Erdoes #495 - working report - Erdős Problem a Day

Erdős problem #495 — wave 9n

Access/research date: 2026-07-28 UTC.

Claims are marked as requested:

(a) elementary-rigorous — a complete argument is given here.
(b) rigorous-modulo-named-theorem — the exact named theorem and a primary source are identified.
(c) plausible/structural-unverified — explicitly not asserted as a theorem.
(d) computational-only — an exact finite computation, reproducible with the supplied checker.

0. Mandatory live-page gate

(d) I fetched the rendered live page erdosproblems.com/495 through the Bright Data browser path on 2026-07-28, rather than trusting the stale YAML or a datacenter curl.

The current statement, copied verbatim from the live page, is:

Let $\alpha,\beta \in \mathbb{R}$. Is it true that \[ > \liminf_{n\to \infty} n \| n\alpha \| \| n\beta\| =0 > \] where $\|x\|$ is the distance from $x$ to the nearest integer?

(d) The complete relevant live-page status/marker check was:

status: OPEN;
page note: “The infamous Littlewood conjecture”;
0 comments on this problem;
0 claimed proofs for this problem;
Interested in collaborating: David_Robey;
Currently working on this problem: None;
This problem looks difficult: Vjeko_Kovac, Dogmachine, ebarschkis;
This problem looks tractable: None;
Likes this problem: Dogmachine;
formalised statement: yes;
I am working on formalising the results on this problem: David_Robey.

The formalisation marker is separate from the page's explicit “Currently working on this problem” field, which says None. Therefore the mandatory claimed-proof/current-worker stop condition did not fire.

(d) Clicking the page's bibliography marker produced the exact citation

[Er61] Erdős, Paul, Some unsolved problems, Magyar Tud. Akad. Mat. Kutató Int. Közl. (1961), 221–254. MR 0177846.

The live page lists no further mathematical results and has no comments.

1. Primary-source literature check

What was verified

(b) In Erdős's cited primary source, problem I.34 on printed pages 238–239 gives this conjecture, observes that only pairs with both coordinates badly approximable can be nontrivial, and says the problem is deep even for $(\sqrt2,\sqrt3)$. Source: Erdős, “Some unsolved problems” (1961), repository PDF. The downloaded PDF used here has SHA-256 6f6dac75cb03edcaf1d7e13509ea9001937248cea1fd5e932b386a6a0a7a1007.

(b) Pollington and Velani, “On a problem in simultaneous Diophantine approximation: Littlewood's conjecture”, Acta Mathematica 185 (2000), 287–306, prove in Theorem 1 that for every fixed badly approximable $\alpha$, a Hausdorff-dimension-one set of badly approximable $\beta$ satisfies a stronger $1/\log q$ conclusion. Their Theorem 2 also constructs, for the convergent denominators $q_j$ of a fixed badly approximable $\alpha$, a positive-dimensional set of badly approximable $\beta$ for which $\|q_j\beta\|$ stays uniformly away from zero. Thus a proof that merely inspects the convergent denominators of one coordinate cannot work uniformly for all pairs.

(b) Einsiedler, Katok, and Lindenstrauss, “Invariant measures and the set of exceptions to Littlewood's conjecture”, Annals of Mathematics 164 (2006), 513–560, arXiv:math/0612721, prove in Theorem 1.5 that the exceptional set has Hausdorff dimension zero and is a countable union of compact sets of box dimension zero. Their Proposition 11.1 gives the exact dynamical equivalence: $(u,v)$ satisfies Littlewood precisely when the $A^+$-orbit of

\tau_{u,v}= \begin{pmatrix} 1&0&0\\u&1&0\\v&0&1 \end{pmatrix}SL(3,\mathbb Z)

is unbounded in $SL(3,\mathbb R)/SL(3,\mathbb Z)$.

(b) Shapira, “A solution to a problem of Cassels and Diophantine properties of cubic numbers”, Annals of Mathematics 173 (2011), 543–557, arXiv:0810.4289, proves a stronger inhomogeneous conclusion when $1,\alpha,\beta$ form a basis of a totally real cubic field. This includes the older homogeneous cubic-field theorem of Cassels and Swinnerton-Dyer. It does not cover $(\sqrt2,\sqrt3)$: their span with $1$ is not a field (it omits $\sqrt6$).

(b) Adamczewski and Bugeaud, “On the Littlewood conjecture in simultaneous Diophantine approximation”, J. London Math. Soc. 73 (2006), 355–366, explicitly construct continuum many suitable badly approximable $\beta$ for every fixed badly approximable $\alpha$. This gives many nontrivial positive cases, not the universal statement.

(b) The same authors' current Chapter 8 notes, §8.6, Exercise 8.3, explicitly label proving the conjecture for $(\sqrt2,\sqrt3)$ as an open problem. This verifies that Erdős's concrete test pair has not quietly become a known special case in the literature located here.

(b) Two recent primary sources were checked to avoid confusing variants with the live problem. De Mathan's arXiv:2402.08850 (2024) still states the classical universal conjecture and studies divisibility variants. Schleischitz's arXiv:2603.12611 (2026) disproves a different uniform conjecture $Q\min_{q\le Q}\|q\xi\|\|q\zeta\|\to0$; its introduction explicitly distinguishes that assertion from the classical liminf in this problem. It is not a disproof of Erdős #495.

(c) Exact-statement, exact-title, arXiv, DOI, and recent-year searches found no primary source claiming a proof or counterexample to the classical universal statement. A search miss is not a theorem; the live page's OPEN status remains the authoritative status check for this run.

2. Elementary reduction of the genuinely open regime

Write

L_{\alpha,\beta}(n)=n\|n\alpha\|\|n\beta\|.

2.1 Both coordinates must be badly approximable

(a) If $\alpha$ is irrational and not badly approximable, there is a sequence $n_j\to\infty$ with $n_j\|n_j\alpha\|\to0$. Since $\|x\|\leq1/2$,

0\leq L_{\alpha,\beta}(n_j) \leq \frac12 n_j\|n_j\alpha\|\longrightarrow0.

If a coordinate is rational, multiples of its denominator make the product zero. Hence a counterexample would require two badly approximable irrationals.

2.2 Explicit construction for every rationally dependent pair

(a) Suppose $1,\alpha,\beta$ are linearly dependent over $\mathbb Q$. After clearing denominators, take

A\alpha+B\beta+C=0,\qquad A,B,C\in\mathbb Z.

If $B=0$, then $\alpha$ is rational and the result is immediate. Otherwise assume $\alpha$ is irrational, and let $p_k/q_k$ be its continued-fraction convergents. Set

n_k=|B|q_k,\qquad \delta_k=\|q_k\alpha\|.

The relation and $\|rx\|\leq |r|\|x\|$ give

\|n_k\alpha\|\leq |B|\delta_k,\qquad \|n_k\beta\|\leq |A|\delta_k.

Since $\delta_k<1/q_{k+1}$ and $q_{k+1}>q_k$,

L_{\alpha,\beta}(n_k) \leq |A||B|^2q_k\delta_k^2 < \frac{|A||B|^2q_k}{q_{k+1}^2} < \frac{|A||B|^2}{q_k}\longrightarrow0.

This is an explicit denominator sequence, not just an existence argument. Thus only the $\mathbb Q$-linearly independent, badly approximable regime can contain counterexamples; many pairs inside that regime are nevertheless known positive cases.

3. Exact Pell-residual reduction for $(\sqrt2,\sqrt3)$

(a) The numbers $1,\sqrt2,\sqrt3$ are linearly independent over $\mathbb Q$. For if $a+b\sqrt2+c\sqrt3=0$, isolating one radical and squaring forces a rational multiple of $\sqrt2$ to be rational; the remaining cases reduce to the irrationality of $\sqrt2,\sqrt3$, or $\sqrt{3/2}$. Both irrational coordinates have periodic continued fractions, so this is a genuinely nontrivial pair of the kind singled out by Erdős.

For $d\in\{2,3\}$, let

m_d(n)=\left\lfloor n\sqrt d+\frac12\right\rfloor,\qquad A_d(n)=|dn^2-m_d(n)^2|.

(a) Difference of squares gives the exact identity

\|n\sqrt d\| =\frac{A_d(n)}{n\sqrt d+m_d(n)}.

Consequently

L(n):=L_{\sqrt2,\sqrt3}(n) =\frac{nA_2(n)A_3(n)} {(n\sqrt2+m_2(n))(n\sqrt3+m_3(n))}. \tag{1}

Because $m_d(n)/n\to\sqrt d$, the denominator in (1), divided by $n^2$, tends to $4\sqrt6>0$. Therefore

\boxed{\quad \liminf_{n\to\infty}L(n)=0 \iff \liminf_{n\to\infty}\frac{A_2(n)A_3(n)}n=0. \quad} \tag{2}

This isolates the concrete remaining arithmetic task: find infinitely good common near-solutions of the two Pell forms $m_2^2-2n^2$ and $m_3^2-3n^2$. Equation (2) is an equivalence, not a claim that the required sequence is known.

4. Exact exhaustive computation through $10^7$

4.1 Why the scan is exact and small

(a) The nearest integer $m=m_d(n)$ is obtained without floating point. If $r=\lfloor n\sqrt d\rfloor=\operatorname{isqrt}(dn^2)$, compare the integers $4dn^2$ and $(2r+1)^2$ to decide on which side of $r+1/2$ the irrational $n\sqrt d$ lies.

(a) Since $m_2(n),m_3(n)\leq2n$, (1) implies

L(n)>\frac{A_2(n)A_3(n)}{16n}. \tag{3}

Also

L(1)=(\sqrt2-1)(2-\sqrt3)<\frac18:

indeed $\sqrt2<17/12$, $\sqrt3>17/10$, and $(5/12)(3/10)=1/8$. Therefore every possible strict record must pass the integer-only sieve

A_2(n)A_3(n)<2n. \tag{4}

Only 119 of the first $10^7$ integers pass (4).

(a) For each survivor, the checker constructs strict rational brackets for $\sqrt2,\sqrt3$ using

r_d=\operatorname{isqrt}(d\,10^{200}),\qquad \frac{r_d}{10^{100}}<\sqrt d< \frac{r_d+1}{10^{100}},

substitutes them into (1), and compares rational endpoints by integer cross multiplication. It aborts rather than guessing if any two enclosures overlap. No overlap occurred.

4.2 Certified record table

(d) The strict records of $L(n)$ for $1\leq n\leq10^7$ are exactly:

| $n$ | $m_2$ | $m_3$ | $2n^2-m_2^2$ | $3n^2-m_3^2$ | certified $L(n)$ | |---:|---:|---:|---:|---:|---:| | 1 | 1 | 2 | 1 | -1 | $0.11098818953188929293353971521937713\ldots$ | | 4 | 6 | 7 | -4 | -1 | $0.09854702572453264020219488512711705\ldots$ | | 7 | 10 | 12 | -2 | 3 | $0.08748860949967484217323954905918255\ldots$ | | 12 | 17 | 21 | -1 | -9 | $0.07608598448071155341153507527778144\ldots$ | | 15 | 21 | 26 | 9 | -1 | $0.06152375232516599122924810959724555\ldots$ | | 41 | 58 | 71 | -2 | 2 | $0.00995678224782801429861448647878187\ldots$ | | 10,864 | 15,364 | 18,817 | 496 | -1 | $0.00465968469939684423255319776005769\ldots$ | | 4,628,523 | 6,545,720 | 8,016,837 | 4,658 | 18 | $0.00184881708844186488185721663662387\ldots$ |

Every true value lies between the displayed lower endpoint and the corresponding upper endpoint obtained by increasing the last displayed digit by one; those displayed endpoints are separated by $10^{-35}$.

(d) In particular, the exact finite theorem is

\min_{1\leq n\leq10^7}L(n)=L(4{,}628{,}523),

with

\begin{aligned} 0.00184881708844186488185721663662387 &<L(4{,}628{,}523)\\ &<0.00184881708844186488185721663662388. \end{aligned}

The same $n$ minimises the normalized residual product on this range:

\min_{1\leq n\leq10^7}\frac{A_2(n)A_3(n)}n =\frac{4658\cdot18}{4{,}628{,}523} =\frac{27948}{1542841}.

(d) Checkpoint minima are:

| $N$ | exact argmin on $1\leq n\leq N$ | |---:|---:| | $10$ | 7 | | $10^2$ | 41 | | $10^3$ | 41 | | $10^4$ | 41 | | $10^5$ | 10,864 | | $10^6$ | 10,864 | | $10^7$ | 4,628,523 |

5. A closed-form 9,332-digit witness

The dense scan is not the best way to obtain a very small explicit value. Use the Pell sequence

p_k+q_k\sqrt3=(2+\sqrt3)^k,

or, entirely in integers,

(p_1,q_1)=(2,1),\qquad (p_{k+1},q_{k+1})=(2p_k+3q_k,\;p_k+2q_k). \tag{5}

(a) The norm in $\mathbb Q(\sqrt3)$ shows $p_k^2-3q_k^2=1$, and hence

\|q_k\sqrt3\|=p_k-q_k\sqrt3 =\frac1{p_k+q_k\sqrt3}.

This turns the $\sqrt3$ factor into an exact Pell error.

(d) Set $k=16{,}317$, $n=q_k$, and let $m=\lfloor q_k\sqrt2+1/2\rfloor$. The standalone checker constructs these integers from (5), verifies the Pell identity and the nearest-integer inequalities, and certifies

\begin{aligned} 0.0000346168532309693360160166635863679995 &<\frac{|2q_k^2-m^2|}{q_k}\\ &<0.0000346168532309693360160166635863679996, \end{aligned}

and

\boxed{ 0.0000035330677881954211069495432621717336 <L(q_k)< 0.0000035330677881954211069495432621717337 <4\cdot10^{-6}.}

Thus (5) is a compact closed-form specification of a verified, explicit denominator giving $L(n)<4\cdot10^{-6}$. It is one denominator, not an infinite limiting argument.

(d) For independent implementation comparison:

$q_{16317}$ has 9,332 decimal digits and SHA-256

2df0222f56511b3289ad0518ebd21a285183dfa3e410f9bdb394b54147346df8;

$m$ has 9,333 digits and SHA-256

804127753ea110cf8301d3b6eafac1660caad5713b0477fe9d258e8439bb6b32;

$|2q_{16317}^2-m^2|$ has 9,328 digits and SHA-256

01aac96badb05ed596c9ca69372400e3893f57a82e464b32ad70f69e72daacf7.

The integers themselves are deliberately regenerated from the 2-by-2 recurrence instead of being trusted as a pasted 9,000-digit fixture.

6. Exact wall: what is still missing

6.1 A precise sufficient lemma for the Erdős test pair

Let

q_k=\frac{(2+\sqrt3)^k-(2-\sqrt3)^k}{2\sqrt3}.

(a) Since $q_k\|q_k\sqrt3\|\to1/(2\sqrt3)$, the following single-orbit statement would prove Littlewood for $(\sqrt2,\sqrt3)$:

\boxed{\liminf_{k\to\infty}\|q_k\sqrt2\|=0.} \tag{6}

Indeed,

q_k\|q_k\sqrt2\|\|q_k\sqrt3\| \sim \frac1{2\sqrt3}\|q_k\sqrt2\|.

(a) Modulo one, $x_k=q_k\sqrt2$ obeys $x_{k+2}=4x_{k+1}-x_k$. Equivalently, $(x_k,x_{k+1})$ is the orbit of $(0,\sqrt2)$ under the hyperbolic toral automorphism

T=\begin{pmatrix}0&1\\-1&4\end{pmatrix}.

Thus (6) asks for zero to be a limit point of the first-coordinate projection of one explicit algebraic starting orbit.

(b) Pollington–Velani Theorem 2 shows why lacunarity of the $q_k$ alone is insufficient: for a general badly approximable second coordinate, one can stay uniformly away from integers along all convergent denominators. A proof of (6) must use special arithmetic of $\sqrt2$, not only the growth $q_{k+1}/q_k>1$.

(c) The verified $k=16{,}317$ witness suggests a small return but does not establish (6). No monotonicity, recurrence of record sizes, or positive-entropy measure is proved for this single orbit.

6.2 Dynamical wall for the full problem

(b) By EKL Proposition 11.1, the universal conjecture is exactly the assertion that no point $\tau_{\alpha,\beta}$ on the indicated two-dimensional horospherical leaf has bounded $A^+$-orbit. EKL's measure rigidity proves that the bounded-orbit intersection has Hausdorff dimension zero; it does not prove that the intersection is empty.

(c) The missing qualitative input is therefore a topological/single-orbit rigidity theorem ruling out every bounded $A^+$-orbit on this leaf (or, for the concrete Pell route, the return statement (6)). Positive-entropy measure classification does not supply such a theorem for a potentially zero-entropy individual orbit.

(c) No finite cutoff can close a liminf assertion. At the measured 18.28 seconds per $10^7$ denominators, a direct Python extension to $10^{12}$ would cost about $508$ core-hours (roughly 21 single-core days, or about USD 20–51 at USD 0.04–0.10/core-hour), while still proving only another finite statement. It was not run.

7. Reproduction

The standalone from-scratch checker is erdos495_wave9n_reverify.py. Its SHA-256 is e55646dbad704894ff975d95205d34f783e3b035a3add5c846821c87b314632c.

Run:

python3 runs/erdos495_wave9n_reverify.py

It uses only the Python standard library, no floating point in any mathematical decision or reported enclosure, no downloaded data, and no precomputed 9,332-digit integer. On this VM the final run ended with:

exact scan range: 1 <= n <= 10,000,000
integer-sieve survivors: 119
strict record indices: (1, 4, 7, 12, 15, 41, 10864, 4628523)
...
Pell witness k=16317:
digits(q_k,m_k,A2)=(9332,9333,9328)
0.0000035330677881954211069495432621717336 < L(q_k)
  < 0.0000035330677881954211069495432621717337
certified simple bound: L(q_k) < 4/1,000,000
PASS: exhaustive exact certificate completed in 18.28 seconds

PARTIAL: proved the rational-dependence subcase constructively, reduced the hard pair $(\sqrt2,\sqrt3)$ to exact Pell residuals, certified its complete record table through $10^7$, and constructed a checked 9,332-digit denominator with $L(n)<4\cdot10^{-6}$; the required infinite return/unbounded-orbit lemma remains open.

This is the AI working report, labelled by outcome — not an independently verified claim unless marked PROVED. ← ledger