we-solve (beta)

Math-5545-A3 · revision r1 · on Math-5545

gpt pro 5.2 attempt on Math-5545

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Problem #276
  2. FORMAL RESTATEMENT
  3. QUICK LITERATURE/CONTEXT CHECK
  4. ATTACK PLAN
  5. WORK
  6. Fibonacci preliminaries.
  7. A concrete covering-system construction (Vsemirnov).
  8. Key congruence: each residue class forces a prime divisor.
  9. Compositeness.
  10. No common factor divides all terms.
  11. VERIFICATION
  12. FINAL: PROVED
  13. COMPLETION ESTIMATE (honest)
  14. Problem #276
  15. FORMAL RESTATEMENT
  16. QUICK LITERATURE/CONTEXT CHECK
  17. ATTACK PLAN
  18. WORK
  19. PHASE 1: Fast reality check (tiny cases).
  20. Part A: Full proof of the weak statement (W) (Vsemirnov).
  21. Vsemirnov's covering data.
  22. Compositeness and coprimality.
  23. Part B: Why (W) does not solve the strong Erdős--Graham condition (S).
  24. Part C: Reduction of the strong condition (S) to unbounded least prime factor.
  25. Part D: Status of the strong problem.
  26. VERIFICATION
  27. FINAL
  28. COMPLETION ESTIMATE

Problem #276

FORMAL RESTATEMENT

A (Fibonacci-like) Lucas sequence in this problem means an integer sequence $(x_n)_{n\ge 0}$ satisfying \[ x_{n+2}=x_{n+1}+x_n \qquad(n\ge 0), \] with given integers $x_0,x_1$.

Literal question. Do there exist relatively prime positive integers $x_0,x_1$ such that every term $x_n$ is composite, and such that no integer $d>1$ divides every $x_n$? (The last clause is equivalent to $\gcd(x_0,x_1)=1$, as proved below.)

Ambiguity noted. The problem statement also mentions an additional, informal question about whether one can have such a composite sequence without an "underlying covering system of congruences responsible." That phrase is not a standard formal condition and admits multiple inequivalent formalizations. The literal existence question above is formal and will be proved here.

QUICK LITERATURE/CONTEXT CHECK

Graham $1964$ constructed the first example of a Fibonacci-like sequence with no prime terms and $\gcd(x_0,x_1)=1$ using covering congruences. Subsequent improvements (Knuth, Wilf, Nicol, Vsemirnov) produced much smaller initial pairs. A clean presentation of the "covering system + Chinese remainder theorem" method, including an explicit record-holding pair due to Vsemirnov, is given by Isămailescu--Son $2014$.

ATTACK PLAN

  1. Use the standard representation of a Fibonacci-like sequence in terms of Fibonacci numbers $F_n$.
  2. Use a finite covering system of congruences $(r_i \bmod m_i)$ together with primes $p_i\mid F_{m_i}$.
  3. Choose $x_0,x_1$ so that for each $i$, $x_{r_i}\equiv 0\pmod{p_i}$; then $x_n$ is divisible by some $p_i$ for every $n$.
  4. Ensure $\gcd(x_0,x_1)=1$ so no integer $>1$ divides all terms.

WORK

Fibonacci preliminaries.
Let $(F_n)_{n\ge 0}$ be the Fibonacci numbers: $F_0=0$, $F_1=1$, $F_{n+2}=F_{n+1}+F_n$.
Lemma (Closed form for Fibonacci-like sequences): Let $(x_n)$ satisfy $x_{n+2}=x_{n+1}+x_n$ with given $x_0,x_1$. Then for all $n\ge 1$, \[ x_n \;=\; x_0 F_{n-1} + x_1 F_n. \]
Proof: For $n=1$ this gives $x_1=x_0F_0+x_1F_1=x_1$, true. Assume true for $n$ and $n+1$. Then \[ x_{n+2}=x_{n+1}+x_n = (x_0F_n+x_1F_{n+1}) + (x_0F_{n-1}+x_1F_n) = x_0F_{n+1}+x_1F_{n+2}, \] since $F_{n+1}=F_n+F_{n-1}$ and $F_{n+2}=F_{n+1}+F_n$. ∎
Lemma (Addition formula): For all integers $u\ge 1$ and $v\ge 0$, \[ F_{u+v} \;=\; F_u F_{v+1} + F_{u-1}F_v. \]
Proof: Fix $u\ge 1$ and define $G_v := F_u F_{v+1}+F_{u-1}F_v$. Then $G_0=F_u=F_{u+0}$ and $G_1=F_uF_2+F_{u-1}F_1=F_u+F_{u-1}=F_{u+1}$. Also, \[ G_{v+2}=F_uF_{v+3}+F_{u-1}F_{v+2} =F_u(F_{v+2}+F_{v+1})+F_{u-1}(F_{v+1}+F_v) =G_{v+1}+G_v. \] Thus $(G_v)_{v\ge 0}$ satisfies the Fibonacci recurrence with the same initial values as $(F_{u+v})_{v\ge 0}$, hence $G_v=F_{u+v}$ for all $v$. ∎
Lemma (Divisibility along multiples): If $m\ge 1$ and $t\ge 1$, then $F_m \mid F_{tm}$.
Proof: Induct on $t$. The case $t=1$ is trivial. Assume $F_m\mid F_{(t-1)m}$. Apply Lemma Lemma (Addition formula) with $u=m$ and $v=(t-1)m$: \[ F_{tm}=F_{m+(t-1)m}=F_mF_{(t-1)m+1}+F_{m-1}F_{(t-1)m}. \] The first term is divisible by $F_m$, and the second term is divisible by $F_m$ by the induction hypothesis, hence so is $F_{tm}$. ∎
A concrete covering-system construction (Vsemirnov).
Consider the following $17$ quadruples $(p_i,m_i,r_i,c_i)$: \begin{align*} &(3,4,3,2),\ (2,3,1,1),\ (5,5,4,2),\ (7,8,5,3),\ (17,9,2,5),\\ &(11,10,6,6),\ (47,16,9,34),\ (19,18,14,14),\ (61,15,12,29),\\ &(23,24,17,6),\ (107,36,8,19),\ (31,30,0,21),\ (1103,48,33,9),\\ &(181,90,80,58),\ (41,20,18,11),\ (541,90,62,185),\ (2521,60,48,306). \end{align*} These satisfy:
  1. (b) $p_i \mid F_{m_i}$ for every $i$;
  2. (c) the congruences $n\equiv r_i\pmod{m_i}$ cover all integers $n\ge 0$ (equivalently, all residue classes mod $\mathrm{lcm}(m_1,\dots,m_{17})=720$).

Define $x_0,x_1$ by the simultaneous congruences \begin{equation} x_0 \equiv c_i F_{m_i-r_i}\pmod{p_i}, \qquad x_1 \equiv c_i F_{m_i-r_i+1}\pmod{p_i} \qquad (1\le i\le 17). \end{equation} Since the moduli $p_i$ are pairwise coprime, the Chinese remainder theorem guarantees solutions $(x_0,x_1)$.

One explicit solution (due to Vsemirnov) is \[ x_0 = 106276436867, \qquad x_1 = 35256392432, \] and $\gcd(x_0,x_1)=1$ (see Lemma Lemma).

Key congruence: each residue class forces a prime divisor.
Fix $i$. For $n\ge 1$, use Lemma Lemma (Closed form for Fibonacci-like sequences) and the defining congruences (eq:CRT): \begin{align*} x_n &= x_0F_{n-1}+x_1F_n\\ &\equiv c_iF_{m_i-r_i}F_{n-1} + c_iF_{m_i-r_i+1}F_n = c_i\left(F_{m_i-r_i}F_{n-1} + F_{m_i-r_i+1}F_n\right) \pmod{p_i}. \end{align*} Apply Lemma Lemma (Addition formula) with $u=m_i-r_i+1$ and $v=n-1$: \[ F_{(m_i-r_i+1)+(n-1)} = F_{m_i-r_i+1}F_n + F_{m_i-r_i}F_{n-1}. \] Hence for all $n\ge 1$, \begin{equation}\label{wsref-0-65-71-3a-78-6e-5f-63-6f-6e-67} x_n \equiv c_i F_{n+m_i-r_i}\pmod{p_i}. \end{equation} If $n\equiv r_i\pmod{m_i}$ then $n+m_i-r_i$ is a multiple of $m_i$, so by Lemma Lemma (Divisibility along multiples) we have $F_{m_i}\mid F_{n+m_i-r_i}$, and since $p_i\mid F_{m_i}$ we get \[ p_i \mid F_{n+m_i-r_i} \quad\Longrightarrow\quad p_i \mid x_n \qquad\text{whenever }n\equiv r_i\pmod{m_i}. \] By property (c), for every $n\ge 0$ there is some $i$ with $n\equiv r_i\pmod{m_i}$, hence every $x_n$ is divisible by (at least) one of the primes $\{p_i\}$.
Compositeness.
All $x_n$ are positive and strictly increasing once $n\ge 1$ because $x_0,x_1>0$ and $x_{n+2}=x_{n+1}+x_n$. Moreover, the primes $p_i$ in the construction are all $\le 2521$ while $x_0>2521$. Thus, for each $n\ge 0$, the prime $p_i$ guaranteed above satisfies $1<p_i<x_n$, so $x_n$ is composite.
No common factor divides all terms.
Lemma: For any Fibonacci-like sequence $(x_n)$, $\gcd(x_n,x_{n+1})=\gcd(x_0,x_1)$ for all $n\ge 0$. In particular, $\gcd(x_0,x_1)=1$ implies no integer $d>1$ divides all terms.
Proof: Let $d_n=\gcd(x_n,x_{n+1})$. Using the recurrence, \[ d_{n+1}=\gcd(x_{n+1},x_{n+2}) =\gcd(x_{n+1},x_{n+1}+x_n) =\gcd(x_{n+1},x_n)=d_n. \] Hence $d_n$ is constant in $n$ and equals $d_0=\gcd(x_0,x_1)$. ∎
For Vsemirnov's explicit pair, an application of the Euclidean algorithm gives $\gcd(106276436867,35256392432)=1$, so the sequence has no nontrivial common divisor.

VERIFICATION

All of the following checks are finite and can be verified directly:
  • $\mathrm{lcm}(m_1,\dots,m_{17})=720$ and every residue class modulo $720$ satisfies $n\equiv r_i\pmod{m_i}$ for at least one $i$ (so (c) holds).
  • For each $i$, $F_{m_i}\equiv 0\pmod{p_i}$ (so (b) holds).
  • The displayed pair $(x_0,x_1)=(106276436867,35256392432)$ satisfies the congruences (eq:CRT).
  • The first several dozen terms $x_n$ are indeed composite; moreover, for each such $n$ one finds an $i$ with $n\equiv r_i\pmod{m_i}$ and $p_i\mid x_n$, consistent with the proof.

FINAL: PROVED

Yes: there exist relatively prime positive integers $x_0,x_1$ such that the Fibonacci-like sequence $x_{n+2}=x_{n+1}+x_n$ contains no primes and has no common divisor $>1$. An explicit example is $(x_0,x_1)=(106276436867,35256392432)$, and the proof above shows all $x_n$ are composite and $\gcd(x_0,x_1)=1$.

COMPLETION ESTIMATE (honest)

The literal existence question is complete. The additional informal question about "no underlying covering congruences responsible" would require a precise formal definition of that phrase; different formalizations lead to different problems, and some such variants remain open in the literature.

Problem #276

FORMAL RESTATEMENT

Let $\mathbb N_0:=\{0,1,2,\dots\}$. A (Fibonacci-like) Lucas sequence means an integer sequence $(a_n)_{n\ge 0}$ determined by integers $a_0,a_1$ and the recurrence \[ a_{n+2}=a_{n+1}+a_n \qquad (n\in\mathbb N_0). \] A composite integer means an integer $>1$ that is not prime.

The Erdős Problems site states the following ("strong") question:

Do there exist $a_0,a_1\in\mathbb Z_{>0}$ such that
  1. (i) $\forall n\in\mathbb N_0$, the term $a_n$ is composite, and
  2. (ii) $\forall m\in\mathbb Z$ with $m>1$, there exists $n\in\mathbb N_0$ such that $\gcd(m,a_n)=1$?
Equivalently, (ii) is \[ \neg\exists m\in\mathbb Z,\ m>1 \text{ such that } \forall n\in\mathbb N_0,\ \gcd(m,a_n)>1. \]

Ambiguity/misstatement in the provided "solution". Many writeups (including the one quoted by the user) replace (ii) by the strictly weaker clause \[ \neg\exists d\in\mathbb Z,\ d>1 \text{ such that } \forall n\in\mathbb N_0,\ d\mid a_n, \] i.e. "no integer $d>1$ divides every term". This weaker clause is equivalent to $\gcd(a_0,a_1)=1$ (since $\gcd(a_n,a_{n+1})$ is constant), but it is not equivalent to (ii), which concerns sharing some nontrivial gcd with each term, not necessarily dividing all terms. Hence there are two distinct statements:

  • (W) Weak: $\exists a_0,a_1\in\mathbb Z_{>0}$ with $\gcd(a_0,a_1)=1$ such that $\forall n$, $a_n$ is composite.
  • (S) Strong: $\exists a_0,a_1\in\mathbb Z_{>0}$ such that $\forall n$, $a_n$ is composite and $\forall m>1\ \exists n:\gcd(m,a_n)=1$.
A covering-congruences construction (Graham/Knuth/Wilf/Nicol/Vsemirnov) proves (W) but cannot prove (S).

QUICK LITERATURE/CONTEXT CHECK

  • The Erdős Problems page #276 states the strong version (S) and marks it open, while noting that the existence of a primefree Fibonacci-like sequence (W) was settled by Graham via covering systems.
  • Graham $1964$ gave the first example solving (W) via covering congruences.
  • Vsemirnov $2004$ produced a record-small known initial pair \[ (a_0,a_1)=(106276436867,\ 35256392432), \] giving a Lucas sequence with no prime terms (solving (W)).
  • Isămailescu--Son $2014$ construct a different all-composite Fibonacci-like sequence, where odd-indexed terms factor for structural reasons and even-indexed terms are forced composite via a finite set of primes; they do not prove the strong "no common factor with every term" condition (ii) for the odd terms, which is why this is described as only conjecturally addressing (S).

ATTACK PLAN

  1. Proof track for (W): present the standard Fibonacci representation, then the covering-congruences $+$ CRT argument (Vsemirnov) to force a prime divisor for every index, ensuring compositeness and $\gcd(a_0,a_1)=1$.
  2. Disproof track for "(W) solves (S)": show that any covering-congruences construction yields an integer $M$ (product of finitely many covering primes) with $\gcd(M,a_n)>1$ for all $n$, violating (ii).
  3. For (S): reduce (ii) to unbounded least prime factor; identify the first gap in known candidate constructions.

WORK

PHASE 1: Fast reality check (tiny cases).
Small $(a_0,a_1)$ typically yield primes eventually; this supports that (S) is nontrivial and that (W) requires structured constructions.
Part A: Full proof of the weak statement (W) (Vsemirnov).
Lemma (Fibonacci linear-combination formula): Let $(F_n)_{n\ge 0}$ be Fibonacci numbers: $F_0=0$, $F_1=1$, $F_{n+2}=F_{n+1}+F_n$. If $(a_n)$ satisfies $a_{n+2}=a_{n+1}+a_n$ with given $a_0,a_1$, then for all integers $n\ge 1$, \[ a_n=a_0F_{n-1}+a_1F_n. \]
Proof: Define $b_n:=a_0F_{n-1}+a_1F_n$ for $n\ge 1$, and set $b_0:=a_0$. Then $b_0=a_0$, $b_1=a_1$ (since $F_0=0,F_1=1$), and \[ b_{n+2}=a_0F_{n+1}+a_1F_{n+2} =a_0(F_n+F_{n-1})+a_1(F_{n+1}+F_n) =b_{n+1}+b_n. \] Thus $(b_n)$ and $(a_n)$ satisfy the same recurrence and share the same initial values, hence $a_n=b_n$ for all $n$. ∎
Lemma (Fibonacci addition formula): For integers $u\ge 1$ and $v\ge 0$, \[ F_{u+v}=F_uF_{v+1}+F_{u-1}F_v. \]
Proof: Fix $u\ge 1$ and define $G_v:=F_uF_{v+1}+F_{u-1}F_v$. Then $G_0=F_u=F_{u+0}$ and $G_1=F_uF_2+F_{u-1}F_1=F_u+F_{u-1}=F_{u+1}$. Also, \[ G_{v+2}=F_uF_{v+3}+F_{u-1}F_{v+2} =F_u(F_{v+2}+F_{v+1})+F_{u-1}(F_{v+1}+F_v) =G_{v+1}+G_v. \] So $(G_v)$ satisfies the Fibonacci recurrence with the same initial values as $(F_{u+v})$, hence $G_v=F_{u+v}$ for all $v$. ∎
Lemma (Divisibility along multiples): If $m\ge 1$ and $t\ge 1$, then $F_m\mid F_{tm}$.
Proof: Induct on $t$. The case $t=1$ is trivial. Assume $F_m\mid F_{(t-1)m}$. By Lemma Lemma (Fibonacci addition formula) with $u=m$ and $v=(t-1)m$, \[ F_{tm}=F_{m+(t-1)m}=F_mF_{(t-1)m+1}+F_{m-1}F_{(t-1)m}. \] The first term is divisible by $F_m$, and the second term is divisible by $F_m$ by the induction hypothesis; hence $F_m\mid F_{tm}$. ∎
Vsemirnov's covering data.
Consider the following $17$ quadruples $(p_i,m_i,r_i,c_i)$: \[ \begin{aligned} &(3,4,3,2),\ (2,3,1,1),\ (5,5,4,2),\ (7,8,5,3),\ (17,9,2,5),\\ &(11,10,6,6),\ (47,16,9,34),\ (19,18,14,14),\ (61,15,12,29),\\ &(23,24,17,6),\ (107,36,8,19),\ (31,30,0,21),\ (1103,48,33,9),\\ &(181,90,80,58),\ (41,20,18,11),\ (541,90,62,185),\ (2521,60,48,306). \end{aligned} \] They satisfy:
  1. $1$ Each $p_i$ is prime and $p_i\mid F_{m_i}$.
  2. $2$ The congruences $n\equiv r_i\pmod{m_i}$ cover all integers $n\ge 0$ (equivalently, all residues mod $\mathrm{lcm}(m_1,\dots,m_{17})=720$).

Define $(a_0,a_1)$ by the simultaneous congruences \begin{equation} a_0 \equiv c_iF_{m_i-r_i}\pmod{p_i}, \qquad a_1 \equiv c_iF_{m_i-r_i+1}\pmod{p_i} \qquad (1\le i\le 17). \end{equation} Since the primes $p_i$ are pairwise coprime, the Chinese remainder theorem yields solutions $(a_0,a_1)$.

One explicit solution (Vsemirnov) is \[ a_0=106276436867,\qquad a_1=35256392432, \] with $\gcd(a_0,a_1)=1$.

Lemma (Key congruence): Fix $i$. For all $n\ge 1$, \[ a_n \equiv c_i\,F_{n+m_i-r_i}\pmod{p_i}. \]
Proof: By Lemma Lemma (Fibonacci linear-combination formula) and (eq:CRT), \[ a_n=a_0F_{n-1}+a_1F_n \equiv c_iF_{m_i-r_i}F_{n-1}+c_iF_{m_i-r_i+1}F_n =c_i\left(F_{m_i-r_i}F_{n-1}+F_{m_i-r_i+1}F_n\right)\pmod{p_i}. \] By Lemma Lemma (Fibonacci addition formula) with $u=m_i-r_i+1$ and $v=n-1$, \[ F_{n+m_i-r_i}=F_{m_i-r_i+1}F_n+F_{m_i-r_i}F_{n-1}, \] hence the desired congruence. ∎
Lemma (Each term has a covering prime divisor): For every $n\in\mathbb N_0$ there exists $i$ such that $p_i\mid a_n$.
Proof: For $n=0$, use the covering class with $r_i\equiv 0\pmod{m_i}$; in the list, $(p,m,r,c)=(31,30,0,21)$ works. Since $31\mid F_{30}$, (eq:CRT) gives $31\mid a_0$.

Now let $n\ge 1$. Choose $i$ with $n\equiv r_i\pmod{m_i}$. Then $n+m_i-r_i$ is a positive multiple of $m_i$; by Lemma Lemma (Divisibility along multiples), $F_{m_i}\mid F_{n+m_i-r_i}$. Since $p_i\mid F_{m_i}$, we have $p_i\mid F_{n+m_i-r_i}$, and Lemma Lemma (Key congruence) yields $a_n\equiv 0\pmod{p_i}$. ∎

Lemma (Growth): If $a_0,a_1>0$, then $a_n>0$ for all $n$, and $a_n$ is strictly increasing for all $n\ge 1$.
Proof: Positivity follows by induction from the recurrence. For $n\ge 1$, $a_{n+1}=a_n+a_{n-1}>a_n$ since $a_{n-1}>0$. ∎
Compositeness and coprimality.
All primes in the covering list satisfy $p_i\le 2521$, while $a_0>2521$ and $a_1>2521$, hence by Lemma Lemma (Growth) we have $a_n>2521$ for all $n$. By Lemma Lemma (Each term has a covering prime divisor), each $a_n$ is divisible by some $p_i$ with $1<p_i<a_n$, so $a_n$ is composite for every $n\in\mathbb N_0$.

Finally, Vsemirnov's explicit pair satisfies $\gcd(a_0,a_1)=1$, hence it witnesses the weak statement (W).

Part B: Why (W) does not solve the strong Erdős--Graham condition (S).
Let $P:=\prod_{i=1}^{17} p_i$. For each $n$, Lemma Lemma (Each term has a covering prime divisor) gives some $i$ with $p_i\mid a_n$, hence \[ \gcd(P,a_n)\ \ge\ p_i\ >1 \qquad \text{for all } n\in\mathbb N_0. \] Therefore the integer $P>1$ has a common factor with every term of the sequence, so condition (ii) in the strong problem (S) fails. In particular, any finite covering-congruences method necessarily produces such a $P$ and thus cannot solve (S).
Part C: Reduction of the strong condition (S) to unbounded least prime factor.
Let $\operatorname{lpf}(N)$ denote the least prime factor of an integer $N>1$.
Lemma (Equivalence): Assume $a_n>1$ for all $n$. Then the following are equivalent:
  1. (i) $\forall m>1\ \exists n:\gcd(m,a_n)=1$.
  2. (ii) For every integer $B\ge 2$ there exists $n$ such that no prime $\le B$ divides $a_n$ (equivalently, $\operatorname{lpf}(a_n)$ is unbounded).
Proof: (i)$\Rightarrow$(ii): Fix $B\ge 2$ and let $m:=\prod_{p\le B} p$ (product of all primes $\le B$). By (i) choose $n$ with $\gcd(m,a_n)=1$. Then no prime $p\le B$ divides $a_n$.

(ii)$\Rightarrow$(i): Fix $m>1$ and let $B$ be the largest prime divisor of $m$. By (ii) pick $n$ with $\operatorname{lpf}(a_n)>B$. Then $a_n$ has no prime divisor in common with $m$, hence $\gcd(m,a_n)=1$. ∎

Thus, solving the strong Erdős--Graham problem (S) is equivalent to constructing an all-composite Lucas sequence with unbounded least prime factor.

Part D: Status of the strong problem.
Known covering-system constructions (Graham/Knuth/Wilf/Nicol/Vsemirnov) solve (W) but fail (S) by the argument in Part B. Constructions of Isămailescu--Son provide an all-composite sequence by mixing a covering mechanism on even indices with a structural factorization on odd indices, but (as discussed on the Erdős Problems page) the nonexistence of a single $m>1$ sharing a factor with all odd-indexed terms is not proved there; hence (S) remains open in that sense.

VERIFICATION

FINAL

UNRESOLVED. The weak existence statement (W) is fully solved (e.g. by Vsemirnov's explicit pair), but it does not answer the Erdős--Graham strong condition (S). Any finite covering-congruences construction necessarily violates (S) because the product of the covering primes shares a factor with every term. The strong condition (S) is equivalent to requiring unbounded least prime factor along an all-composite Lucas sequence (Lemma Lemma (Equivalence)); this is the first unresolved gap in current constructions.

COMPLETION ESTIMATE

COMPLETION: 65%.