{"id":"4f9b02ef-6545-4c98-8060-221e18688fff","arxiv_id":"2506.14379","paper_version":4,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":5.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"For m >= 2, the equation (L_m)^n + (L_m)^(n+k) = L_r has no positive integer solutions in r, m, n, and k.","lead":"This paper proves that no Lucas number can be written as one power of a smaller Lucas number plus another power of the same smaller Lucas number. The result is a complete answer to a specific Diophantine equation, using standard tools from number theory.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Inequality (3.12) is not justified: with only n ≤ 61 available it should contain (61 + k), not (28 + k), and the J^(m)(S) values used to conclude k ≤ 8 are not listed; the reduction to the finite check is therefore unverified.","rationale":"The paper's central claim is plausible, and the n = 1 case appears correct. The n ≥ 2 proof follows a standard Matveev-plus-continued-fractions template, with clean bounds leading to m ≤ 55 and n ≤ 61. The load-bearing problem is exactly where the reader located it: inequality (3.12) does not follow from (3.10) and (3.11). The correct combined inequality would have (n + k) on the right, and since only n ≤ 61 is known, the printed (28 + k) is unsupported. This is not a matter of disagreement with consensus; it is an internal derivational gap. Moreover, the step 'using the corresponding values of J^(m)(S), we find that k ≤ 8' supplies no numerical evidence in the text. The Zenodo code may settle it, but the written proof is incomplete without either the J values or the code output. I did not make Lemma 2.4 the central objection: it is unproved in this paper but cited to a published source, and its use is natural since the second factor L_m^k + 1 is coprime to L_m. If the corrected computation still yields k ≤ 8, the paper is readily repairable; if not, the claimed reduction collapses. Because this matches the reader's conditional assessment, I recommend no change to the verdict.","tokens_in":6098,"tokens_out":9735,"duration_ms":100572,"concrete_test":"Recompute (3.12) with the correct factor (61 + k) instead of (28 + k), using the actual continued-fraction values J^(m)(S) for m = 2, ..., 55 and S = 61 + 1.71 × 10^11, either from the Zenodo code or by independently computing partial quotients up to q_N > S. For each m, compute the maximal k satisfying L_m^k < 1.0025 (J^(m)(S) + 2) (61 + k) / log α; if the maximum exceeds 8 for any m, the reduction to k ≤ 8 and hence to r ≤ 3865 fails. Also verify the same computation with (28 + k) to check whether the printed claim at least holds for the stated constants.","verdict_should_be":"UNCHANGED","load_bearing_attack":"The critical reduction in the n ≥ 2 case rests on inequality (3.12). Combining (3.10) and (3.11) gives L_m^k < 1.0025 (J^(m)(S) + 2) (n + k) / log α. At that point the paper has established only n ≤ 61, so n + k ≤ 61 + k. The printed inequality (3.12) replaces this with (28 + k), which is not justified by any bound in the paper. This makes the subsequent claim, that (3.12) forces k ≤ 8 for all m = 2, ..., 55, depend on an inequality weaker than what the proof has shown. The conclusion k ≤ 8 is also asserted without listing any of the J^(m)(S) values, and without showing the code output, so the reader cannot independently verify the numerical step that shrinks the search space to r ≤ 3865. This may be a repairable typo or a residue of an earlier stronger bound on n, but as written the proof does not establish the reduction. Lemma 2.4, the other candidate weak point, is imported from a published source and its application here is legitimate because gcd(L_m, L_m^k + 1) = 1, so the exact divisibility condition L_m^n || L_r does follow from (1.1). Thus the primary load-bearing gap is the unjustified inequality and the missing numerical data.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper claims to settle the Diophantine equation L_m^{n+k} + L_m^n = L_r completely: for positive integers r, m, n, k with m >= 2, the authors assert there are no solutions. The proof splits into the case n = 1, handled by an exact factorization of L_r - L_m together with Carmichael's primitive divisor theorem, and the case n >= 2, handled by an exact divisibility lemma, Matveev's lower bound for linear forms in logarithms, a Legendre/continued-fraction reduction, and a final finite computer check. The n = 1 argument is clean and self-contained. For n >= 2, the structure is standard, but the reduction step contains several unverified and, as written, incorrect ingredients.","tokens_in":6400,"tokens_out":9412,"duration_ms":94032,"significance":"If the proof were correct, the result would be a definitive non-existence theorem for a natural exponential Diophantine equation involving Lucas numbers, complementing recent results on Fibonacci and Lucas equations. The n = 1 part is a strong, complete contribution. The n >= 2 part uses standard and appropriate tools (Matveev, exact divisibility, Legendre) with no fitted parameters, and the authors provide a Zenodo link for the computational part. However, the current manuscript does not establish the key reduction to a finite search, because a central inequality is unjustified and the stated continued-fraction lemma is false in the form used.","major_comments":[{"comment":"Inequality (3.12) is not justified. At that point in the proof the only established bound is n <= 61 from (3.8), so combining (3.10) and (3.11) yields L_m^k < 1.0025 (J^{(m)}(S)+2) (n+k)/log alpha, and hence at best L_m^k < 1.0025 (J^{(m)}(S)+2) (61+k)/log alpha. The printed inequality (3.12) contains (28+k), which is not derivable from any bound appearing before it. Since the subsequent conclusion k <= 8 depends precisely on this inequality, the reduction to the finite range r <= 3865 is not established as written. The authors must either prove a bound n <= 28 at that stage or redo the numerical step with (61+k) and report the resulting bounds.","section":"Section 3, Eq. (3.12)"},{"comment":"Lemma 2.5 as stated is false. The lower bound is asserted to hold with J(S) := max{a_i : i = 0, 1, ..., N}, where q_N > S, but the correct continued-fraction lower bound requires the next partial quotient a_{N+1}, not the maximum of the previous ones. For example, for x = pi = [3; 7, 15, 1, 292, ...], take N = 1, S = 10, q_1 = 7 > S, and p/q = 22/7. Then |pi - 22/7| ≈ 0.00126, which is less than 1/(2*7^2), yet the claimed bound with J(S) = max{3,7} = 7 would give |pi - 22/7| > 1/((7+2)*7^2) ≈ 0.00227, a contradiction. Thus inequality (3.11), which is essential for deriving (3.12) and ultimately k <= 8, rests on an invalid lemma. The authors must correct the statement of Lemma 2.5, recompute the partial-quotient maxima accordingly, and verify that the numerical conclusion k <= 8 still follows.","section":"Lemma 2.5 and Eq. (3.11)"},{"comment":"The step \"By taking m = 2, 3, ..., 55 in (3.12), and using the corresponding values of J^{(m)}(S), we find that k <= 8\" is not verifiable from the manuscript. No values of N_m or J^{(m)}(S) are listed, and no code output or table is provided. This numerical step is load-bearing: it shrinks the search space from k < 1.71*10^11 to k <= 8, which then yields r <= 3865 and the final finite range. The authors should include a table of the relevant continued-fraction data (or the code output) so that the reduction can be checked independently, especially because the two preceding issues affect exactly this computation.","section":"Section 3, finite verification after (3.12)"}],"minor_comments":[{"comment":"The displayed Matveev inequality contains a missing closing parenthesis: the exponent reads \"(1 + log(2r) logLm\" and should be \"(1 + log(2r)) logLm\" or similar; please correct the typo.","section":"Section 3, Matveev application"},{"comment":"The sentence \"It is clear that L_r is even, and therefore 3|r\" would benefit from a brief justification: since L_m^{k+1} + L_m is even, L_r is even, and Lucas numbers are even exactly when their index is divisible by 3.","section":"Section 3, n = 1 case"},{"comment":"Remark 3.1 is confusing: the phrase \"the upper bound for r in inequality (3.6) is rounded up, so the inequality may not hold for values of r near this bound\" could be read as suggesting that a legitimate bound may fail; since all subsequent functions are increasing, the remark is unnecessary and should be removed or rewritten to state the simple monotonicity argument.","section":"Remark 3.1"},{"comment":"The notation in Lemma 2.5 is ambiguous because J(S) is defined in terms of N inside the same lemma; if the lemma is corrected, please define N as the index satisfying q_N > S and state clearly whether J(S) includes a_{N+1}.","section":"General presentation"}],"recommendation":"major_revision","confidential_remarks":"The central claim is plausible and the n = 1 proof is solid, but the n >= 2 reduction currently rests on an unjustified inequality and a false stated form of Legendre's lemma. These are repairable, so I am not recommending rejection, but the authors must correct the continued-fraction lemma, redo the numerical reduction with the correct bounds, and provide the numerical data needed for verification."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"The short version: the theorem is new, the n=1 case is genuinely clean, and the overall strategy is standard-but-competent. The n≥2 proof, however, has a real gap at inequality (3.12), and unless the authors supply the missing numerical data, the paper does not establish the claimed reduction.\n\nWhat the paper does well: the n=1 argument using the factorization identity and Carmichael's primitive divisor theorem is neat and correct. The exceptional cases are handled honestly. The application of Lemma 2.4 is legitimate, since gcd(L_m, L_m^k+1)=1, so L_m^n || L_r follows from the equation. The authors also ship Python code on Zenodo, which is more than many papers in this area do.\n\nThe soft spot is exactly where the stress test points. Combining (3.10) and (3.11) gives L_m^k < 1.0025 (J^(m)(S)+2)(n+k)/log α. At that point in the proof the only available bound is n≤61, so the right side should be (61+k), not (28+k). The number 28 appears without justification. If this is a typo, the intended argument might still work, but as printed, (3.12) is stronger than what has been proved. The subsequent conclusion k≤8 also depends on unlisted J^(m)(S) values. The Zenodo code may contain them, but the paper does not show the output or describe how the N_m were selected, so a reader cannot independently verify the reduction to r≤3865. This is a load-bearing gap, not a cosmetic omission.\n\nThe rest of the argument—the Matveev bounds, the derivation of m≤55 and n≤61, and the final finite check—looks plausible. Remark 3.1 is odd but harmless. The earlier concern about Lemma 2.4 does not hold up; that lemma is applied correctly.\n\nWho is this for? Specialists in Diophantine equations with Lucas and Fibonacci sequences. They will want to know this equation is settled, and the n=1 proof is worth reading. But the paper currently needs one substantive correction and one transparency fix. I would send it to a serious referee with instructions to check the (3.12) step carefully and to request the code output for the J values.","headline":"New result, clean n=1 case, but the n≥2 reduction has an unjustified inequality at (3.12) and missing numerical data; deserves a careful referee.","tokens_in":6977,"tokens_out":7808,"would_cite":false,"duration_ms":73929,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11B39","11D61","11B83"],"pacs":[],"model":"deepseek-v4-flash","headline":"No Lucas number is a sum of two powers of a single Lucas number.","keywords":["Lucas numbers","Diophantine equations","exponential Diophantine equations","linear forms in logarithms","continued fractions","exact divisibility","primitive divisors","computer-assisted proof"],"falsifier":"Run the finite check the proof reduces to -- $2\\le m\\le 12$, $2\\le n\\le 16$, $1\\le k\\le 8$, which force $r\\le 313$ -- looking for $(m,n,k,r)$ with $L_m^{n+k}+L_m^n=L_r$; one hit refutes the theorem, while confirming the full search closes the proof.","tokens_in":5876,"feed_emoji":"➕","tokens_out":11532,"duration_ms":105200,"temperature":0.7,"pith_summary":"The paper tries to establish a complete nonexistence result: the exponential Diophantine equation $L_m^{n+k}+L_m^n = L_r$, where $L_t$ is the $t$-th Lucas number, has no positive-integer solutions $r,m,n,k$ with $m\\ge 2$. If true, this settles the question of when a Lucas number can be written as a sum of two powers of the same Lucas number. The proof splits into two regimes with different tools: $n=1$ is handled by an exact factorization of differences of Lucas numbers together with the primitive divisor theorem, while $n\\ge 2$ uses an exact divisibility lemma, a lower bound for linear forms in logarithms, a continued-fraction criterion, and a finite computer search. The result matters because it closes this whole family of equations rather than merely bounding the solutions.","feed_headline":"No Lucas number is a sum of two powers of a single Lucas number","feed_subtitle":"Settles the exponential Diophantine equation completely by mixing divisibility, log bounds, and a finite search.","key_machinery":"The engine of the proof is a chain of effective estimates. The starting point is the closed form $L_t=\\alpha^t+\\beta^t$ with $\\alpha=(1+\\sqrt5)/2$, which yields the inequalities $\\alpha^{t-1}\\le L_t < \\alpha^{t+1}$. The $n\\ge 2$ branch begins with the exact divisibility statement that $L_m^n\\parallel L_r$ forces $L_m^{n-1}\\parallel r/m$, giving the inequality $\\log m + (n-1)(m-1)\\log\\alpha \\le \\log r$; this is what bounds $m$ and $n$. The equation itself is then rewritten as $0<\\alpha^r L_m^{-(n+k)} - 1 < 1.0025/L_m^k$, a small nonzero linear form in logarithms, to which an explicit lower bound is applied, yielding crude upper bounds on $k$ and $r$. A continued-fraction criterion then sharpens the bound to $k\\le 8$, and the remaining finite box is enumerated by computer.","core_discovery":"The central claim is Theorem 3.1: for every $m\\ge 2$, no positive integers $n,k,r$ satisfy $L_m^{n+k}+L_m^n=L_r$. In the $n=1$ case the proof shows that $L_m$ must divide the difference $L_r-L_m$ and uses a factorization depending on $r-m \\bmod 4$ to force either a factor of $5$ dividing a Lucas number (impossible) or a primitive prime divisor of a large Lucas factor dividing a smaller one (also impossible), leaving only two finite pairs to check. For $n\\ge 2$, the proof uses an exact divisibility property to bound $m$ and $n$, then an explicit lower bound for linear forms in logarithms to bound $k$, then a continued-fraction criterion to push $k$ down to at most $8$, and finally an explicit computer search. The theorem is presented as complete: after the reductions only a finite range remains, and the authors report checking it exhaustively.","pith_inferences":["An extension implicit in the method is that the analogous Fibonacci equation $F_m^{n+k}+F_m^n=F_r$ should be approachable by the same route, since primitive divisors, exact divisibility, and continued-fraction reduction all have Fibonacci counterparts.","The only externally imported ingredient in the $n\\ge 2$ branch is the exact divisibility lemma; if a comparable lemma is proved for other Lucas sequences of the first and second kinds, the paper's open equations (4.1) and (4.2) would likely be solvable for fixed parameters $a,b$ by the same scheme.","The numerical constants in the linear-forms bound are very conservative, so the final search range could probably be shrunk well below $r\\le 313$ by a sharper computation; the structural nonexistence result would be unchanged."],"forward_implications":["The equation $L_m^{n+k}+L_m^n=L_r$ is completely resolved: no solution exists for any base $m\\ge 2$, so no search for large exotic solutions is needed.","In the $n=1$ subcase, the only candidate pairs surviving the divisibility analysis are $(m,r)=(2,10)$ and $(3,9)$, and the latter fails by direct evaluation.","When $L_m$ happens to be prime, the theorem gives a complete existence answer for a special case of equations of the form $w_t=p^a+p^b$ with $p$ prime.","The proof yields explicit effective bounds: after reduction, any hypothetical solution must satisfy $m\\le 12$, $n\\le 16$, $k\\le 8$, and $r\\le 313$, making the claim checkable by a finite computation."],"supporting_citations":[{"why":"Supplies the inequalities $\\alpha^{t-1}\\le L_t<\\alpha^{t+1}$ and the divisibility criterion $L_m\\mid L_r$ iff $r=m(2s-1)$.","marker":"[4]"},{"why":"The primitive divisor theorem for Lucas numbers, used in the $n=1$ case to show a prime divisor of $L_{(r+m)/2}$ cannot divide $L_m$.","marker":"[1]"},{"why":"The exact divisibility lemma $L_m^{n-1}\\parallel r/m$ when $L_m^n\\parallel L_r$, which drives the size bounds for $n\\ge 2$.","marker":"[11]"},{"why":"An explicit lower bound for nonzero linear forms in logarithms of algebraic numbers, which produces the first upper bound on $k$.","marker":"[10]"},{"why":"The continued-fraction criterion used to replace the huge bound on $k$ by the small bound $k\\le 8$.","marker":"[6]"}],"fun_headline_variants":["L_m^n and L_m^{n+k} never sum to a Lucas number","No solutions: L_m^{n+k}+L_m^n = L_r for m≥2","Complete proof: Lucas power-sum equation has zero solutions","Two powers of a single Lucas number can't equal a Lucas","Exponential Lucas equation solved: no positive integers exist"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The whole $n\\ge 2$ argument rests on an imported lemma, unproved in this paper, saying that if $L_m^n$ exactly divides $L_r$, then $L_m^{n-1}$ exactly divides $r/m$.","fun_headline_variants_meta":{"raw":{"variants":["L_m^n and L_m^{n+k} never sum to a Lucas number","No solutions: L_m^{n+k}+L_m^n = L_r for m≥2","Complete proof: Lucas power-sum equation has zero solutions","Two powers of a single Lucas number can't equal a Lucas","Exponential Lucas equation solved: no positive integers exist"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.00078,"raw_usage":{"total_tokens":3405,"prompt_tokens":864,"completion_tokens":2541,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":480,"completion_tokens_details":{"reasoning_tokens":2448}},"tokens_in":480,"tokens_out":2541,"duration_ms":17501,"temperature":1.0,"reasoning_tokens":2448,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-07T00:18:50.709245+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Run the finite check the proof reduces to -- $2\\le m\\le 12$, $2\\le n\\le 16$, $1\\le k\\le 8$, which force $r\\le 313$ -- looking for $(m,n,k,r)$ with $L_m^{n+k}+L_m^n=L_r$; one hit refutes the theorem, while confirming the full search closes the proof.","supporting_citations":[{"cited_title":"Wiley, New York (2001)","cited_arxiv_id":null,"evidence_quote":"Supplies the inequalities $\\alpha^{t-1}\\le L_t<\\alpha^{t+1}$ and the divisibility criterion $L_m\\mid L_r$ iff $r=m(2s-1)$."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"The primitive divisor theorem for Lucas numbers, used in the $n=1$ case to show a prime divisor of $L_{(r+m)/2}$ cannot divide $L_m$."},{"cited_title":"AIMS Math.6(11), 11733–11748 (2021)","cited_arxiv_id":null,"evidence_quote":"The exact divisibility lemma $L_m^{n-1}\\parallel r/m$ when $L_m^n\\parallel L_r$, which drives the size bounds for $n\\ge 2$."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"An explicit lower bound for nonzero linear forms in logarithms of algebraic numbers, which produces the first upper bound on $k$."},{"cited_title":"Duprat, Paris (1798)","cited_arxiv_id":null,"evidence_quote":"The continued-fraction criterion used to replace the huge bound on $k$ by the small bound $k\\le 8$."}],"review_version":1}