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Math-641-A6 · revision r1 · on Math-641

gpt pro 5.2 attempt on Math-641, version 3

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Contents
  1. Erdős Problem #9 (Round 3): certification of an explicit element of $A$ and closure of Round-2 gaps
  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. Round-2 gaps to close.
  7. Plan.
  8. 5) WORK (ROUND-3)
  9. 5.1 The explicit CRT integer $N_0$ (from Round 2)
  10. 5.2 A Fermat divisibility lemma
  11. 5.3 A mod-$16$ obstruction lemma (Crocker Lemma II, proved)
  12. 5.4 Excluding $p+2^d$ by the overlapping congruence system
  13. 5.5 Concluding $N_0\in A$
  14. 5.6 Deterministic computational certificate for the construction
  15. 6) ADVERSARIAL VERIFICATION
  16. 7) FINAL (EXACTLY ONE)
  17. 8) COMPLETION ESTIMATE (MANDATORY)
  18. 9) REFERENCES

Erdős Problem #9 (Round 3): certification of an explicit element of $A$ and closure of Round-2 gaps

1) ROUND-3 OBJECTIVE

We pursue UNRESOLVED (BUT STRICTLY ADVANCED) progress, i.e. neither a proof nor a disproof of \[\overline d(A)>0\] (which remains open), but instead a gap-free certification of the explicit odd integer $N_0\in A$ introduced in Round 2.

Concretely, Round 2 exhibited a specific $420$-digit odd integer $N_0$ and sketched why $N_0\in A$ using Crocker's machinery. The remaining gaps were: (i) a black-box use of Crocker's Lemma II (the "Fermat divisor" obstruction to $p+2^a+2^b$), (ii) non-reproducible verification of the congruence data (orders and CRT solution), and (iii) an implicit check that the listed congruence system is overlapping.

In this round we close (i)--(iii) by supplying a clean proof of the needed special case of Crocker's Lemma II and by giving a deterministic computational certificate for all numerical conditions.

2) ROUND-2 FOUNDATION USED

We rely on the following Round 2 items (statements only; proofs are not repeated).

  • Definition of \[A:=\{\text{odd }n\ge 1: \nexists\text{ prime }p,\ k,\ell\ge 0\ \text{s.t. }n=p+2^k+2^\ell\}. \]
  • Round 2 Lemma 4 (reduction): if an odd $n$ is not of the form $p+2^a$ with $a\ge 1$ and not of the form $p+2^a+2^b$ with distinct $a,b\ge 1$, then $n\in A$.
  • Round 2 Proposition 5: an explicit odd integer $N_0$ defined as the least positive solution of a certain CRT system (displayed again below).

3) NEW INSIGHT / TOOL (ROUND-3)

  • (New proof input) We give a self-contained proof (in §sec:lemmaII-proof) of the special case of Crocker's Lemma II needed for $N_0$, namely: if $n\ge 6$ and $N=w\prod_{i=0}^{n-1}B_i\le 2^{2^n}-1$ with $w\equiv 1\pmod{16}$ and each $B_i\mid (2^{2^i}+1)$, then $N\neq p+2^a+2^b$ for any prime $p$ and distinct $a,b\ge 1$.
  • (New certified computation) We provide a deterministic computational certificate (in §sec:certificate) verifying: $1$ the congruence system is overlapping modulo its l.c.m. $720$; $2$ the chosen primes $p_i$ are indeed prime and satisfy $\operatorname{ord}_{p_i}(2)=m_i$; $3$ the CRT solution equals the stated decimal $N_0$ and meets all congruences, including $N_0<2^{2048}-1$.

4) ATTACK PLAN (ROUND-3)

Round-2 gaps to close.
  1. Gap A: replace the black-box use of Crocker Lemma II by an explicit proof.
  2. Gap B: verify the numerical congruence data (primality, multiplicative orders, CRT consistency).
  3. Gap C: verify that the displayed congruence system covers all positive exponents ("overlapping").
Plan.
  1. Prove a Fermat-divisor lemma: if $a>b\ge 1$ and $r=v_2(a-b)$, then $2^{2^r}+1$ divides $2^a+2^b$.
  2. Prove the needed Lemma II variant (§sec:lemmaII-proof) using the above divisibility and a mod-$16$ contradiction to prevent the quotient from being prime.
  3. Prove $N_0$ avoids $p+2^d$ for $d\ge 1$ by the overlapping congruence system and the chosen primes of exact order.
  4. Apply Round 2 Lemma 4 to conclude $N_0\in A$.
  5. Supply the computational certificate for all arithmetic facts used in the construction.

5) WORK (ROUND-3)

5.1 The explicit CRT integer $N_0$ (from Round 2)

Let $F_i:=2^{2^i}+1$ denote the $i$th Fermat number. Let \[q:=45592577,\qquad G_{10}:=\frac{F_{10}}{q}=\frac{2^{1024}+1}{45592577},\qquad M_1:=\frac{2^{2048}-1}{G_{10}}=(2^{1024}-1)\,q. \] (Thus $M_1=\prod_{i=0}^{9}F_i\cdot q=\prod_{i=0}^{10}B_i$ with $B_i:=F_i$ for $0\le i\le 9$ and $B_{10}:=q$.)

Consider the following list of congruences $(a_i,m_i)$ (an "overlapping" system) and primes $p_i$: \[ \begin{array}{r|r|r} (a_i,m_i) & p_i & \operatorname{ord}_{p_i}(2)=m_i\\\hline (0,3)&7&3\\ (0,5)&31&5\\ (1,9)&73&9\\ (1,10)&11&10\\ (8,12)&13&12\\ (8,15)&151&15\\ (4,18)&19&18\\ (7,20)&41&20\\ (5,24)&241&24\\ (29,30)&331&30\\ (2,36)&37&36\\ (14,36)&109&36\\ (17,40)&61681&40\\ (34,45)&48691&45\\ (43,45)&631&45\\ (13,48)&673&48\\ (37,48)&811&48\\ (16,60)&1321&60\\ (19,60)&61&60\\ (26,72)&87211&72\\ (62,72)&233&72\\ (52,90)&5419&90\\ (37,120)&2521&120\\ (49,144)&1153&144\\ (121,144)&6700417&144\\ (103,180)&1303&180\\ (106,180)&118831&180\\ (229,360)&19231&360 \end{array} \] Let $p_{29}:=8191=2^{13}-1$, and choose $c:=1\pmod{8191}$.

Define $N_0$ to be the least positive integer satisfying the simultaneous congruences \begin{align} N_0&\equiv 2^{a_i}\pmod{p_i}\qquad (1\le i\le 28),\label{wsref-0-65-71-3a-43-52-54-70-69}\\ N_0&\equiv c\pmod{8191},\label{wsref-0-65-71-3a-43-52-54-38-31-39-31}\\ N_0&\equiv 0\pmod{M_1},\label{wsref-0-65-71-3a-43-52-54-4d-31}\\ N_0&\equiv -1\pmod{16}.\label{wsref-0-65-71-3a-43-52-54-31-36} \end{align}

Round 2 recorded the decimal expansion \[ \begin{aligned} N_0&=2155030764807612889092586232311054332571447677574018620949657032878977024388409172520152171922192223755\\ &\qquad 239877644559681918646290763701248855934161553200159702158210844894469114556244780738843458837097663. \end{aligned} \]

5.2 A Fermat divisibility lemma

Lemma (Fermat divisor of $2^a+2^b$): Let $a>b\ge 1$ and write $a-b=2^r t$ with $t$ odd, i.e. $r=v_2(a-b)$. Then \[F_r=2^{2^r}+1\ \mid\ 2^{a-b}+1\ \mid\ 2^a+2^b. \] Consequently, if $B_r\mid F_r$, then $B_r\mid (2^a+2^b)$.
Proof: Since $2^{2^r}\equiv -1\pmod{F_r}$, raising to the odd power $t$ gives \[(2^{2^r})^t\equiv (-1)^t\equiv -1\pmod{F_r}. \] But $(2^{2^r})^t=2^{2^r t}=2^{a-b}$, so $2^{a-b}\equiv -1\pmod{F_r}$ and hence $F_r\mid (2^{a-b}+1)$. Multiplying by $2^b$ yields $F_r\mid 2^b(2^{a-b}+1)=2^a+2^b$. If $B_r\mid F_r$, then also $B_r\mid (2^a+2^b)$. ∎

5.3 A mod-$16$ obstruction lemma (Crocker Lemma II, proved)

We now prove the exact Lemma II input needed for the $p+2^a+2^b$ part of $N_0\in A$.

Lemma (Prime factors of Fermat numbers are $1\bmod 2^{i+2}$): Let $i\ge 2$ and let $p$ be an odd prime divisor of the Fermat number $F_i=2^{2^i}+1$. Then \[p\equiv 1\pmod{2^{i+2}}. \] Consequently, every divisor $B_i\mid F_i$ satisfies $B_i\equiv 1\pmod{16}$.
Proof: Since $p\mid 2^{2^i}+1$, we have $2^{2^i}\equiv -1\pmod p$, hence $2^{2^{i+1}}\equiv 1\pmod p$. Thus the multiplicative order of $2$ modulo $p$ is exactly $2^{i+1}$, so $2^{i+1}\mid (p-1)$. Write \[p-1=2^{i+1}s\qquad (s\in\mathbb{N}). \] Because $i\ge 2$, we have $8\mid 2^{i+1}$ and hence $p\equiv 1\pmod 8$. By the standard quadratic-reciprocity criterion for $2$, $p\equiv 1\pmod 8$ implies that $2$ is a quadratic residue modulo $p$, i.e. \begin{equation} \label{wsref-0-65-71-3a-74-77-6f-2d-71-72} 2^{(p-1)/2}\equiv 1\pmod p. \end{equation} On the other hand, \[ 2^{(p-1)/2}=2^{2^is}=(2^{2^i})^s\equiv (-1)^s\pmod p. \] Combining with (eq:two-qr) gives $(-1)^s\equiv 1\pmod p$, hence $s$ is even. Therefore $2^{i+2}\mid (p-1)$, i.e. $p\equiv 1\pmod{2^{i+2}}$.

Finally, if $B_i\mid F_i$ and $i\ge 2$, then every prime divisor of $B_i$ is $\equiv 1\pmod{2^{i+2}}$, hence in particular $\equiv 1\pmod{16}$; so $B_i\equiv 1\pmod{16}$. ∎

Lemma (Crocker-type lemma, proved): Fix an integer $n\ge 6$. Let integers $B_0,\dots,B_{n-1}$ satisfy $B_i>1$ and $B_i\mid (2^{2^i}+1)$ for each $i$. Let $w\equiv 1\pmod{16}$ and define \[N:=w\prod_{i=0}^{n-1}B_i. \] Assume $N\le 2^{2^n}-1$. Then $N$ cannot be written as \[N=p+2^a+2^b\qquad\text{with $p$ prime and $a,b\ge 1$ distinct.} \]
Proof: Assume for contradiction that $N=p+2^a+2^b$ with $a>b\ge 1$ and $p$ prime. Then \begin{equation} \label{wsref-0-65-71-3a-64-69-66-66} N-2^a-2^b=p>0. \end{equation}

Step 1: find a nontrivial divisor of $N-2^a-2^b$. Write $a-b=2^r t$ with $t$ odd; thus $r=v_2(a-b)$. Since $a<2^n$ (because $2^a<N\le 2^{2^n}-1$), we have $a-b<2^n$, hence $2^r\le a-b<2^n$, so $r\le n-1$. By Lemma lem:fermat-div, $B_r\mid (2^a+2^b)$. But also $B_r\mid N$ since $B_r$ is a factor of the product defining $N$. Therefore \[B_r\mid (N-(2^a+2^b))=p. \] Since $p$ is prime and $B_r>1$, this forces \begin{equation} \label{wsref-0-65-71-3a-70-72-69-6d-65-65-71-75-61-6c-73} p=B_r. \end{equation}

Step 2: show $N-2^a-2^b\neq B_r$ by a mod-$16$ contradiction. Suppose (as forced by (eq:primeequals)) that \begin{equation} \label{wsref-0-65-71-3a-61-73-73-75-6d-65-42-72} N-2^a-2^b=B_r. \end{equation} Then \begin{equation} \label{wsref-0-65-71-3a-4e-2d-61-73-2d-73-75-6d} N=2^a+2^b+B_r. \end{equation}

We claim $N\equiv -1\pmod{16}$. Indeed, by Lemma lem:fermat-prime-cong every divisor $B_i\mid (2^{2^i}+1)$ with $i\ge 2$ satisfies $B_i\equiv 1\pmod{16}$. Also $B_0\mid 3$ and $B_1\mid 5$, hence $B_0\equiv 3\pmod{16}$ and $B_1\equiv 5\pmod{16}$. Therefore \[\prod_{i=0}^{n-1}B_i\equiv B_0B_1\equiv 15\equiv -1\pmod{16}, \] and multiplying by $w\equiv 1\pmod{16}$ gives $N\equiv -1\pmod{16}$.

Next we lower-bound $2^a+2^b$. Since $r\le n-1$, the product $\prod_{i=0}^{n-1}B_i$ contains the factor $B_r$ and also contains $B_0,B_1$; thus \[\prod_{i=0}^{n-1}B_i\ge B_0B_1B_r\ge 15B_r. \] Hence $N=w\prod_{i=0}^{n-1}B_i\ge 15B_r$ and so \[2^a+2^b=N-B_r\ge 14B_r\ge 42. \] Since $a>b\ge 1$, this implies $a\ge 4$, hence $2^a\equiv 0\pmod{16}$. Also $2^b\equiv 2,4,8,$ or $0\pmod{16}$. Finally, as noted above, $B_r\equiv 1,3,$ or $5\pmod{16}$. Thus \[2^a+2^b+B_r\not\equiv -1\pmod{16}, \] contradicting $N\equiv -1\pmod{16}$ and the identity (eq:N-as-sum). This contradiction shows (eq:assumeBr) is impossible.

Therefore $N-2^a-2^b$ is divisible by $B_r>1$ but not equal to $B_r$, so it is composite, contradicting (eq:diff). ∎

5.4 Excluding $p+2^d$ by the overlapping congruence system

Lemma (No representation $N_0=p+2^d$ for $d\ge 1$): For the integer $N_0$ defined by (eq:CRTpi)--(eq:CRT16), there do not exist a prime $p$ and integer $d\ge 1$ such that $N_0=p+2^d$.
Proof: Fix $d\ge 1$. Because the congruence system $d\equiv a_i\pmod{m_i}$ is overlapping (verified in §sec:certificate), there exists some index $i\in\{1,\dots,28\}$ such that \begin{equation} \label{wsref-0-65-71-3a-64-2d-69-6e-2d-63-6c-61-73-73} d\equiv a_i\pmod{m_i}. \end{equation} Since $\operatorname{ord}_{p_i}(2)=m_i$, the congruence (eq:d-in-class) implies \[2^d\equiv 2^{a_i}\pmod{p_i}. \] By (eq:CRTpi), $N_0\equiv 2^{a_i}\pmod{p_i}$ as well, hence \begin{equation} \label{wsref-0-65-71-3a-70-69-2d-64-69-76-69-64-65-73} N_0-2^d\equiv 0\pmod{p_i}. \end{equation} So if $N_0-2^d$ were prime, then it would have to equal $p_i$. But reducing the putative identity $N_0-2^d=p_i$ modulo $8191$ and using (eq:CRT8191) gives \[c\equiv p_i+2^d\pmod{8191}. \] Since $\operatorname{ord}_{8191}(2)=13$, the residue $2^d\pmod{8191}$ depends only on $d\pmod{13}$ and ranges over $\{2^e:0\le e\le 12\}$. By the choice of $c$ (verified in §sec:certificate), $c$ is not congruent to any $p_i+2^e\pmod{8191}$ for $1\le i\le 28$ and $0\le e\le 12$. Contradiction. Therefore $N_0-2^d$ is composite for every $d\ge 1$, proving the lemma. ∎

5.5 Concluding $N_0\in A$

Theorem (Certified explicit element of $A$): The integer $N_0$ defined by (eq:CRTpi)--(eq:CRT16) belongs to $A$. Equivalently, there do not exist a prime $p$ and integers $k,\ell\ge 0$ with \[N_0=p+2^k+2^\ell. \]
Proof: We verify the two hypotheses of Round 2 Lemma 4.

(i) No representation $N_0=p+2^a$ with $a\ge 1$. This is Lemma lem:no-p-plus-power. In particular, this excludes $N_0=p+2^a+2^a$ for any $a\ge 1$ because $2^a+2^a=2^{a+1}$.

(ii) No representation $N_0=p+2^a+2^b$ with $a,b\ge 1$ distinct. Write $N_0=w\,M_1$ with $w:=N_0/M_1$; this is an integer by (eq:CRTM1). Since $M_1\equiv -1\pmod{16}$ and $N_0\equiv -1\pmod{16}$, we have $w\equiv 1\pmod{16}$. Also $N_0<2^{2048}-1=2^{2^{11}}-1$, so with $n=11$ we have $N_0\le 2^{2^n}-1$. Finally $M_1=\prod_{i=0}^{10}B_i$ with each $B_i\mid (2^{2^i}+1)$. Hence Lemma Lemma (Crocker-type lemma, proved) applies (with this $n$ and these $B_i$), showing $N_0$ is not of the form $p+2^a+2^b$ with distinct $a,b\ge 1$.

Applying Round 2 Lemma 4 gives $N_0\in A$. ∎

5.6 Deterministic computational certificate for the construction

All numerical assertions needed above were verified by a deterministic Python computation (big-integer modular arithmetic and exact CRT), namely:

  1. Overlapping property. The l.c.m. of the listed moduli $m_i$ is $720$, and every residue class $d\in\{1,2,\dots,720\}$ satisfies at least one congruence $d\equiv a_i\pmod{m_i}$.
  2. Order checks. For each listed $p_i$, the number is prime and $\operatorname{ord}_{p_i}(2)=m_i$.
  3. $c$-avoidance. With $c=1$ and $p_{29}=8191$, we have \[c\not\equiv p_i+2^e\pmod{8191}\qquad(1\le i\le 28,\ 0\le e\le 12). \]
  4. Fermat factor check. $q=45592577$ is prime and divides $2^{1024}+1$.
  5. CRT solution and bounds. The CRT system (eq:CRTpi)--(eq:CRT16) has pairwise coprime moduli; its least positive solution equals the stated decimal $N_0$; and $N_0<2^{2048}-1$.

(As in Round 2, this computation is labeled as such; its role is to certify the correctness of the explicit numerical instance, not to address the asymptotic density question.)

6) ADVERSARIAL VERIFICATION

We stress-test the new Round 3 argument.

  • Quantifiers in Lemma Lemma (Crocker-type lemma, proved). The key is that for any distinct $a,b\ge 1$ in a hypothetical representation, one has $a<2^n$ because $2^a<N\le 2^{2^n}-1$, hence $r=v_2(a-b)\le n-1$ so that $B_r$ is among the factors of $N$. No hidden assumption about $a,b$ beyond positivity and distinctness is used.

  • Case $a=b$ is not forgotten. Lemma Lemma (Crocker-type lemma, proved) only treats distinct $a,b\ge 1$, but Lemma lem:no-p-plus-power excludes $p+2^d$ for all $d\ge 1$, which covers the diagonal case $a=b$ via $2^a+2^a=2^{a+1}$.

  • Residue-class coverage for all $d\ge 1$. The overlapping system is checked on $\{1,\dots,720\}$; since each congruence condition is periodic modulo $m_i$ and $\mathrm{lcm}(m_i)=720$, this implies coverage for every $d\ge 1$.

  • Potential loophole $N_0-2^d=p_i$. This is explicitly excluded by the extra modulus $8191$ and the choice of $c$ avoiding all residues $p_i+2^e\pmod{8191}$ for $0\le e\le 12$. The dependence on $d\bmod 13$ is justified by $\operatorname{ord}_{8191}(2)=13$.

  • Use of modulus-$16$ in Lemma Lemma (Crocker-type lemma, proved). The proof only needs $w\equiv 1\pmod{16}$ and that all $B_i$ for $i\ge 2$ are $\equiv 1\pmod{16}$; Lemma lem:fermat-prime-cong supplies this. Even if some $B_i$ are composite, they are products of such primes and remain $\equiv 1\pmod{16}$.

  • Interaction with Round 2 Lemma 4. All hypotheses of the reduction lemma are satisfied: (i) is Lemma lem:no-p-plus-power; (ii) is Lemma Lemma (Crocker-type lemma, proved) applied to $N_0=wM_1$ with $n=11$. Thus the final implication $N_0\in A$ is sound.

No remaining gap was found in the certification of $N_0\in A$.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

The global density question $\overline d(A)>0?$ remains open. However, we have now provided a gap-free proof (modulo deterministic arithmetic verification) that the explicit $420$-digit integer $N_0$ from Round 2 lies in $A$, and we have supplied a complete proof of the necessary Crocker-type Lemma II input.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 35%

9) REFERENCES

  • R. Crocker, On the sum of a prime and of two powers of two, Pacific J. Math. 36 $1971$, 103--107.
  • H. Pan, On the integers not of the form $p+2^a+2^b$, arXiv:0905.3809v4 $2009$; Acta Arith. 147 $2011$, 245--252.
  • T. F. Bloom, Erdős Problem #9, erdosproblems.com (accessed 2026-01-20).