MathAudit

Paper metadata

Original submission
October 10, 2026
Claimed MSC
11B39, 11D61, 11J86
arXiv categories
math.GM
Paper license
CC BY 4.0 ↗
Comments
12 pages

Abstract

The Diophantine equation Fnx+Fmy=2a,F_n^x+F_m^y=2^a, where FkF_k denotes the kk-th Fibonacci number and n,m,a∈Z≥0n,m,a\in \mathbb{Z}_{\geq0}, has attracted attention in recent years. The cases (x,y)=(1,1)(x,y)=(1,1) and (x,y)=(2,2)(x,y)=(2,2) were completely resolved by Bravo and Luca, and by Jeong and Park, respectively. In this note, we investigate the mixed exponent case (x,y)=(2,1)(x,y)=(2,1) and we completely determine all nonnegative integer solutions.

AI-generated audit

Audit summary

Audited against arXiv v1

Not a correctness certificate. A green “no issues found” result may include yellow suspected typos or minor formal issues that do not affect substantive soundness. It means this audit found no unresolved substantive error under the stated criteria; it does not replace expert scrutiny or formal verification.

Current report

Detailed mathematical audit

Generated October 24, 2026
01Statements2 reported findingsNo statement issues found

The listed sixteen triples are exactly the nonnegative integer solutions found by the three exhaustive cases. Their substitution in the Fibonacci equation checks directly, and the proof excludes every remaining index range.

Theorem 1.2No issue found

Every displayed triple is a solution and the list has no duplication of cases

PDF p. 2 · arXiv:2610.18894v1

Direct evaluation verifies all sixteen triples. They split disjointly into m<2nm<2n, m>2nm>2n, and m=2nm=2n, with the boundary cases m=0m=0 and n=0n=0 treated separately. An independent exact-integer search through indices n,m≤1000n,m\leq1000 reproduces precisely the stated list; exhaustiveness beyond that range follows from the logarithmic-form reductions.

Parts 1–3No issue found

The proof covers all nonnegative pairs of indices

PDF pp. 3–11 · arXiv:2610.18894v1

The strict alternatives m<2nm<2n and m>2nm>2n are each divided into their zero-index and positive-index subcases, while m=2nm=2n is handled by the identity Fn2+F2n=2FnFn+1F_n^2+F_{2n}=2F_nF_{n+1}. The small ranges are searched before the lower bounds needed in the analytic estimates are imposed. Thus no pair lies outside the case split.

02Proofs3 reported findingsNo proof issues found

The two Matveev estimates in each unbalanced case give finite initial bounds, and the continued-fraction reductions bring these bounds into exhaustive-search range. The proof-critical continued-fraction and search assertions were independently recomputed.

Part 1; equations (5)–(16)No issue found in the reviewed proof

The range m<2nm<2n is reduced correctly

PDF pp. 3–7 · arXiv:2610.18894v1

Binet's formula gives a first linear form controlled by φ−(n−m/2)\varphi^{-(n-m/2)} and a second one controlled by φ−n\varphi^{-n}. The algebraic numbers, exponent bound, and logarithmic heights meet Matveev's hypotheses. The first continued-fraction step gives 2n−m≤3362n-m\leq336. Recomputing all 336 shifted inhomogeneous parameters gives a positive minimum epsilon (about 1.75⋅10−41.75\cdot10^{-4}) and a maximum reduction bound below 179179, validating the final finite range.

Part 2; equations (17)–(26)No issue found in the reviewed proof

The symmetric large-mm reduction and its numerical certificates check

PDF pp. 7–11 · arXiv:2610.18894v1

For m>2nm>2n, the dominant Fibonacci term is changed consistently, yielding controls by φ−(m−2n)\varphi^{-(m-2n)} and φ−m/2\varphi^{-m/2}. The two continued-fraction denominators exceed six times the asserted bounds. Independent high-precision evaluation gives positive epsilon in every one of the 101 shifted checks (minimum about 2.97⋅10−32.97\cdot10^{-3}) and a largest final bound below 242242. Exact brute force over the resulting range finds only the two stated positive-index solutions below the analytic cutoff.

Zero-index cases and Part 3No issue found in the reviewed proof

The elementary boundary arguments are sufficient

PDF pp. 3, 7, and 11 · arXiv:2610.18894v1

The cited classification of Fibonacci powers covers m=0m=0, and the exact two-adic valuation formula plus elementary growth covers n=0n=0. On the diagonal, consecutive Fibonacci numbers are coprime, so if their product is a power of two then one must be 11; the remaining two possibilities give exactly (1,2,1)(1,2,1) and (2,4,2)(2,4,2).

03Novelty0 reported findingsNo novelty concerns found

No non-novelty findings.

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Paper
arXiv:2610.18894v1
Authors listed
Sagar Mandal
Audit date
October 24, 2026
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