The collection of best constants for the Diophantine approximation problem of finding infinitely many rational solutions to the inequality

with is encoded by the so-called *Lagrange spectrum* .

In a similar vein, the Markov spectrum encodes best constants for a Diophantine problem involving indefinite binary quadratic real forms.

These spectra were first studied in a systematic way by A. Markov in 1880, and, since then, their structures attracted the attention of several mathematicians (including Hurwitz, Perron, etc.).

Among the basic properties of these spectra, it is worth mentioning that are closed subsets of the real line. Moreover, the works of Markov from 1880 and Hall from 1947 imply that

is a increasing sequence of quadratic surds converging to , and

On the other hand, it took some time to decide whether . Indeed, Freiman proved in 1968 that by exhibiting a countable (infinite) collection of isolated points in . After that, Freiman constructed in 1973 an element of which was shown to be a non-isolated point of by Flahive in 1977.

A common feature of these examples of elements in is the fact that they occur before In 1975, Cusick conjectured that there were no elements in beyond .

In our preprint uploaded to arXiv a couple of days ago, Gugu and I provide the following negative answer to Cusick’s conjecture:

Below the fold, we give an outline of the proof of this theorem.

Remark 1The basic reference for this post is the classical book of Cusick and Flahive.

**1. Description of the key ideas**

Recall that and , where

and is the usual continued fraction expansion.

We consider the finite word of odd length: this is a *non* semi-symmetric word in the sense of Flahive (i.e., it can *not* be decomposed into two palindromes).

Remark 2The examples constructed by Freiman of elements in were based on two non semi-symmetric words, and Flahive showed in her paper that “an element of is often associated to non semi-symmetric words”.

The periodic sequence obtained by infinite concatenation of has Markov value

If we try to glue the word on the *right* of , we get a new sequence with Markov value

On the other hand, if we try to glue the word on the *left* of in a way to obtain the smallest possible change in the Markov value, then the best choice is the sequence whose Markov value is

Hence, the cost of gluing to the left of is always higher than the cost of . This indicates that the Markov value of *doesn’t* belong to because any attempt to reproduce as the of for some sequence would force the appearance of large chunks of at arbitrarily large positions in , so that would contain a subword (which is essentially) arbitrarily close to and, thus, its Markov value would be *at least* the Markov value of , a contradiction (since ).

Closing this section, we remind for later use the following standard comparison lemma for continued fractions.

**2. “Big” words and self-replication**

The following lemma provides a list of words whose appearance in a sequence forces its Markov value to be

Lemma 3If contain any of the words

- (1)
- (2)
- (3)
- (4)
- (5)
- (6)
- (7)
- (8)
- (9)
- (10)
- (11)
- (12)
- (13)
- (14) or
- (15) or
- (16) or
- (17)
- (18) or
- (19)
- (20)
- (21)
- (22)
- (23)

and is the position in asterisk, then

for some .

*Proof:* We prove this lemma by straightforward calculations using the standard comparison Lemma 2.

We verify the items in their order of appearance: for example, , , etc.

Sometimes, we need to use the previous items ((1), (2) and/or (5)) and the assumption that for all to get the desired conclusion for a given item: for instance, by (1) and the fact that , we obtain (3):

See our original article for more details or this Mathematica notebook here.

A direct consequence of this lemma is the fact that the word

must extend as

whenever the Markov value is .

Corollary 4Let where is the position in asterisk. If for all , then extends as

and the vicinity of the position is .

*Proof:* By Lemma 3 (3) and (19), extends to the left as

By Lemma 3 (5) and (20), continues to the right as

By Lemma 3 (6) and (7), extends to the right as

By Lemma 3 (1) and (8), continues to the right as

By Lemma 3 (1) and (21), extends to the left as

By Lemma 3 (10) and (22), continues to the left as

By Lemma 3 (11) and (23), extends to the left as

By Lemma 3 (15), continues to the left as

By Lemma 3 (1), (2) and (4), extends to the left as

This completes the proof.

**3. “Small” words and local uniqueness**

The next lemma gives a list of words associated to positions where a Markov value is *not* reached.

Lemma 5If contain any of the words

- (24)
- (25)
- (26)
- (27) or
- (28)
- (29) or
- (30)
- (31)
- (32) or
- (33)
- (34) or
- (35)
- (36)
- (37) or
- (38) or

and is the position in asterisk, then

*Proof:* The proof of this lemma is a straightforward calculation based on the standard comparison Lemma 2: see our original article or this Mathematica notebook here for more details.

By putting together Lemmas 3 and 5, we show below that a Markov value between and is necessarily attained at positions whose vicinities are

and for all , then, up to transposition about the th position,

where the asterisk indicates the th position.

*Proof:* By Lemma 3 (1) and (2), and Lemma 5 (24), we have that, up to transposition,

By Lemma 3 (3) and Lemma 5 (25), extends to the right as

By Lemma 3 (1) and (4), extends to the left as

By Lemma 3 (5) and Lemma 5 (26), continues to the right as

By Lemma 5 (27), is forced to extend to the left as

By Lemma 3 (1) and Lemma 5 (28), continues to the left as

By Lemma 3 (6), (7) and (8), is forced to continue on the right as

By Lemma 3 (10) and Lemma 5 (30), extends on the left as

By Lemma 5 (31) and Lemma 3 (1), (2), continues as

or

We claim that the first possibility can’t occur. Indeed, by Lemma 5 (32) and Lemma 3 (1), (2) and (4), this word would extend to the left as

so that its left side would contain the word , a contradiction with Lemma 3 (5). Therefore, we have to analyse the word

By Lemma 3 (11) and Lemma 5 (33), extends on the left as

By Lemma 3 (1) and (13), continues on the right as

By Lemma 3 (15), is forced to extend on the left as

By Lemma 3 (1), (2) and (4), continues on the left as

This ends the proof.

**4. Conclusions**

The first consequence of our previous discussion is the absence of Lagrange spectrum in the interval with extremities and .

Proposition 7

*Proof:* Suppose that , say for some .

By definition, we can select and an increasing sequence such that

for all , and

for each .

By applying Corollary 6 at the positions and Corollary 4 at adequate positions of the form with , we deduce that

a contradiction. (Here denotes the periodic sequence obtained by infinite concatenation of the word .)

Hence, any Markov value in the interval with extremities and does not belong to the Lagrange spectrum. As it turns out, it is not hard to construct a whole Cantor set of such values: in fact, it is not difficult to check that the Markov values of the sequences

form a Cantor set isomorphic to such that

In particular, we deduce that

and the Hausdorff dimension of is equal to the Hausdorff dimension of . This completes the proof of Theorem 1 in view of the recent results of Jenkinson and Pollicott on the Hausdorff dimension of .

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