gpt pro 5.4 attempt on Math-5545, version 2

Erdős Problem #276

1) ROUND-2 OBJECTIVE

We pursue path (C): obstruction/correction. Round 1 already separated the weak Graham--Vsemirnov statement from the stronger Erdős problem, but it left one conceptual gap: the informal phrase

"without an underlying system of covering congruences responsible"

was not yet formalized exactly.

The goal of Round 2 is to close that gap. We prove that for the recurrence \[ a_{n+2}=a_{n+1}+a_n, \] the natural precise meaning of "a finite covering system is responsible" is equivalent to the literal strong condition \[ \forall m>1\ \exists n\ge 0:\ \gcd(m,a_n)=1. \] So the intended conjecture is not a different one: after the minimal natural formalization, it is the same conjecture.

2) Round-1 FOUNDATION USED

We use the following Round 1 results without re-proving them.

• $F1$ The weak statement is solved: there exist relatively prime positive integers $a_0,a_1$ such that every term of the corresponding Lucas sequence is composite (Graham; record-small known example due to Vsemirnov).
• $F2$ Any covering-congruence construction of Graham type cannot solve the strong Erdős problem, because the product of the finitely many covering primes has a nontrivial gcd with every term.
• $F3$ For Fibonacci numbers $F_n$, one has the addition formula \[ F_{u+v}=F_uF_{v+1}+F_{u-1}F_v \] and the divisibility property \[ F_m\mid F_{tm}\qquad (m,t\ge 1). \]
• $F4$ For a Fibonacci-like sequence, the gcd of consecutive terms is constant: \[ \gcd(a_n,a_{n+1})=\gcd(a_0,a_1)\qquad (n\ge 0). \] In particular, any sequence satisfying the strong Erdős condition automatically has $\gcd(a_0,a_1)=1$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new structural fact is the following.

Single-class zero theorem modulo a prime. Fix a prime $p$ and a Fibonacci-like sequence $(a_n)_{n\ge 0}$. Then the set \[ Z_p:=\{n\ge 0: p\mid a_n\} \] is of one of the following three types:

1. (i) $Z_p=\varnothing$;
2. (ii) $Z_p=\{0,1,2,\dots\}$;
3. (iii) $Z_p$ is a single arithmetic progression \[ Z_p=\{n\ge 0: n\equiv r_p\pmod{z(p)}\}, \] where $z(p)$ is the least positive integer with $p\mid F_{z(p)}$.

For sequences relevant to the strong Erdős problem, case (ii) cannot occur because then $p$ would divide $a_0$ and $a_1$, contradicting $\gcd(a_0,a_1)=1$.

Consequently, failure of the strong condition \[ \forall m>1\ \exists n:\ \gcd(m,a_n)=1 \] is equivalent to the existence of a finite family of primes whose corresponding single congruence classes cover all indices. This gives the exact mathematical meaning of "an underlying covering system of congruences responsible."

4) ATTACK PLAN (ROUND-2)

Round 1 already showed that the strong problem was not settled by Graham/Vsemirnov-type constructions. The exact remaining gap was:

Does the literal Erdős condition merely suggest the absence of a covering system, or is it actually equivalent to the absence of a finite prime-based congruence covering?

To close this gap, we prove four claims.

1. (A) For every prime $p$, the Fibonacci numbers have a well-defined rank of apparition $z(p)$, and \[ p\mid F_n \iff z(p)\mid n. \]
2. (B) For every Fibonacci-like sequence and every prime $p$, the set of indices where $p$ divides the sequence is either empty, all of $\mathbf{N}_0$, or a single residue class modulo $z(p)$.
3. (C) Therefore a finite common-factor obstruction \[ \exists m>1\ \forall n:\ \gcd(m,a_n)>1 \] is exactly the same as a finite prime-covering congruence system.
4. (D) For any fixed finite set of primes, deciding whether that set covers the sequence reduces to a finite computation.

This is strict progress beyond Round 1 because it turns the informal "covering congruences responsible" phrase into a rigorous equivalent condition.

5) WORK (ROUND-2)

Let $F_0=0$, $F_1=1$, and $F_{n+2}=F_{n+1}+F_n$ be the Fibonacci numbers.

Claim A. For every prime $p$, there exists a least positive integer $z(p)$ such that $p\mid F_{z(p)}$. Moreover, \[ p\mid F_n \iff z(p)\mid n \qquad (n\ge 0). \]

Proof. Let \[ A=\left(\begin{array}{cc}0&1\\ 1&1\end{array}\right). \] Modulo $p$, the matrix $A$ lies in the finite group $\mathrm{GL}_2(\mathbf{F}_p)$ because $\det A=-1\not\equiv 0\pmod p$. Hence $A$ has finite order $T_p>0$. Since \[ A^n\left(\begin{array}{c}0\\ 1\end{array}\right)= \left(\begin{array}{c}F_n\\ F_{n+1}\end{array}\right), \] we get \[ A^{T_p}\left(\begin{array}{c}0\\ 1\end{array}\right)= \left(\begin{array}{c}0\\ 1\end{array}\right), \] so $p\mid F_{T_p}$. Therefore the least positive such index $z(p)$ exists.

If $z(p)\mid n$, then by Foundation $F3$ we have $F_{z(p)}\mid F_n$, hence $p\mid F_n$.

Conversely, suppose $p\mid F_n$. If $n<z(p)$, then by minimality of $z(p)$ we must have $n=0$, so certainly $z(p)\mid n$. Now assume $n\ge z(p)$ and write \[ n=qz(p)+r,\qquad q\ge 1,\qquad 0\le r<z(p). \] By the addition formula from Foundation $F3$, \[ F_n=F_{qz(p)+r}=F_{qz(p)}F_{r+1}+F_{qz(p)-1}F_r. \] Since $z(p)\mid qz(p)$, the first term is $0\pmod p$, so \[ F_n\equiv F_{qz(p)-1}F_r\pmod p. \] Now use Cassini's identity: \[ F_{qz(p)+1}F_{qz(p)-1}-F_{qz(p)}^2=(-1)^{qz(p)}. \] Because $p\mid F_{qz(p)}$, this gives \[ F_{qz(p)+1}F_{qz(p)-1}\equiv (-1)^{qz(p)}\not\equiv 0\pmod p, \] so $F_{qz(p)-1}\not\equiv 0\pmod p$. Therefore $p\mid F_n$ implies $p\mid F_r$. By the minimality of $z(p)$, this forces $r=0$. Hence $z(p)\mid n$.

This proves Claim A. $\square$

Claim B. Let $(a_n)_{n\ge 0}$ satisfy $a_{n+2}=a_{n+1}+a_n$, and fix a prime $p$. Then the set \[ Z_p=\{n\ge 0: p\mid a_n\} \] is either empty, all of $\mathbf{N}_0$, or a single residue class modulo $z(p)$.

Proof. First note that if $p\mid a_r$ and $p\mid a_{r+1}$ for some $r$, then the recurrence gives $p\mid a_{r+2}$, and the reversible form \[ a_{r-1}=a_{r+1}-a_r \] gives $p\mid a_{r-1}$. Inducting forward and backward, $p$ divides every term. This is case (ii).

Now suppose case (ii) does not hold, but $Z_p\neq \varnothing$. Pick $r\ge 0$ with $p\mid a_r$. Then necessarily $p\nmid a_{r+1}$, for otherwise $p$ would divide all terms.

Work modulo $p$, and extend the recurrence uniquely to all integers by the rule \[ a_{n-1}=a_{n+1}-a_n. \] Also extend the Fibonacci sequence to all integers by \[ F_{-t}=(-1)^{t+1}F_t\qquad (t\ge 1), \] so that the recurrence still holds for every integer index.

Let $b$ be the residue class of $a_{r+1}$ modulo $p$. Consider the bi-infinite sequence \[ u_n:=bF_{n-r}\qquad (n\in \mathbf{Z}). \] It satisfies the same recurrence as $(a_n)$ modulo $p$, and \[ u_r=bF_0=0\equiv a_r\pmod p, \] \[ u_{r+1}=bF_1=b\equiv a_{r+1}\pmod p. \] Since a second-order linear recurrence is uniquely determined by two consecutive values, we have \[ a_n\equiv u_n\equiv bF_{n-r}\pmod p \qquad (n\in \mathbf{Z}). \] Because $b\not\equiv 0\pmod p$, this yields \[ p\mid a_n \iff p\mid F_{n-r}. \] By Claim A, \[ p\mid F_{n-r} \iff z(p)\mid (n-r). \] Therefore \[ Z_p=\{n\ge 0: n\equiv r\pmod{z(p)}\}. \] So in the nontrivial case, $Z_p$ is a single residue class modulo $z(p)$.

This proves Claim B. $\square$

Definition. Let $(a_n)$ be a Fibonacci-like sequence with $\gcd(a_0,a_1)=1$. A finite prime-covering congruence system for $(a_n)$ means a finite set of primes $p_1,\dots,p_t$ and residues $r_i$ modulo $z(p_i)$ such that \[ p_i\mid a_n \iff n\equiv r_i\pmod{z(p_i)} \qquad (1\le i\le t), \] and the residue classes \[ n\equiv r_i\pmod{z(p_i)} \] cover all integers $n\ge 0$.

Claim C. Assume $\gcd(a_0,a_1)=1$. Then 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. $2$ The sequence $(a_n)$ admits a finite prime-covering congruence system.

Proof. Assume $1$. Let $S$ be the finite set of prime divisors of $m$. Because $\gcd(a_0,a_1)=1$, no prime can divide every term of the sequence. Hence for each $p\in S$, Claim B applies in the nontrivial case: either $p$ divides no term, or there is a unique residue class \[ n\equiv r_p\pmod{z(p)} \] on which $p$ divides $a_n$. Since $\gcd(m,a_n)>1$ for every $n$, for each $n$ some prime $p\in S$ divides $a_n$. Thus the finitely many residue classes coming from those primes cover all $n\ge 0$. This is exactly $2$.

Conversely, assume $2$, and set \[ m:=\prod_{i=1}^t p_i. \] Given any $n\ge 0$, the covering property yields an index $i$ with \[ n\equiv r_i\pmod{z(p_i)}. \] By definition of the prime-covering congruence system, $p_i\mid a_n$. Therefore \[ \gcd(m,a_n)\ge p_i>1. \] So $\gcd(m,a_n)>1$ for every $n\ge 0$, proving $1$.

This proves Claim C. $\square$

Claim D. Let $S$ be a fixed finite set of primes. Then deciding whether \[ \forall n\ge 0\ \exists p\in S:\ p\mid a_n \] holds is a finite computation.

Proof. For each $p\in S$, Claim B shows that the set of indices $n$ with $p\mid a_n$ is either empty or a single class modulo $z(p)$ (under the nontrivial coprime assumption relevant to the strong problem). Let \[ L:=\mathrm{lcm}\{z(p): p\in S\}. \] Then the condition \[ \forall n\ge 0\ \exists p\in S:\ p\mid a_n \] is equivalent to the statement that every residue class modulo $L$ lies in at least one of the lifted congruence classes coming from the primes in $S$. This requires checking only the finitely many residues $0,1,\dots,L-1$.

This proves Claim D. $\square$

Consequence. The literal Erdős condition \[ \forall m>1\ \exists n\ge 0:\ \gcd(m,a_n)=1 \] is exactly equivalent to saying that no finite prime-covering congruence system exists. Hence the phrase

"without an underlying system of covering congruences responsible"

has a canonical precise formalization for this problem, and that formalization is the strong Erdős statement itself.

6) ADVERSARIAL VERIFICATION

We check the new work against the main possible failure modes.

• Could $Z_p$ be a union of two residue classes? No. Claim B shows that once a single zero $a_r\equiv 0\pmod p$ occurs and $p$ does not divide all terms, the entire sequence modulo $p$ is forced to be a scalar multiple of $F_{n-r}$, so the zero set is exactly one class modulo $z(p)$.

• Did the proof of Claim A assume a theorem not justified here? No. Existence of $z(p)$ came from the finite order of the companion matrix in $\mathrm{GL}_2(\mathbf{F}_p)$; the divisibility criterion $p\mid F_n\iff z(p)\mid n$ was proved directly from the Round 1 addition formula, the Round 1 divisibility lemma, and Cassini's identity.

• What about the case that a prime divides all terms? Claim B isolates that as a separate case. In the actual Erdős problem, any admissible sequence must satisfy $\gcd(a_0,a_1)=1$, so this case cannot occur.

• Did Claim C accidentally use the weak statement instead of the strong one? No. Claim C is purely structural: it does not assume all terms are composite. It only compares a fixed finite gcd-obstruction with a finite prime-covering congruence system.

• Does Claim D solve the whole problem by finite computation? No. It only says that for a prescribed finite set of primes $S$, checking whether $S$ covers the sequence is finite. The open difficulty is that there is no known a priori bound on which primes could appear in such a set.

No contradiction with the Round 1 foundation appears.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 proves a new equivalence that was not available in Round 1:

For Fibonacci-like sequences, the literal strong Erdős condition \[ \forall m>1\ \exists n\ge 0:\ \gcd(m,a_n)=1 \] is equivalent to the absence of any finite prime-covering congruence system.

Therefore the intended actual conjecture is not different from the literal one. After the minimal natural formalization of "an underlying covering system is responsible", they are the same statement.

This does not solve Problem #276, but it sharpens the remaining obstruction to one precise gap:

Construct an all-composite Fibonacci-like sequence for which no finite family of primes gives a covering by the single congruence classes $n\equiv r_p\pmod{z(p)}$.

Equivalently, one must prove that for every finite set of primes $S$, there exists an index $n$ such that no prime in $S$ divides $a_n$.

8) COMPLETION ESTIMATE

COMPLETION: 80%.

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. Ismailescu 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.