GPT 6 Astra Ultra attempt on Math-5545

Erdős Problem #276: independent continuation

Date: 2026-09-23. Research attempt; no claim of a new published result.

1) TARGET AND SOURCE AUDIT

Construct positive integers $a_0,a_1$ such that $a_{n+2}=a_{n+1}+a_n$ is always composite and, for every integer $m>1$, some term is coprime to $m$.

All supplied versions considered: 276.tex, 276_v2.tex, 276_v3.tex. Version 1 confuses "no common divisor of all terms" with the stated stronger quantifier. Its assertion that an even-term prime cover obstructs the strong condition is false: odd terms can supply the coprime witness. Versions 2 and 3 correctly use the strong condition. Version 3 has a CRT coprimality omission and a claimed finite covering check; I repair both below.

2) WORK

Repairing CRT coprimality.

Suppose compatible local initial data are imposed modulo $Q$, a product of distinct covering primes, and at each such prime the two initial residues are not both zero. Choose a positive representative $a_0$ of its prescribed class. For every prime $\ell\mid a_0$ with $\ell\nmid Q$, additionally require $a_1\equiv1\pmod\ell$. These conditions are compatible with the prescribed $a_1\pmod Q$ by CRT. Choosing a large positive solution gives $\gcd(a_0,a_1)=1$. Thus arbitrary CRT representatives need not be coprime, but suitable representatives exist.

Replacing the six-prime computation.

For a coprime recurrence, each prime that divides some term divides exactly one index class modulo its Fibonacci rank $z(p)$: at a zero index $r$ the recurrence is a nonzero scalar multiple of $F_{n-r}$ modulo $p$. Hence a finite prime cover requires $\sum_p1/z(p)\ge1$. The smallest ranks for distinct primes are $3,4,5,7$; their reciprocal sum is $389/420<1$. A five-prime cover would have largest rank at most $420/31<14$. Factoring $F_1,\ldots,F_{13}$ gives the possible ranks $3,4,5,7,8,9,10,11,13$, each attached to one prime in this range. The only five-element selections with reciprocal sum at least one are \[ (3,4,5,7,t)\ (t=8,9,10,11,13),\quad(3,4,5,8,9),\quad(3,4,5,8,10). \] A residue class whose modulus is coprime to the least common multiple of all other moduli cannot complete a cover unless the other classes already cover: if they miss $r\pmod L$, CRT supplies an integer congruent to $r\pmod L$ but outside that single class. For the first five selections delete modulus $7$; the remaining reciprocal sum is at most $1/3+1/4+1/5+1/8<1$. For the last two delete respectively $5$ and $3$; again the remaining sum is below one. None covers. Thus at least six primes are necessary, with no exhaustive computation.

3) ADVERSARIAL VERIFICATION

The rank list is only claimed for ranks at most thirteen; distinct primes can have equal larger ranks. An even-subsequence cover and absence of an odd-subsequence cover together imply absence of a whole-sequence finite cover, rather than obstruct it. These repairs certify structural lemmas only.

4) FINAL

UNRESOLVED. No all-composite recurrence escaping every finite prime cover is constructed. Necessary covering-size bounds do not resolve the infinite avoidance condition.

5) SOURCES CHECKED

All three supplied versions read. Actual quantifiers checked in https://raw.githubusercontent.com/google-deepmind/formal-conjectures/main/FormalConjectures/ErdosProblems/276.lean, 2026-09-23. The rank-class fact is reproved locally in the supplied versions; no claim of a new literature result is made.

6) COMPLETION ESTIMATE

This is a conservative subjective measure of progress toward the original question, not a probability of correctness or a claim that the remaining work is routine. COMPLETION: 15%