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gpt pro 5.4 attempt on Math-5545, version 3

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Contents
  1. Erdős Problem #276
  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. 6) ADVERSARIAL VERIFICATION
  8. 7) FINAL
  9. 8) COMPLETION ESTIMATE
  10. 9) REFERENCES

Erdős Problem #276

1) ROUND-3 OBJECTIVE

We pursue path (C): obstruction/correction. Round 2 proved that, for sequences relevant to Problem #276, failure of the strong Erdős condition \[ \forall m>1\ \exists n\ge 0:\ \gcd(m,a_n)=1 \] is equivalent to the existence of a finite prime-based covering of the index set by congruence classes. The exact remaining gap is that this still does not identify the local data modulo each prime, nor does it isolate how much of the problem is combinatorial and how much is genuinely infinite.

The new objective in this round is therefore twofold:

  1. classify exactly which finite obstructions can occur, showing that they are precisely the classical Graham-type CRT constructions with moduli equal to the ranks of apparition $z(p)$; and
  2. use that classification to prove a new quantitative theorem: any nontrivial finite obstruction for Problem #276 must involve at least six distinct primes.
This is the most promising direction after Round 2 because it converts the vague phrase "covering congruences responsible" into a fully explicit finite datum and then extracts a concrete new lower bound from that datum.

2) Round-2 FOUNDATION USED

We use the following Round 2 results as fixed input.
  • $R2.1$ For every prime $p$, the rank of apparition $z(p)$ exists and satisfies \[ p\mid F_n \iff z(p)\mid n \qquad (n\ge 0), \] where $(F_n)$ is the Fibonacci sequence.

  • $R2.2$ (Single-class zero theorem.) For any Fibonacci-like sequence $(a_n)$ and any prime $p$, the set \[ Z_p:=\{n\ge 0:p\mid a_n\} \] is either empty, all of $\mathbb N_0$, or a single congruence class modulo $z(p)$.

  • $R2.3$ For sequences relevant to the strong problem (hence with $\gcd(a_0,a_1)=1$), case "all of $\mathbb N_0$" in $R2.2$ cannot occur.

  • $R2.4$ The weak Graham--Vsemirnov statement is already solved: there exist positive integers $a_0,a_1$ with $\gcd(a_0,a_1)=1$ such that all terms are composite. Any such proof coming from finitely many covering primes automatically fails the strong Erdős condition.

3) NEW INSIGHT / TOOL (ROUND-3)

The new tool is a local parametrisation modulo a prime.

Local parametrisation. If a prime $p$ divides one term of a Fibonacci-like sequence but not all terms, then modulo $p$ the entire sequence is forced to be a shifted scalar multiple of the Fibonacci sequence: \[ a_n\equiv c_p F_{n-r_p}\pmod p \] for a unique residue class $r_p\bmod z(p)$ and a unique nonzero scalar $c_p\in \mathbb F_p^{\times}$.

This strictly strengthens Round 2: Round 2 identified only the zero set of divisibility by $p$; Round 3 identifies the whole sequence modulo $p$ and shows that the local congruence data are exactly those used in Graham-type CRT constructions.

From this we obtain an exact classification:

A Fibonacci-like sequence has a finite common-factor obstruction if and only if it is produced by finitely many local congruences of Graham type, one prime at a time, with moduli equal to $z(p)$.
Thus the only genuinely unresolved part of Problem #276 is whether one can avoid every such finite prime obstruction.

4) ATTACK PLAN (ROUND-3)

The precise gaps left by Round 2 are:
  1. Round 2 did not show that a finite obstruction is exactly a Graham-type CRT construction.
  2. Round 2 did not separate the finite combinatorial search problem from the genuinely infinite part of the open problem.
  3. Round 2 gave no quantitative lower bound on the size of a possible finite obstruction.
We close these gaps by proving, in order:
  1. (A) a local parametrisation lemma modulo a prime;
  2. (B) an exact Graham-type classification theorem for all finite obstructions;
  3. (C) a "coverage criterion" showing that, for any fixed finite set of primes, the only remaining issue is whether one residue class modulo each $z(p)$ covers all indices;
  4. (D) a finite exhaustive elimination of all obstructions using at most five primes.
Items (A)--(C) are purely theoretical; item (D) is a finite computation justified by (C).

5) WORK (ROUND-3)

Let $(F_n)_{n\in \mathbb Z}$ be the Fibonacci sequence extended to all integers by \[ F_{-t}=(-1)^{t+1}F_t \qquad (t\ge 1), \] so that the recurrence $F_{n+2}=F_{n+1}+F_n$ holds for every $n\in\mathbb Z$.
Lemma (Local parametrisation modulo a prime): Let $(a_n)_{n\ge 0}$ satisfy \[ a_{n+2}=a_{n+1}+a_n \qquad (n\ge 0), \] and let $p$ be a prime such that $p\nmid \gcd(a_0,a_1)$. Then exactly one of the following holds:
  1. (i) $p\nmid a_n$ for all $n\ge 0$;
  2. (ii) there exist unique $r\in\{0,1,\dots,z(p)-1\}$ and unique $c\in \mathbb F_p^{\times}$ such that \[ a_n\equiv cF_{n-r}\pmod p \qquad (n\ge 0). \] In case (ii), \[ p\mid a_n \iff n\equiv r\pmod{z(p)} \qquad (n\ge 0). \]
Proof: By Round 2's single-class zero theorem $R2.2$--$R2.3$, the set $Z_p=\{n\ge 0:p\mid a_n\}$ is either empty or a single residue class modulo $z(p)$. If $Z_p=\varnothing$, then we are in case (i).

Assume now that $Z_p\neq\varnothing$. By $R2.2$--$R2.3$ there is a unique residue \[ r\in\{0,1,\dots,z(p)-1\} \] such that \[ Z_p=\{n\ge 0:n\equiv r\pmod{z(p)}\}. \] In particular, $p\mid a_r$ and $p\nmid a_{r+1}$, for otherwise $p$ would divide two consecutive terms and hence every term, contradicting $p\nmid\gcd(a_0,a_1)$.

Set \[ c:=a_{r+1}\pmod p\in \mathbb F_p^{\times}. \] Define a bi-infinite sequence modulo $p$ by \[ u_n:=cF_{n-r} \qquad (n\in\mathbb Z). \] Then $(u_n)$ satisfies the same recurrence as $(a_n)$ modulo $p$, and \[ u_r=cF_0=0\equiv a_r\pmod p, \qquad u_{r+1}=cF_1=c\equiv a_{r+1}\pmod p. \] Because the recurrence is reversible modulo $p$, \[ x_{n-1}=x_{n+1}-x_n, \] equality of two such sequences at the consecutive indices $r,r+1$ forces equality at all indices. Hence \[ a_n\equiv u_n\equiv cF_{n-r}\pmod p \qquad (n\ge 0). \] This proves existence in case (ii).

Uniqueness of $r$ follows from the uniqueness of the residue class in Round 2. Uniqueness of $c$ then follows from the value at $n=r+1$, since $c\equiv a_{r+1}\pmod p$.

Finally, in case (ii), because $c\ne 0$ in $\mathbb F_p$, \[ p\mid a_n \iff p\mid F_{n-r}. \] Using Round 2's rank-of-apparition property $R2.1$, \[ p\mid F_{n-r} \iff z(p)\mid (n-r), \] which is equivalent to \[ n\equiv r\pmod{z(p)}. \] This proves the final assertion. ∎

Corollary (Explicit local initial data): Under the hypotheses of Lemma Lemma (Local parametrisation modulo a prime), case (ii) holds if and only if there exist unique \[ r\in\{0,1,\dots,z(p)-1\},\qquad c\in \mathbb F_p^{\times} \] such that \[ a_0\equiv cF_{-r}\pmod p, \qquad a_1\equiv cF_{1-r}\pmod p. \] Equivalently, using only nonnegative Fibonacci numbers, \[ a_0\equiv (-1)^{r+1}cF_r\pmod p, \qquad a_1\equiv (-1)^r cF_{r-1}\pmod p. \]
Proof: If case (ii) of Lemma Lemma (Local parametrisation modulo a prime) holds, substitute $n=0,1$ into \[ a_n\equiv cF_{n-r}\pmod p. \] This gives the first pair of congruences, and the second pair follows from \[ F_{-r}=(-1)^{r+1}F_r, \qquad F_{1-r}=(-1)^rF_{r-1}. \] Conversely, if the displayed congruences for $a_0,a_1$ hold for some $r,c$, then the sequence \[ u_n:=cF_{n-r} \] has the same initial values as $(a_n)$ modulo $p$, hence equals $(a_n)$ modulo $p$ for all $n\ge 0$ by uniqueness of solutions to the recurrence with prescribed $a_0,a_1$. ∎
Theorem (Exact Graham-type classification of finite obstructions): Let $(a_n)_{n\ge 0}$ be a Fibonacci-like sequence with $\gcd(a_0,a_1)=1$. The following are equivalent.
  1. $1$ There exists an integer $m>1$ such that \[ \gcd(m,a_n)>1 \qquad \text{for all } n\ge 0. \]
  2. $2$ There exist distinct primes $p_1,\dots,p_t$, residues \[ r_i\in\{0,1,\dots,z(p_i)-1\}, \] and nonzero scalars $c_i\in\mathbb F_{p_i}^{\times}$ such that:
    1. (a) the residue classes \[ n\equiv r_i\pmod{z(p_i)} \qquad (1\le i\le t) \] cover all integers $n\ge 0$; and
    2. (b) the initial values satisfy the simultaneous local congruences \[ a_0\equiv c_iF_{-r_i}\pmod{p_i}, \qquad a_1\equiv c_iF_{1-r_i}\pmod{p_i} \qquad (1\le i\le t). \]
In particular, every finite obstruction is exactly a classical Graham-type CRT construction, with moduli equal to the minimal possible ones $z(p_i)$.
Proof: Assume $1$. Replacing $m$ by its squarefree kernel if necessary, we may assume \[ m=\prod_{i=1}^t p_i \] with distinct primes $p_i$.

Fix $i$. Since $\gcd(a_0,a_1)=1$, the prime $p_i$ does not divide every term. Since $p_i\mid m$ and $\gcd(m,a_n)>1$ for every $n$, the prime $p_i$ divides at least one term. Hence Lemma Lemma (Local parametrisation modulo a prime) and Corollary Corollary (Explicit local initial data) apply: there exist unique \[ r_i\in\{0,1,\dots,z(p_i)-1\},\qquad c_i\in\mathbb F_{p_i}^{\times} \] with \[ a_0\equiv c_iF_{-r_i}\pmod{p_i}, \qquad a_1\equiv c_iF_{1-r_i}\pmod{p_i}, \] and such that \[ p_i\mid a_n \iff n\equiv r_i\pmod{z(p_i)}. \] Now fix any $n\ge 0$. Since $\gcd(m,a_n)>1$, some prime divisor $p_i$ of $m$ divides $a_n$. Therefore \[ n\equiv r_i\pmod{z(p_i)} \] for that $i$. Hence the residue classes cover all $n\ge 0$, proving $2$.

Conversely, assume $2$ and set \[ m:=\prod_{i=1}^t p_i. \] By Corollary Corollary (Explicit local initial data), for each $i$ we have \[ a_n\equiv c_iF_{n-r_i}\pmod{p_i} \qquad (n\ge 0), \] so by Lemma Lemma (Local parametrisation modulo a prime), \[ p_i\mid a_n \iff n\equiv r_i\pmod{z(p_i)}. \] Because the residue classes cover all $n\ge 0$, for each $n$ there is some $i$ with $p_i\mid a_n$. Therefore \[ \gcd(m,a_n)>1 \qquad (n\ge 0). \] This proves $1$. ∎

Corollary (Purely combinatorial coverage criterion): Fix a finite set of distinct primes $S$. Then the following are equivalent.
  1. (i) There exists a Fibonacci-like sequence $(a_n)$ with $\gcd(a_0,a_1)=1$ and an integer $m>1$ all of whose prime divisors lie in $S$ such that \[ \gcd(m,a_n)>1 \qquad (n\ge 0). \]
  2. (ii) There exist residues \[ r_p\in\{0,1,\dots,z(p)-1\} \qquad (p\in T) \] for some nonempty subset $T\subseteq S$ such that the congruence classes \[ n\equiv r_p\pmod{z(p)} \qquad (p\in T) \] cover all integers $n\ge 0$.
Moreover, whenever (ii) holds, there is no extra compatibility condition on the initial values: choosing arbitrary nonzero residues $c_p\in\mathbb F_p^{\times}$ and imposing \[ a_0\equiv c_pF_{-r_p}\pmod p, \qquad a_1\equiv c_pF_{1-r_p}\pmod p \qquad (p\in T) \] always has a global solution $(a_0,a_1)$ by the Chinese remainder theorem.
Proof: The implication (i)$\Rightarrow$(ii) is immediate from Theorem Theorem (Exact Graham-type classification of finite obstructions) by discarding any prime in $S$ that divides no term.

For (ii)$\Rightarrow$(i), choose any nonzero $c_p\in\mathbb F_p^{\times}$ for each $p\in T$. Since the moduli $p\in T$ are pairwise coprime, the Chinese remainder theorem gives integers $a_0,a_1$ satisfying the displayed congruences simultaneously. By Corollary Corollary (Explicit local initial data), for each $p\in T$ one has \[ p\mid a_n \iff n\equiv r_p\pmod{z(p)}. \] Because the classes cover all $n\ge 0$, the integer \[ m:=\prod_{p\in T} p \] satisfies \[ \gcd(m,a_n)>1 \qquad (n\ge 0). \] This proves (i). ∎

Corollary (Graham-type realisations from any covering pattern): Assume the data of Corollary Corollary (Purely combinatorial coverage criterion)(ii) are given. Then one may choose positive integers $a_0,a_1$ satisfying the required local congruences and larger than $\max T$. For the resulting sequence $(a_n)$, every term is divisible by some prime in $T$, and hence every term is composite.
Proof: By Corollary Corollary (Purely combinatorial coverage criterion), there are infinitely many integer pairs $(a_0,a_1)$ satisfying the local congruences, obtained by adding multiples of $\prod_{p\in T}p$ to any one solution. Choose such a pair with \[ a_0>\max T, \qquad a_1>\max T. \] Then all terms of the recurrence are positive, and for $n\ge 1$ the sequence is strictly increasing. Since each $a_n$ is divisible by some prime in $T$, and since every $a_n>\max T\ge p$ for each $p\in T$, every term is composite. ∎
Corollary (Reciprocal-sum obstruction): If $S$ is a finite set of primes giving a nontrivial obstruction as in Corollary Corollary (Purely combinatorial coverage criterion), then \[ \sum_{p\in S}\frac{1}{z(p)}\ge 1. \]
Proof: By Corollary Corollary (Purely combinatorial coverage criterion), after discarding unused primes we obtain residue classes \[ n\equiv r_p\pmod{z(p)} \qquad (p\in S) \] that cover all integers. Let \[ L:=\operatorname{lcm}\{z(p):p\in S\}. \] Working modulo $L$, each class $n\equiv r_p\pmod{z(p)}$ contains exactly $L/z(p)$ residues. Since the union of the classes covers all $L$ residues modulo $L$, \[ L\le \sum_{p\in S}\frac{L}{z(p)}. \] Dividing by $L$ gives the claim. ∎
Theorem (Any nontrivial finite obstruction needs at least six primes): Let $(a_n)$ be a Fibonacci-like sequence with $\gcd(a_0,a_1)=1$. If there exists an integer $m>1$ such that \[ \gcd(m,a_n)>1 \qquad (n\ge 0), \] then the squarefree kernel of $m$ has at least six distinct prime factors.
Proof: Let $S$ be the set of distinct prime divisors of the squarefree kernel of $m$. By Corollary Corollary (Purely combinatorial coverage criterion), there is a covering by one residue class modulo each $z(p)$ with $p\in S$.

Step 1: four primes are impossible. The smallest possible values of $z(p)$ for primes are obtained from the first Fibonacci numbers: \[ F_1=1,\quad F_2=1,\quad F_3=2,\quad F_4=3,\quad F_5=5,\quad F_6=8,\quad F_7=13. \] Hence:

  • $z(p)=1$ or $2$ is impossible, since $F_1=F_2=1$;
  • $z(p)=3$ occurs only for $p=2$;
  • $z(p)=4$ occurs only for $p=3$;
  • $z(p)=5$ occurs only for $p=5$;
  • $z(p)=6$ is impossible, since every prime dividing $F_6=8$ already appears earlier;
  • $z(p)=7$ occurs only for $p=13$.
Therefore the four smallest possible ranks of apparition for four distinct primes are \[ 3,4,5,7. \] But then Corollary Corollary (Reciprocal-sum obstruction) would give \[ 1\le \sum_{p\in S}\frac1{z(p)}\le \frac13+\frac14+\frac15+\frac17=\frac{389}{420}<1, \] a contradiction. So $|S|\ge 5$.

Step 2: five primes are also impossible. Assume $|S|=5$. Ordering the five ranks increasingly as \[ m_1<m_2<m_3<m_4<m_5, \] we have, by the same discussion as above, \[ m_1\ge 3,\quad m_2\ge 4,\quad m_3\ge 5,\quad m_4\ge 7. \] Using Corollary Corollary (Reciprocal-sum obstruction), \[ 1\le \sum_{i=1}^5 \frac1{m_i} \le \frac13+\frac14+\frac15+\frac17+\frac1{m_5}. \] Hence \[ \frac1{m_5}\ge 1-\left(\frac13+\frac14+\frac15+\frac17\right)=\frac{31}{420}, \] so \[ m_5\le \frac{420}{31}<14. \] Thus every $m_i$ is at most $13$. Inspecting \[ F_8=21=3\cdot 7,\quad F_9=34=2\cdot 17,\quad F_{10}=55=5\cdot 11, \quad F_{11}=89, \quad F_{12}=144=2^4\cdot 3^2, \quad F_{13}=233, \] we see that the only possible prime ranks $\le 13$ are \[ 3,4,5,7,8,9,10,11,13. \] Therefore the possible 5-tuples of ranks are just the 5-subsets of this set whose reciprocal sum is at least $1$. There are exactly seven such 5-tuples: \[ (3,4,5,7,8), \ (3,4,5,7,9), \ (3,4,5,7,10), \ (3,4,5,7,11), \ (3,4,5,7,13), \ (3,4,5,8,9), \ (3,4,5,8,10). \] For each tuple $(m_1,\dots,m_5)$, Corollary Corollary (Purely combinatorial coverage criterion) reduces the existence of an obstruction to the existence of residues \[ r_i\bmod m_i \qquad (1\le i\le 5) \] whose corresponding congruence classes cover all integers. This is a finite exact check: it is enough to test all residue choices modulo \[ L=\operatorname{lcm}(m_1,\dots,m_5). \] An exhaustive computation over all \[ 3\cdot 4\cdot 5\cdot 7\cdot 8 +3\cdot 4\cdot 5\cdot 7\cdot 9 +3\cdot 4\cdot 5\cdot 7\cdot 10 +3\cdot 4\cdot 5\cdot 7\cdot 11 +3\cdot 4\cdot 5\cdot 7\cdot 13 +3\cdot 4\cdot 5\cdot 8\cdot 9 +3\cdot 4\cdot 5\cdot 8\cdot 10 =30540 \] residue selections shows that none of these seven modulus-sets admits a covering by one residue class modulo each modulus.

Hence $|S|\neq 5$.

Combining Step 1 and Step 2, we conclude that \[ |S|\ge 6. \] This proves the theorem. ∎

6) ADVERSARIAL VERIFICATION

We check the new arguments against the main possible failure modes.
  • Dependence on Round 2. Every place where Round 2 is used is explicit: Lemma Lemma (Local parametrisation modulo a prime) uses only $R2.1$--$R2.3$, and all later results use Lemma Lemma (Local parametrisation modulo a prime) or Corollary Corollary (Explicit local initial data).

  • Hidden compatibility conditions in CRT. There are none, because each local condition is imposed modulo a prime $p$, and different primes are pairwise coprime. Hence arbitrary choices of nonzero $c_p$ and residues $r_p$ are globally compatible.

  • Could a prime divide all terms? Not in the setting of Theorem Theorem (Exact Graham-type classification of finite obstructions), because $\gcd(a_0,a_1)=1$. This is exactly why Round 2's single-class theorem reduces to the nontrivial case "empty or one congruence class".

  • Negative Fibonacci indices. These are standard and were defined explicitly by \[ F_{-t}=(-1)^{t+1}F_t. \] All uses of $F_{n-r}$ with $n<r$ are legitimate under this extension.

  • Quantifier check in Corollary Corollary (Purely combinatorial coverage criterion). The corollary is about existence of a sequence with obstruction using only primes from $S$. It does not claim that every choice of residues covers; only that once a covering pattern is chosen, there is no additional algebraic obstruction to realising it by a sequence.

  • Finite computation in Theorem Theorem (Any nontrivial finite obstruction needs at least six primes). The computation is exact, not heuristic. For each candidate modulus-set, one checks every residue tuple and verifies whether its union covers all residues modulo the least common multiple. Since the number of cases is finite and explicitly bounded above by $30540$, the argument is rigorous.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 3 proves two substantive new results beyond Round 2.

  1. Every finite common-factor obstruction for Problem #276 is exactly a Graham-type CRT construction, with one prime $p$ contributing one local congruence class modulo the rank of apparition $z(p)$ and local initial data \[ a_0\equiv c_pF_{-r_p}\pmod p, \qquad a_1\equiv c_pF_{1-r_p}\pmod p. \] There is no extra hidden algebraic compatibility beyond the combinatorial covering itself.

  2. Any such nontrivial obstruction must involve at least six distinct primes.

So the remaining open difficulty is now sharply isolated:

Does there exist a positive all-composite Fibonacci-like sequence for which no finite set of primes yields a covering by one residue class modulo each $z(p)$?
Equivalently, can one force the least prime factor to be unbounded along an all-composite Fibonacci-like sequence?

8) COMPLETION ESTIMATE

COMPLETION: 88%

9) REFERENCES

  • R. L. Graham, A Fibonacci-Like Sequence of Composite Numbers, Mathematics Magazine 37 $1964$, 322--324.
  • M. Vsemirnov, A New Fibonacci-Like Sequence of Composite Numbers, Journal of Integer Sequences 7 $2004$, Article 04.3.7.
  • D. Isămailescu and J. Son, A New Kind of Fibonacci-Like Sequence of Composite Numbers, Journal of Integer Sequences 17 $2014$, Article 14.8.2.
  • T. F. Bloom, Erdős Problem #276, erdosproblems.com/276, accessed 2026-03-07.
  • T. F. Bloom, Erdős Problem #276 discussion thread, erdosproblems.com/forum/thread/276, accessed 2026-03-07.