \] This is an integer because $G\mid (2^{1024}+1)=F_{10}$ and \[ 2^{2^\nu}-1=\prod_{j=0}^{\nu-1}F_j\qquad (\nu\ge 11). \] Equivalently, \[ M_\nu=q\prod_{\substack{0\le j\le \nu-1\\ j\neq 10}}F_j. \]

Now define $N_\nu$ to be the least positive integer satisfying the simultaneous congruences \begin{align} N_\nu&\equiv 0\pmod{M_\nu},\label{wsref-0-65-71-3a-6e-75-2d-4d}\\ N_\nu&\equiv -1\pmod{16},\label{wsref-0-65-71-3a-6e-75-2d-31-36}\\ N_\nu&\equiv 1\pmod{8191},\label{wsref-0-65-71-3a-6e-75-2d-38-31-39-31}\\ N_\nu&\equiv 2^{a_i}\pmod{p_i}\qquad (1\le i\le 28).\label{wsref-0-65-71-3a-6e-75-2d-70-69} \end{align}

Lemma 5.2. For every $\nu\ge 11$, the congruence system (eq:nu-M)--(eq:nu-pi) has a unique solution modulo $LM_\nu$, and the least positive solution satisfies \[ M_\nu\le N_\nu<LM_\nu. \] In particular, \[ N_\nu<2^{2^\nu}-1. \]

Proof. We first show $\gcd(L,M_\nu)=1$.

The factors $16$ and $8191$ are coprime to $M_\nu$: clearly $M_\nu$ is odd, and if $8191\mid M_\nu$, then $8191\mid (2^{2^\nu}-1)$, which would force $\operatorname{ord}_{8191}(2)\mid 2^\nu$, i.e. $13\mid 2^\nu$, impossible.

Next fix one of the primes $p_i$. If $p_i\mid M_\nu$, then again $p_i\mid (2^{2^\nu}-1)$, so $\operatorname{ord}_{p_i}(2)\mid 2^\nu$. But $\operatorname{ord}_{p_i}(2)=m_i$, and every $m_i$ in the list has an odd factor greater than $1$. Therefore $m_i\nmid 2^\nu$, contradiction. Hence $\gcd(p_i,M_\nu)=1$ for all $i$.

So all moduli in (eq:nu-M)--(eq:nu-pi) are pairwise coprime, and the Chinese remainder theorem gives a unique residue class modulo $LM_\nu$.

The least positive solution is divisible by $M_\nu$, so $N_\nu\ge M_\nu$. Since it is the least positive element of its CRT class, we also have $N_\nu<LM_\nu\cdot L$. Because $L<G$ by the finite certificate, \[ N_\nu<LM_\nu L < M_\nu G = 2^{2^\nu}-1. \] This proves the claim. $\square$

5.4. Excluding representations $p+2^d$

Lemma 5.3. For every $\nu\ge 11$, there are no prime $p$ and integer $d\ge 1$ such that \[ N_\nu=p+2^d. \]

Proof. Fix $\nu\ge 11$ and $d\ge 1$. Because the residue system $(a_i,m_i)$ is overlapping, there exists an index $i$ such that \[ d\equiv a_i\pmod{m_i}. \] Since $\operatorname{ord}_{p_i}(2)=m_i$, this implies \[ 2^d\equiv 2^{a_i}\pmod{p_i}. \] Together with (eq:nu-pi), we obtain \[ N_\nu-2^d\equiv 0\pmod{p_i}. \] So if $N_\nu-2^d$ were prime, then necessarily \[ N_\nu-2^d=p_i. \] Reducing modulo $8191$ and using (eq:nu-8191), we would get \[ 1\equiv p_i+2^d\pmod{8191}. \] Since $\operatorname{ord}_{8191}(2)=13$, the residue $2^d\pmod{8191}$ depends only on $d\bmod 13$, so \[ 1\equiv p_i+2^e\pmod{8191} \] for some $0\le e\le 12$, contradicting the finite certificate from § 5.2. Thus $N_\nu-2^d$ is never prime. $\square$

5.5. Excluding representations $p+2^a+2^b$ with $a\neq b$

Lemma 5.4. For every $\nu\ge 11$, there do not exist a prime $p$ and distinct integers $a,b\ge 1$ such that \[ N_\nu=p+2^a+2^b. \]

Proof. By (eq:nu-M), the quotient \[ w_\nu:=\frac{N_\nu}{M_\nu} \] is an integer. Now \[ M_\nu=q\prod_{\substack{0\le j\le \nu-1\\ j\neq 10}}F_j. \] Since $q\equiv 1\pmod{16}$, $F_0=3$, $F_1=5$, and $F_j\equiv 1\pmod{16}$ for all $j\ge 2$, we have \[ M_\nu\equiv 3\cdot 5\equiv -1\pmod{16}. \] Together with (eq:nu-16), this gives \[ w_\nu\equiv 1\pmod{16}. \]

Write \[ B_j= \begin{cases} F_j,&0\le j\le \nu-1,\ j\neq 10,\\ q,&j=10. \end{cases} \] Then each $B_j>1$ divides $F_j=2^{2^j}+1$, and \[ N_\nu=w_\nu\prod_{j=0}^{\nu-1}B_j. \] By Lemma 5.2 we have $N_\nu<2^{2^\nu}-1$. Therefore all hypotheses of the Round 3 Crocker-type lemma are satisfied (with $n=\nu$), and it follows that \[ N_\nu\neq p+2^a+2^b \] for all primes $p$ and all distinct $a,b\ge 1$. $\square$

5.6. The corrected explicit infinite family in $A$

Theorem 5.5. For every integer $\nu\ge 11$, one has \[ N_\nu\in A. \] Hence $A$ contains an explicit infinite family.

Proof. By Lemma 5.3, $N_\nu$ is not of the form $p+2^d$ with $d\ge 1$. Therefore it is not of the diagonal form $p+2^a+2^a$, since $2^a+2^a=2^{a+1}$.

By Lemma 5.4, $N_\nu$ is also not of the form $p+2^a+2^b$ with distinct $a,b\ge 1$.

Applying the Round 2 / Round 3 reduction lemma, we conclude $N_\nu\in A$. $\square$

Corollary 5.6. The family $\{N_\nu\}_{\nu\ge 11}$ is infinite and consists of distinct elements of $A$.

Proof. By Lemma 5.2, $N_\nu\ge M_\nu$ for each $\nu$, and $M_\nu$ is strictly increasing in $\nu$. Hence the $N_\nu$ are distinct. $\square$

5.7. Quantitative size of the corrected family

Theorem 5.7. Define \[ \mathcal{N}(X):=\#\{\nu\ge 11: N_\nu\le X\}. \] Then \[ \mathcal{N}(X)=\log_2\log_2 X+O(1)\qquad (X\to\infty). \] More explicitly, for all sufficiently large $X$, \[ \left\lfloor \log_2\log_2\!\left(\frac{G}{L}X+1\right)\right\rfloor-10 \le \mathcal{N}(X) \le \left\lfloor \log_2\log_2(GX+1)\right\rfloor-10. \]

Proof. From Lemma 5.2 we have \[ M_\nu\le N_\nu<LM_\nu=\frac{L}{G}\left(2^{2^\nu}-1\right). \]

For the upper bound, if $N_\nu\le X$, then $M_\nu\le X$, so \[ \frac{2^{2^\nu}-1}{G}\le X, \] hence \[ 2^{2^\nu}\le GX+1. \] Therefore \[ \nu\le \log_2\log_2(GX+1), \] which gives the displayed upper bound.

For the lower bound, if \[ \frac{L}{G}\left(2^{2^\nu}-1\right)\le X, \] then $N_\nu<X$ by Lemma 5.2. This is implied by \[ 2^{2^\nu}\le \frac{G}{L}X+1. \] So every integer $\nu\ge 11$ satisfying \[ \nu\le \log_2\log_2\!\left(\frac{G}{L}X+1\right) \] contributes one element $N_\nu\le X$, giving the displayed lower bound.

The two bounds differ by at most an additive constant, since $G/L$ and $G$ are fixed positive constants. Hence \[ \mathcal{N}(X)=\log_2\log_2 X+O(1). \] $\square$

Corollary 5.8 (explicit lower bound inside $A$). For all sufficiently large $X$, \[ |A\cap[1,X]|\ge \mathcal{N}(X)\gg \log\log X. \]

Proof. This is immediate from Theorem 5.5 and Theorem 5.7. $\square$

Corollary 5.9 (construction-level obstruction). The corrected fixed-data CRT construction from Rounds 2--4 cannot by itself prove \[ \overline d(A)>0. \] Indeed, the family it produces has counting function only $\log_2\log_2 X+O(1)$, and therefore has density $0$.

Proof. This is immediate from Theorem 5.7. $\square$

6) ADVERSARIAL VERIFICATION

I now try to break the new argument.

1. Did the correction really fix the Round 3 flaw? Yes. Proposition 5.1 explicitly shows that the Round 3 numerical table is invalid. The new family uses the valid Round 2 table instead.
2. Could some $p_i$ divide $M_\nu$ for large $\nu$? No. If $p_i\mid M_\nu$, then $\operatorname{ord}_{p_i}(2)\mid 2^\nu$. But each $\operatorname{ord}_{p_i}(2)=m_i$ has an odd factor greater than $1$, impossible.
3. Could the congruence argument for $p+2^d$ fail when $d$ is huge? No. The overlap property is periodic modulo $720$, and the reduction modulo $8191$ is periodic modulo $13$; both are uniform in $d$.
4. Could the family collapse to finitely many repeated numbers? No. Lemma 5.2 gives $N_\nu\ge M_\nu$, and $M_\nu$ strictly increases with $\nu$.
5. Is Corollary 5.9 overstated? The corollary rules out only the fixed corrected CRT family constructed here. It does not rule out all conceivable covering-system or sieve arguments for positive density.
6. Did we re-use a broken Round 3 step? Only the abstract Round 3 lemmas were re-used. The concrete Round 3 number-theoretic data were discarded and replaced.

No further gap remains in the corrected family construction.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

The upper-density problem remains unresolved here, but Round 4 makes genuine progress in two ways:

1. it identifies and repairs a concrete mathematical error in Round 3; and
2. it upgrades the corrected construction to an explicit infinite family $\{N_\nu\}_{\nu\ge 11}\subseteq A$ with \[ \#\{\nu\ge 11:N_\nu\le X\}=\log_2\log_2 X+O(1). \]

So the minimal corrected statement extracted from the Round 2 / Round 3 CRT method is:

There is an explicit infinite family in $A$, generated by a fixed corrected CRT recipe, but this family is necessarily sparse (size $\asymp \log\log X$ up to $X$), so this route cannot by itself prove positive upper density.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 50%

9) REFERENCES

1. R. Crocker, On the sum of a prime and of two powers of two, Pacific J. Math. 36 $1971$, 103--107.
2. H. Pan, On the integers not of the form $p+2^a+2^b$, Acta Arith. 148 $2011$, no. 1, 55--61.
3. T. F. Bloom, Erdős Problem #9, erdosproblems.com/9.

10) OUTPUT FORMAT

This file is the requested export for problem number $x=9$, namely 9_solutions.tex.