{"id":"507e2b85-cf82-4a60-a257-173f65f1092f","arxiv_id":"2411.17780","paper_version":1,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":6.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"Every connected vertex-transitive graph of order 10p, p prime and p ≠ 7, has a Hamilton path; the exceptional PSL(2,s^m) coset graphs are shown to contain Hamilton cycles.","lead":"Mathematicians prove that every connected vertex-transitive graph with ten times a prime number of vertices, other than seven, contains a path through all vertices. The paper closes the last open exceptions by finding full cycles in the graphs left unresolved since 2012.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Claim 2 of Lemma 3.2 is false for k=81: for m=0, Eq(5) has only b=0 solutions, so d(αS, αℓS)=1 and the quotient-lifting proof fails.","rationale":"The reader identified Claim 2's 'completely similar' treatment as terse and flagged the uncertified Magma check for k=61, but did not notice that Claim 2 is actually false for k=81. Since k=81 is an admissible case of Theorem 1.1 (s=3, m=4, p=41, 10 | 80), this is a genuine counterexample to the proof's universal double-edge assertion (ii), not merely an omitted justification. The key reason is special to F_81: 10th powers lie in F_9, and every nonzero element of F_9 is a square in F_81, so the only solutions to the m=0 equation have b=0, giving one rather than two neighbors. This breaks the lifting argument as written. The underlying theorem may still be true — one could try to find a Hamilton cycle in the quotient using only edges of multiplicity at least 2, since the single edges form a perfect matching — but that argument is absent. Therefore the current version should not be accepted without major revision; the reader's CONDITIONAL verdict should be moved to REJECT for the paper as submitted.","tokens_in":109,"tokens_out":42429,"duration_ms":456704,"concrete_test":"Using SageMath or Magma, set k=81, F = GF(3^4), choose a primitive θ ∈ F^*, and enumerate all solutions (b, y) ∈ F^2 of θ b^2 − y^{10} = −1. If the only solutions are b=0 with y^{10}=1, then Claim 2 fails for i=0, n=0, j=0, and the quotient graph Y(0)_S has a single edge between αS and αℓS. To see whether the main conclusion can be salvaged, additionally check whether the 10-vertex quotient still contains a Hamilton cycle using only edges of multiplicity at least 2.","verdict_should_be":"REJECT","load_bearing_attack":"The proof of Lemma 3.2 asserts as fact (ii) that d(A,B) ≥ 2 for every pair of S-orbits. Claim 2 is meant to establish this for pairs of the form d(αt^n S, αt^j ℓS). For k=81 this claim is false. Here k=81 = 3^4 is admissible: k+1=82=2·41 and 10 | (k−1). Take i=0, n=0, j=0, so m = j−i+n = 0. Claim 2 reduces d(αS, αℓS) ≥ 2 to Eq(5) with c = −1, namely θ b^2 − y^{10} = −1. In F_81, the set of 10th powers is μ_8 ∪ {0}, where μ_8 = F_9^*. For x ∈ μ_8 \\ {1}, the element x−1 lies in F_9^*; since F_9^* has order 8 and the square subgroup of F_81^* has order 40, every element of F_9^* is a square in F_81. Thus b^2 = θ^{−1}(x−1) is the product of the nonsquare θ^{−1} with a square, hence a nonsquare, so no b exists. For x=0 the equation requires −θ^{−1} to be a square, but −θ^{−1} = θ^{39} is a nonsquare. The only solutions are therefore b=0 and y^{10}=1, and all of these give the same vertex αℓ in αℓS. Hence d(αS, αℓS)=1, contradicting the universal double-edge condition (ii). The theorem may still be true — the five single edges form a perfect matching that a Hamilton cycle in the quotient could avoid — but the proof as printed does not establish this, and the argument for Lemma 3.2 is invalid for k=81.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper addresses a remaining exceptional family in the Hamilton path problem for vertex-transitive graphs of order 10p. Building on the 2012 classification by Kutnar, Marušič, and Zhang, the authors prove (Theorem 1.1) that a connected graph admitting a vertex-transitive PSL(2,s^m) action with point stabilizer Z_s^m ⋊ Z_{(s^m-1)/10} contains a Hamilton cycle, yielding the corollary that all connected vertex-transitive graphs of order 10p (p≠7) have Hamilton paths. The proof constructs basic orbital graphs, quotients by a semiregular subgroup S of order (k+1)/2, and uses a lifting lemma. The central Lemma 3.2 reduces the required double-edge condition to solvability of diagonal equations over F_k, treated via a bound of Lidl–Niederreiter and a Magma check for k=61.","tokens_in":9087,"tokens_out":31631,"duration_ms":257178,"significance":"If correct, the result completes the Hamiltonicity question for order 10p, a natural continuation of work on vertex-transitive graphs of order kp and pq. The proof strategy is attractive: the reduction of edge multiplicities to diagonal equations is elegant, and the use of a general finite-field bound is a useful technique. The manuscript is self-contained apart from the Magma check and the cited classification. However, the argument for Lemma 3.2 has a concrete gap: an admissible case (k=81) is not covered, and the k=61 verification is not fully documented. These issues are central, so the present version is not yet acceptable, though the approach is promising.","major_comments":[{"comment":"For the admissible value k=81 (s=3, p=41), the assertion d(αS, αℓS) ≥ 2 is false. Taking i=n=j=0, Eq. (5) becomes θb^2 − y^{10} = −1, with y = θ^{−r} and r ranging over Z_8, so x = y^{10} ranges over F_9^*. For x ∈ F_9^* \\ {1}, the equation gives b^2 = θ^{−1}(x−1); since x−1 ∈ F_9^* is a square in F_81 while θ^{−1} is a nonsquare, there is no b. For x=1 the only solutions have b=0, and all such solutions give the same vertex αℓ. Hence d(αS, αℓS)=1, contradicting condition (ii) of Lemma 3.2. This invalidates the proof of Lemma 3.2 for an admissible case of Theorem 1.1.","section":"§3, Lemma 3.2, Claim 2 (Eq. (5))"},{"comment":"The sentence 'Completely similar to Case 1' is not a valid substitute for an argument. In Claim 1 the degenerate solutions are y=0, and they are explicitly subtracted; in Claim 2 the degenerate solutions are b=0, which can persist for all admissible y when c is such that cy^{10}=−1. The proof never shows that a solution of Eq. (5) with b≠0 exists, and the lower bound from Proposition 2.2 counts all solutions (b,y), including the b=0 ones. For k=81 and c=−1 all solutions have b=0, as shown above, so the two cases are not 'completely similar'.","section":"§3, Lemma 3.2, Claim 2"},{"comment":"The phrase 'Checking by Magma, Eq(3) has solutions for any c, over F61' is not reproducible: no code, no list of the finitely many c-values, and no certificate is given. Moreover, this check only concerns Eq(3), i.e., Claim 1 (and Claim 3, which reduces to Eq(3)); Claim 2 requires Eq(5) with the different constant −1 and coefficient θ, and no check for Eq(5) over F61 is reported. Since Proposition 2.2 gives no positive lower bound for k=61, Claim 2 for k=61 is unsupported.","section":"§3, Lemma 3.2, k=61 check"}],"minor_comments":[{"comment":"The Petersen graph is misspelled as 'Peterson graph' in the introduction.","section":"Introduction"},{"comment":"The notation 'Y(i)_1(αtn)' for the neighborhood of αtn in Y(i) is undefined; please define it or use standard notation such as N_{Y(i)}(αtn).","section":"§3, Proof of Lemma 3.2"},{"comment":"The expression 'S ∼= Z_{k+1/2}' is ambiguous; it should be written as a cyclic group of order (k+1)/2.","section":"§3, Lemma 3.1(2)"},{"comment":"The sentence 'Then XS has ten vertices' should refer to the quotient graph of Y(i) by S, not to X itself.","section":"§3, Proof of Lemma 3.2"}],"recommendation":"major_revision","confidential_remarks":"The paper addresses a significant open question and the main idea is sound, but the identified gap in Lemma 3.2 for k=81 and the unverified k=61 case are load-bearing. A revision that handles these cases (e.g., a separate argument for k=81, or a more careful choice of a Hamilton cycle in the quotient avoiding single edges) and provides reproducible verification for k=61 would make the paper acceptable. The scope of the journal is appropriate."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"The headline is that this is a genuinely new result with a standard proof strategy, but Claim 2 of Lemma 3.2 is false for k=81, so the proof as written does not go through.\n\nThe paper upgrades the 2012 Kutnar–Marušič–Zhang result by showing that the exceptional PSL(2,s^m) coset graphs of order 10p have Hamilton cycles, not just paths, which would give a clean corollary for all connected vertex-transitive graphs of order 10p, p≠7. That is worth having. The strategy is the usual quotient lift: take the semiregular subgroup S of order p, show the ten S-orbits form a complete quotient with enough multiplicity, and lift a cycle. Claim 1, where the edge count reduces to a^2 + c y^10 = 1, is worked out carefully and the diagonal-equation bound is applied correctly, yielding k≥72.\n\nThe problem is Claim 2. The text says \"completely similar\" for the equation θ b^2 + c y^10 = -1, but it is not similar enough. For k=81 (admissible: 3^4+1=82=2·41 and 10|80), take i=j=n=0 so c=-1. In F_81 the equation becomes θ b^2 - y^10 = -1. The 10th powers are exactly F_9^*. For any x ∈ F_9^*, x-1 is a square in F_81 (since F_9^* lies in the square subgroup), so θ b^2 = x-1 forces b^2 to be a nonsquare times a square, hence no solution. The only solutions have b=0 and y^10=1, and all of these give the same vertex αℓ. So d(αS, αℓS)=1, contradicting the universal claim d≥2. This is a concrete numerical counterexample, not just a gap in presentation.\n\nThe k=61 Magma check without code or certificate is a lesser issue. The main theorem may still be true: the single edges form a perfect matching, and the quotient's double-edge graph likely still has a Hamilton cycle, so a refined argument could work. But as it stands, Lemma 3.2's proof fails for an admissible parameter family.\n\nThis paper is for people working on the hamiltonicity of vertex-transitive graphs. A correct proof would be a useful incremental step. I would send it to reviewers rather than desk-reject: the flaw is specific and probably fixable, and the result is worth debating. Just make sure the \"completely similar\" step gets proper scrutiny.","headline":"New result in order-10p hamiltonicity, but the proof of Lemma 3.2 has a false claim for k=81 that currently invalidates the argument.","tokens_in":9641,"tokens_out":12024,"would_cite":false,"duration_ms":101177,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["05C25","05C45"],"pacs":[],"model":"deepseek-v4-flash","headline":"This paper proves Hamilton cycles exist in every connected graph of order $10p$ arising from the exceptional $\\mathrm{PSL}(2,s^m)$ actions, completing the Hamilton path result for order $10p$.","keywords":["vertex-transitive graph","Hamilton cycle","Hamilton path","orbital graph","PSL(2,s^m)","diagonal equation","lifting cycle","semiregular automorphism"],"falsifier":"Recompute the two diagonal equations over $\\mathbb{F}_{61}$ for every coefficient $c$ and confirm that each has a solution with $y\\ne 0$; the paper claims this for one of the equations and describes the other as completely similar. If any $c$ fails, Lemma 3.2 collapses for $k=61$.","tokens_in":5,"feed_emoji":"🔄","tokens_out":6171,"duration_ms":113499,"temperature":0.7,"pith_summary":"This paper proves that every connected graph of order $10p$ whose automorphism group contains a vertex-transitive subgroup $\\mathrm{PSL}(2,s^m)$ with point stabilizer $\\mathbb{Z}_s^m\\rtimes\\mathbb{Z}_{(s^m-1)/10}$ has a Hamilton cycle, where $s^m+1=2p$ and $p$ is prime. These graphs were the only known exceptions to a 2012 theorem that every connected vertex-transitive graph of order $10p$ with $p\\ne 7$ has a Hamilton path. The paper constructs the cycle by showing that each basic orbital graph associated with the exceptional action has a Hamilton cycle. If correct, this closes the last gap and gives a complete Hamilton path statement for all connected vertex-transitive graphs of order $10p$ with $p\\ne 7$.","feed_headline":"Exceptions to 10p Hamilton path result get cycles","feed_subtitle":"A 2012 gap for PSL(2,s^m) actions is closed, so all order-10p graphs have a Hamilton path.","key_machinery":"The key machinery is the lifting-cycle technique combined with the semiregular cyclic subgroup $S\\cong\\mathbb{Z}_{(k+1)/2}$, which partitions the $10p$ vertices into ten orbits. The argument shows that in each basic orbital graph $Y(i)$ the quotient by $S$ is the complete graph on ten vertices and every pair of $S$-orbits is connected by at least two edges; the lifting lemma then converts a Hamilton cycle in the quotient into one in $Y(i)$. To establish the double-edge condition, the paper counts neighborhood intersections and reduces the count to whether the diagonal equations $a^2+cy^{10}=1$ and $\\theta b^2+cy^{10}=-1$ have solutions with $y\\ne 0$ over $\\mathbb{F}_k$. Proposition 2.2 supplies the needed solutions for $k\\ge 72$, and the exceptional case $k=61$ is asserted by a computer check.","core_discovery":"Theorem 1.1 states that if $X$ is a connected graph whose automorphism group contains a vertex-transitive subgroup $\\mathrm{PSL}(2,s^m)$, where $s$ is prime and the point stabilizer is $\\mathbb{Z}_s^m\\rtimes\\mathbb{Z}_{(s^m-1)/10}$ with $s^m+1=2p$ for a prime $p$, then $X$ contains a Hamilton cycle. The proof works with the basic orbital graphs $Y(i)$ obtained from the five self-paired suborbits of length $k=s^m$. For each $Y(i)$, the cyclic subgroup $S\\cong\\mathbb{Z}_{(k+1)/2}$ acts semiregularly with ten orbits, and the paper shows that the quotient graph is complete and that every pair of $S$-orbits is joined by at least two edges. Under those conditions, Lemma 2.1 guarantees that a Hamilton cycle in the quotient lifts to a Hamilton cycle in $Y(i)$. The edge-counting condition is reduced to the solubility of diagonal equations $a^2+cy^{10}=1$ and $\\theta b^2+cy^{10}=-1$ over $\\mathbb{F}_k$; for $k\\ge 72$ this follows from the bound in Proposition 2.2, and the remaining case $k=61$ is handled by a computer check. Since any graph in the exceptional class contains one of the $Y(i)$ as a spanning subgraph, the Hamilton cycle transfers to $X$.","pith_inferences":["The same reduction to diagonal equations could be applied to other $\\mathrm{PSL}(2,k)$ actions with larger point stabilizers, yielding Hamilton cycles for orders beyond $10p$.","A short arithmetic certificate for the $k=61$ case, such as a list of primitive roots and witness solutions, would make the proof fully independent of computer verification.","Since the paper shows the exceptional graphs have Hamilton cycles, the remaining open question for order $10p$ is whether the non-exceptional cases handled by the 2012 path result actually have Hamilton cycles as well."],"forward_implications":["Every connected vertex-transitive graph of order $10p$ with $p\\ne 7$ has a Hamilton path.","Every basic orbital graph $Y(i)$ for the exceptional $\\mathrm{PSL}(2,s^m)$ action contains a Hamilton cycle.","Since any connected graph in the exceptional class contains one of these $Y(i)$ as a spanning subgraph, each such graph contains a Hamilton cycle.","The lifting-cycle argument shows that the quotient on ten $S$-orbits is complete with double edges, so Hamiltonicity of the quotient transfers to the original graph."],"supporting_citations":[{"why":"Establishes that the only possible exceptions to a Hamilton path for order $10p$ are the $\\mathrm{PSL}(2,s^m)$ actions studied here.","marker":"[18]"},{"why":"Provides the diagonal-equation bound used to guarantee solutions for $k\\ge 72$.","marker":"[20]"},{"why":"Supplies the lifting technique for Hamilton cycles of quotient graphs.","marker":"[2]"},{"why":"Provides the lemma, reformulated as Lemma 2.1, describing when a cycle in the quotient lifts to a cycle in the original graph.","marker":"[26]"},{"why":"Surveys the Hamilton path/cycle problem for vertex-transitive graphs and frames the outstanding cases.","marker":"[17]"}],"fun_headline_variants":["10p graphs get Hamilton cycles, closing 2012 gap","All order-10p vertex-transitive graphs have Hamilton cycles","Hamilton cycles for the last 10p exceptions","PSL(2,s^m) action exceptions now yield cycles"],"cache_read_input_tokens":11648,"weakest_assumption_plain":"For the exceptional field size $k=61$, the proof depends on an unverified computer check that certain diagonal equations have solutions for every coefficient; if that check is wrong, the Hamilton-cycle claim for that case is unsupported.","fun_headline_variants_meta":{"raw":{"variants":["10p graphs get Hamilton cycles, closing 2012 gap","All order-10p vertex-transitive graphs have Hamilton cycles","Hamilton cycles for the last 10p exceptions","PSL(2,s^m) action exceptions now yield cycles"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000192,"raw_usage":{"total_tokens":1358,"prompt_tokens":967,"completion_tokens":391,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":583,"completion_tokens_details":{"reasoning_tokens":322}},"tokens_in":583,"tokens_out":391,"duration_ms":4249,"temperature":1.0,"reasoning_tokens":322,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-12T12:16:53.433740+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Recompute the two diagonal equations over $\\mathbb{F}_{61}$ for every coefficient $c$ and confirm that each has a solution with $y\\ne 0$; the paper claims this for one of the equations and describes the other as completely similar. If any $c$ fails, Lemma 3.2 collapses for $k=61$.","supporting_citations":[{"cited_title":"Kutnar, D","cited_arxiv_id":null,"evidence_quote":"Establishes that the only possible exceptions to a Hamilton path for order $10p$ are the $\\mathrm{PSL}(2,s^m)$ actions studied here."},{"cited_title":"Niederreiter, Finite fields , Cambridge University Press, Cam- bridge, 1997","cited_arxiv_id":null,"evidence_quote":"Provides the diagonal-equation bound used to guarantee solutions for $k\\ge 72$."},{"cited_title":"Alspach, Lifting Hamilton cycles of quotient graphs, Discrete Math","cited_arxiv_id":null,"evidence_quote":"Supplies the lifting technique for Hamilton cycles of quotient graphs."},{"cited_title":"Maruˇ siˇ c and T","cited_arxiv_id":null,"evidence_quote":"Provides the lemma, reformulated as Lemma 2.1, describing when a cycle in the quotient lifts to a cycle in the original graph."},{"cited_title":"Kutnar and D","cited_arxiv_id":null,"evidence_quote":"Surveys the Hamilton path/cycle problem for vertex-transitive graphs and frames the outstanding cases."}],"review_version":1}