{"id":"90352761-d174-4a43-b2a5-fc6357c6c60a","arxiv_id":"2607.09105","paper_version":1,"verdict":"ACCEPT","confidence":"HIGH","novelty_score":5.5,"correctness_risk":"low","formal_verification":"none","parameter_count":0,"one_line_summary":"For Lucas sequences U_n(x,±1) the equation ∑_{j=1}^k j U_j^p = U_n^q has only the listed solutions when k=2 (any p,q) and none when k≥3 and max{p,q}≤11 under mild conditions on x.","lead":"The paper completely solves a power-sum Diophantine equation for a broad family of Lucas sequences U_n(x, ±1) when the exponents are at most 11, and lists every solution when the sum has only two terms. It unifies and extends earlier Fibonacci and Pell results, showing almost no solutions exist beyond a short explicit list.","discovery_kind":"extension","skeptic_critique":{"model":"grok-4.5","headline":"No significant objection identified","rationale":"The reader correctly isolates the finite SageMath check as the sole non-formal step and correctly judges its risk low: after the analytic reduction the search space is tiny (29 tuples) and fully explicit. Re-running that check with independent exact arithmetic is the natural verification; nothing deeper in the analytic scaffolding (bounds, Zsigmondy/Primitive-Divisor arguments for k=2, or the norm argument leading to (4.4)) shows a load-bearing flaw inside the paper’s stated hypotheses. Consequently the ACCEPT verdict stands without adjustment.","tokens_in":17316,"tokens_out":383,"duration_ms":4596,"concrete_test":"Independently re-implement the final search of §4: for each of the 29 residual (k,n,p,q,x,y) tuples, evaluate both sides of (1.4) with exact integer arithmetic (e.g., Python/gmpy2 or Magma) and confirm that equality never holds. If any equality is found, the non-existence claim fails; otherwise the computer-assisted step is corroborated.","verdict_should_be":"UNCHANGED","load_bearing_attack":"The central claim of Theorem 1.1 rests on standard Binet-type estimates (Lemmas 2.2–2.5), a norm argument that produces an explicit upper bound k<838 after the x-restriction of Lemma 4.4, and a finite SageMath enumeration of the remaining 29 candidate tuples. Those steps are correctly derived within the stated range max{p,q}≤11 and y=±1; the only non-formal ingredient is a completely finite computer search whose outcome is stated clearly. No hidden analytic gap or circularity appears that would undermine the non-existence assertion for k≥3.","agreement_with_reader":"agree"},"referee_report":{"model":"grok-4.5","summary":"The paper solves the Diophantine equation ∑_{j=1}^k j U_j(x,y)^p = U_n(x,y)^q for Lucas sequences of the first kind with y=±1, positive integers x,p,q,k,n and max{p,q}≤11. Theorem 1.1 completely classifies the solutions of the k=2 case 1+2x^p=U_n(x,y)^q (five families) and proves that no solutions exist for k≥3 under the stated size restrictions on x relative to y. The argument proceeds via Binet-type growth bounds (Lemmas 2.2–2.5), a norm argument in Q(√D) that yields an explicit upper bound k<838 after the polynomial-divisibility restriction of Lemma 4.4, and a finite SageMath enumeration that eliminates the remaining 29 candidate tuples.","tokens_in":17460,"tokens_out":890,"duration_ms":9680,"significance":"The work cleanly unifies and extends the earlier Fibonacci (Soydan–Németh–Szalay, Gueth–Luca–Szalay) and Pell (Tchammou–Togbé) results to the full two-parameter family U_n(x,±1). The k=2 classification is unconditional on the exponents and relies on classical tools (Mihăilescu, Zsigmondy, Carmichael/Bilu–Hanrot–Voutier). The computer-assisted non-existence proof for k≥3 is fully rigorous once the analytic bound is accepted; the finite search is small (k≤838, x≤10) and therefore reproducible. The paper also supplies a transparent explanation why the same method fails for |y|>1, which is useful for future work.","major_comments":[],"minor_comments":[{"comment":"The abstract sentence is missing the word “in” (“We solve the equation … positive integers”). A quick grammatical pass would catch several similar slips (e.g., “EQUA TION”, “F AMILY” in the title).","section":null},{"comment":"In the statement of Theorem 1.1 the phrase “with x≥2 when y=−1 and x≥3 when y=1” appears only in the second half; it would be clearer to list the precise range of x for each y at the beginning of the theorem.","section":null},{"comment":"Lemma 2.7 is stated only for y=−1 and odd p; a one-sentence remark that the even-p case is already settled by the earlier norm argument would help the reader.","section":null},{"comment":"The final computer search (end of §4) reports that 29 candidates remain and that none work, but does not list them. A short table or a link to a SageMath worksheet would make the verification completely transparent.","section":null},{"comment":"References [7] and [8] are cited for the Fibonacci case with max{p,q}≤10; the paper claims the extension to exponent 11 is routine, yet never records the new numerical bound obtained after changing the constants. A single sentence would close the gap.","section":null}],"recommendation":"accept","confidential_remarks":"The manuscript is a solid, carefully written contribution that sits comfortably within the scope of a number-theory journal that publishes Diophantine equations involving linear recurrences. The only non-formal step is a finite, easily reproducible computer search; I see no reason to request major changes. The authors already acknowledge the referee of an earlier version, so the present text appears to be a polished revision."},"author_rebuttal":null,"desk_editor":{"model":"grok-4.5","letter":"This paper does exactly what the title claims: it takes the sum equation that Gueth–Luca–Szalay and Tchammou–Togbé solved for Fibonacci and Pell, and settles it for every Lucas sequence U_n(x,±1) when max{p,q}≤11 and k≥3, plus a complete solution list for the k=2 case with no exponent restriction.\n\nWhat is new is the uniform treatment. The k=2 analysis (Proposition 3.1) is self-contained and uses Zsigmondy plus Carmichael/primitive-divisor theorems in a way that genuinely needs y=±1; the five listed solutions look exhaustive. For k≥3 the argument re-uses the familiar Binet-plus-norm-plus-computer strategy, but the authors carefully re-derive the necessary growth lemmas (2.2–2.5) and the p-adic valuation control (via Sanna) so that the same bound machinery works for every x. The final computer reduction to 29 candidates under k<838 is finite and stated clearly; the stress-test correctly finds no analytic gap.\n\nSoft spots are minor and proportional. The bound on k is looser than the pure Fibonacci case because of the need for uniformity, and the SageMath enumeration is not machine-checked, so an undetected overflow could in principle leave an extra solution. The authors themselves explain why |y|>1 breaks the method (constant-term divisibility becomes exponential in k), so the restriction is honest rather than artificial. Citations are appropriate and non-circular.\n\nThis is for people who work on Diophantine equations involving linear recurrences or power sums of special sequences. It is solid enough that a serious editor should send it to referees; the result is modest in scope but rigorously established inside its stated limits. I would cite the k=2 list if I needed it, and I would bring the paper to a reading group if we were discussing these techniques.","headline":"Clean, correctly executed extension of known power-sum Diophantine results from Fibonacci/Pell to the full family U_n(x,±1), with a new complete k=2 analysis.","tokens_in":18058,"tokens_out":537,"would_cite":true,"duration_ms":6012,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11B39","11D61","11D45"],"pacs":[],"model":"grok-4.5","headline":"The weighted power-sum equation for Lucas sequences U_n(x, ±1) has no solutions when k ≥ 3 and max{p, q} ≤ 11, and only five solutions when k = 2.","keywords":["Lucas sequences","Fibonacci numbers","Pell numbers","Diophantine equation","power sums","Binet formula","primitive divisors"],"falsifier":"Exhibit any single sextuple (n, k, p, q, x, y) with k ≥ 3, max{p, q} ≤ 11, y = ±1 and x large enough that satisfies the sum equation, or exhibit an extra solution of the two-term equation beyond the five listed tuples.","tokens_in":18253,"feed_emoji":"🔢","tokens_out":797,"duration_ms":8058,"temperature":0.7,"pith_summary":"The paper asks when a weighted sum of p-th powers of terms from a Lucas sequence of the first kind can itself be a pure q-th power of a later term. The sequences considered are those with the second parameter fixed at y = ±1, so they include the Fibonacci and Pell numbers as special cases and many others. For the two-term sum (k = 2) the authors list every positive-integer solution without restricting the exponents. For longer sums they prove there are no solutions at all once both exponents are at most 11 and the first parameter x is large enough relative to y. The result therefore settles, inside a uniform family, several earlier Diophantine questions that had been treated only for single sequences, and it shows that the phenomenon of “almost no solutions” is not special to Fibonacci or Pell numbers.","feed_headline":"Lucas power-sum equation has almost no solutions","feed_subtitle":"Only five solutions when k=2; none at all for longer sums with exponents up to 11","key_machinery":"Binet-form approximation of the Lucas terms together with the p-adic valuation formula for U_n and a short computer search that reduces the analytic bound k < 838 to a finite list of 29 candidate tuples, all of which are then checked by hand.","core_discovery":"Theorem 1.1 asserts two things. First, the only positive-integer solutions of 1 + 2x^p = U_n(x, y)^q with y = ±1 are the five explicit tuples (4, p, 1, 1, -1), (1 + 2^{p+1}, p, 1, 2, 1), (3, 2, 2, 2, 1), (3, 1, 1, 2, -1) and (5, 3, 1, 3, 1). Second, when k ≥ 3, max{p, q} ≤ 11 and x is at least 2 (respectively 3) according as y = -1 (respectively +1), the full equation ∑_{j=1}^k j U_j(x, y)^p = U_n(x, y)^q has no solutions whatever.","pith_inferences":[],"forward_implications":[],"fun_headline_variants":["Only five k=2 solutions for Lucas power-sum equation","No solutions for k≥3 Lucas sums with powers ≤11","Lucas power-sum equation has five solutions only when k=2","Longer Lucas power sums never equal a higher power","Generalized Lucas equation ∑j Uj^p = Un^q nearly solution-free"],"cache_read_input_tokens":128,"weakest_assumption_plain":"The final computer search that discards the last 29 candidate sextuples after the analytic bound has already been reduced must be free of overflow or incomplete enumeration.","fun_headline_variants_meta":{"raw":{"variants":["Only five k=2 solutions for Lucas power-sum equation","No solutions for k≥3 Lucas sums with powers ≤11","Lucas power-sum equation has five solutions only when k=2","Longer Lucas power sums never equal a higher power","Generalized Lucas equation ∑j Uj^p = Un^q nearly solution-free"]},"model":"grok-4.5","effort":"low","cost_usd":0.007042,"raw_usage":{"total_tokens":1789,"prompt_tokens":827,"num_sources_used":0,"completion_tokens":93,"cost_in_usd_ticks":70420000,"prompt_tokens_details":{"text_tokens":827,"audio_tokens":0,"image_tokens":0,"cached_tokens":256},"completion_tokens_details":{"audio_tokens":0,"reasoning_tokens":869,"accepted_prediction_tokens":0,"rejected_prediction_tokens":0}},"tokens_in":827,"tokens_out":93,"duration_ms":9441,"temperature":1.0,"reasoning_tokens":869,"cache_read_input_tokens":256,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-07-13T05:21:47.606957+00:00","model_set":{"reader":"grok-4.5"},"falsifier":"Exhibit any single sextuple (n, k, p, q, x, y) with k ≥ 3, max{p, q} ≤ 11, y = ±1 and x large enough that satisfies the sum equation, or exhibit an extra solution of the two-term equation beyond the five listed tuples.","supporting_citations":[],"review_version":1}