Abstract

Continued fractions have been generalized over the field of pp-adic numbers, where it is still not known an analogue of the famous Lagrange's Theorem. In general, the periodicity of pp-adic continued fractions is well studied and addressed as a hard problem. In this paper, we show a strong connection between periodic pp--adic continued fractions and the convergence to real quadratic irrationals. In particular, in the first part we prove that the convergence in R\mathbb{R} is a necessary condition for the periodicity of the continued fractions of a quadratic irrational in Qp\mathbb{Q}_p. Moreover, we leave several conjectures on the converse, supported by experimental computations. In the second part of the paper, we exploit these results to develop a probabilistic argument for the non-periodicity of Browkin's pp-adic continued fractions. The probabilistic results are conditioned under the assumption of uniform distribution of the pp-adic digits of a quadratic irrational, that holds for almost all pp-adic numbers.

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Audited against arXiv v3

Not a correctness certificate. A “Correct” result may include yellow typos or minor formal corrections 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.

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Detailed mathematical audit

Generated August 19, 2026
01Statements4 reported findingsContains wrong statements

Proposition 1, the necessary real-convergence condition for periodic Browkin-type expansions, is correct. The probabilistic part is not supported by a defined probability law on quadratic irrationals, and Theorem 26's conditional and unconditional expectation formulas are false even under the independent uniform-digit model used implicitly in its proof.

Proposition 1Correct

Eventual periodicity forces convergence to a real conjugate

Pages 3 and 12-14 · Proposition 1 and its proof · arXiv:2410.09215v3

For a purely periodic tail of period kk, the two real conjugates are the fixed points of the period matrix. Lane's criterion reduces possible Euclidean nonconvergence to equality of their distances from Ak1/Bk1A_{k-1}/B_{k-1}, since a finite convergent is rational and cannot equal either quadratic fixed point. The equality is equivalent to Ak1+Bk2=0A_{k-1}+B_{k-2}=0. Substitution into the discriminant and the determinant identity gives D=4(1)k1>0D=4(-1)^{k-1}>0, so it can occur only for odd kk. Browkin II has even period because its integer and noninteger partial quotients alternate. For Browkin I and Algorithm (12), the valuation identities imply that the equality would require vp(a0)+vp(ak1)=0v_p(a_0)+v_p(a_{k-1})=0, while the algorithmic valuation conditions make this sum negative in the only remaining odd-period cases. Reattaching a finite preperiod is a rational linear-fractional substitution and sends the two real tail values to the two real conjugates of the original quadratic irrational.

Full paper, version 3
Assumption 3 and Propositions 22-24Not able to verify

The claimed probability law on quadratic irrationals is not defined

Page 4 · Assumption 3; pages 17-19 · Remark 21 and Propositions 22-24 · arXiv:2410.09215v3

The set of pp-adic numbers quadratic over Q\mathbb Q is countable, because it is contained in the union of the root sets of the countably many quadratic polynomials in Q[x]\mathbb Q[x]. It consequently has Haar measure zero in Qp\mathbb Q_p, so restricting Haar measure to quadratic irrationals does not yield the probability measure invoked in Assumption 3. Moreover, the displayed assumption specifies only the one-coordinate marginals P(cn=j)=1/p\mathbb P(c_n=j)=1/p; it specifies neither joint independence of digit blocks nor an invariant law for the complete quotients produced by Browkin I. Thus the random variables in Propositions 22-24 have no stated probability law in the advertised quadratic-irrational setting, and their distributional claims cannot be verified from Assumption 3.

Full paper, version 3
Proposition 24Typo

The expected valuation has the wrong sign

Pages 18-19 · Proposition 24 and its first displayed line of proof · arXiv:2410.09215v3

Proposition 22 assigns the value vp(αn)=kv_p(\alpha_n)=-k with probability (p1)/pk(p-1)/p^k. Therefore its own distribution gives E(vp(αn))=k1kp1pk=pp1,\mathbb E(v_p(\alpha_n))=-\sum_{k\geq1}k\frac{p-1}{p^k}=-\frac{p}{p-1}, not p/(p1)p/(p-1). Insert the missing minus sign in the statement and throughout the calculation. The result is not used in the proof of Theorem 26, and Remark 25 already describes the valuation as negative, so this correction is mechanically determined and has no downstream effect.

Full paper, version 3
Theorem 26Incorrect

Both expectation formulas use the wrong conditional sample space

Pages 19-21 · Theorem 26 and proof · arXiv:2410.09215v3

Even under the independent balanced-digit model implicitly used by the proof, conditioning on vp(a)=kv_p(a)=-k requires the leading digit ckc_k in a=c0+c1/p++ck/pka=c_0+c_1/p+\cdots+c_k/p^k to be nonzero. The proof instead averages over all pk+1p^{k+1} digit strings, including ck=0c_k=0. For the decisive case p=3p=3 and k=1k=1, the six conditional values are 4/3,2/3,1/3,1/3,2/3,4/3-4/3,-2/3,-1/3,1/3,2/3,4/3, so E(avp(a)=1)=79,\mathbb E(|a|\mid v_p(a)=-1)=\frac79, whereas the theorem gives 20/2720/27. Direct summation for general pp and kk gives E(avp(a)=k)=p2k+1+14p2k,\mathbb E(|a|\mid v_p(a)=-k)=\frac{p^{2k+1}+1}{4p^{2k}}, and, with the distribution asserted in Proposition 22, E(a)=p4+14(p2+p+1).\mathbb E(|a|)=\frac p4+\frac{1}{4(p^2+p+1)}. These differ from both printed formulas, so Theorem 26 is false under the natural strengthened model as well as unsupported under Assumption 3.

Full paper, version 3
02Proofs4 reported findingsContains incorrect or incomplete proofs

The proof of Proposition 1 is correct after immediate-consequence closure. The probabilistic proof chain is not valid: Proposition 22 uses independence absent from Assumption 3, and Theorem 26 averages over a sample space inconsistent with its conditioning event. Two local sign or index defects are separately identified as typographical.

Proof of Proposition 1Correct and complete

The period-matrix and valuation argument closes

Pages 12-14 · proof of Proposition 1, equations (20)-(23) · arXiv:2410.09215v3

The proof excludes the equimodular fixed-point case through the determinant identity and then uses exactly the parity and valuation restrictions supplied by each of the three algorithms. The additional clause in Lane's criterion is automatic because every finite convergent is rational while the fixed points are quadratic irrationals. The finite preperiod is handled by the same rational linear-fractional relation used to define complete quotients, an immediate formal consequence that preserves the two conjugate values.

Full paper, version 3
Proof of Proposition 22Incomplete as written

Marginal digit uniformity is used as block independence

Page 18 · proof of Proposition 22, immediately before equation (25) · arXiv:2410.09215v3

The step P(c1==ck1=0)=p(k1)\mathbb P(c_1=\cdots=c_{k-1}=0)=p^{-(k-1)} does not follow from the only stated hypothesis P(ci=j)=1/p\mathbb P(c_i=j)=1/p. For example, if a single uniform variable UU in {0,,p1}\{0,\ldots,p-1\} is used for every digit, then every marginal is uniform but the displayed block probability is 1/p1/p, not p(k1)p^{-(k-1)}. The proof also multiplies that block probability by P(ck0)\mathbb P(c_k\neq0) without establishing independence, and it does not show that the Browkin complete-quotient transformation preserves any proposed law. Repair classification: No repair supplied. A repair would require a precisely defined probability space, the necessary joint digit law at every complete quotient, and a justification connecting that law to quadratic irrationals; none follows from Assumption 3.

Full paper, version 3
Proof of Theorem 26Incorrect as written

The recurrence computes an unconditional digit average

Pages 20-21 · proof of Theorem 26, recurrence for EkE_k and the final averaging step · arXiv:2410.09215v3

The recurrence defines EkE_k by averaging c0+c1/p++ck/pk|c_0+c_1/p+\cdots+c_k/p^k| over every choice of c0,,ckc_0,\ldots,c_k. The conditional expectation stated in the theorem is instead over the strict subset ck0c_k\neq0, because that condition is exactly vp(a)=kv_p(a)=-k. The base case already exposes the mismatch: for p=3p=3 and k=1k=1, excluding c1=0c_1=0 gives 7/97/9, while the recurrence including it gives 20/2720/27. Every subsequent substitution into the sum over kk therefore propagates the wrong quantity. Repair classification: Verified repair under an expressly independent uniform-digit model is E(avp(a)=k)=p2k+1+14p2k\mathbb E(|a|\mid v_p(a)=-k)=\frac{p^{2k+1}+1}{4p^{2k}} and E(a)=p4+14(p2+p+1).\mathbb E(|a|)=\frac p4+\frac{1}{4(p^2+p+1)}. This calculation does not repair the separate absence of such a model for quadratic irrationals.

Full paper, version 3
Proof of Proposition 1Typo

One denominator in the reformulation of equation (20) has the wrong index

Page 13 · display immediately before equation (22) · arXiv:2410.09215v3

The displayed fraction (Ak1+Bk2)/Bk+1(A_{k-1}+B_{k-2})/B_{k+1} must have denominator Bk1B_{k-1}. This is forced by the two fractions in the preceding line. Only the vanishing of the numerator is used afterward, so the uniquely determined correction does not affect the proof.

Full paper, version 3
03Novelty0 reported findingsNo non-novelty findings

No non-novelty findings.

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Paper
arXiv:2410.09215v3
Authors listed
Giuliano Romeo
Audit date
August 19, 2026
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