Rational approximation of an integer combination of two irrationals

Let $$x$$ be an irrational number, and $$\beta$$ strictly larger than its irrationality index, which means that for some $$C>0$$, for all $$n\in \mathbb{Z}^*$$, $$d(nx,\mathbb{Z})>C n^{-\beta}.$$ It is known that for a.e. irrational number $$x$$, the irrationality index is $$1$$. It is even known that some numbers satisf the above for $$\beta=1$$ (for instance, $$x=\sqrt{2}$$).

By arguments from measure theory, I have been able to prove that if $$a_n$$ satisfies $$\sum_{n=1}^{\infty}na_n<\infty,$$ almost every couple $$(x,y)$$ of $$\mathbb{R}^2$$ satisfies for some $$C>0$$ $$d(nx+my,\mathbb{Z})>Ca_{|n|+|m|},n,m\in \mathbb{Z}.$$ Ideally, I would like, as for a single number $$x$$, find irrational numbers $$(x,y)$$ such that this holds in the limiting case $$a_n=n^{-2}$$.

Has anyone an idea? Or has anyone a useful reference for such things?

• I suspect the answer is no. Almost all real numbers $x$ have irrationality index $2$, and it seems very likely that one can a sum of two such numbers which is a Liouville number, for example. – Greg Martin Nov 12 '19 at 21:20
• E.g., if $\alpha$ is Liouville, then chances are that $\alpha+\sqrt2$ and $\alpha-\sqrt2$ are both index $2$ while their sum is $2\alpha$. – Gerry Myerson Nov 12 '19 at 22:41
• Ok thanks. So it is likely that even if I carefully choose $x$ and $y$, there will be no better bound than $0$? – kaleidoscop Nov 13 '19 at 5:39
• I completely modified the question as I have made some progress on my side. – kaleidoscop Nov 27 '19 at 19:28
• We must assume $(m,n) \neq (0,0)$, else $d(nx+my, {\bf Z}) = 0$ . . . – Noam D. Elkies Nov 28 '19 at 17:44

Yes, such $$(x,y)$$ exist; for example, $$x = \root 3 \of 2$$ and $$y = x^2$$. For $$l,m,n \in \bf Z$$, define $$N(l,m,n) := l^3 + 2m^3 + 4n^3 - 6lmn \in {\bf Z};$$ this is the algebraic norm $$(l + mx + nx^2) (l + m\rho x + n \bar\rho x^2) (l + m\bar\rho x + n \rho x^2)$$ of $$l + mx + nx^2$$, where $$\rho$$ is the cube root of unity $$e^{2\pi i/3} = (-1 + \sqrt{-3})/2$$. But if $$(m,n) \neq (0,0)$$ then $$l + mx + nx^2 \neq 0$$, so $$\left|N(l,m,n)\right| \geq 1$$ and $$\left|l + mx + nx^2\right| \gg (|l|+|m|+|n|)^{-2}$$. Taking for $$l$$ the integer nearest to $$-(mx+nx^2)$$ we deduce that $$d(mx+nx^2,{\bf Z}) \gg (|m|+|n|)^{-2}$$, as claimed.
The same argument (which generalizes the familiar one for $$\sqrt 2$$) shows that in general if $$x$$ is an algebraic number of degree $$D$$ then $$d\Bigl(\sum_{j=1}^{D-1} n_j x^j, {\bf Z}\Bigr) \gg \left( \sum_{j=1}^{D-1} \left| n_j \right| \right)^{1-D}$$ for $$n_1,\ldots,n_{D-1} \in \bf Z$$ not all zero. This is best possible up to the value of the implicit constant, because Dirichlet's celebrated "pigeonhole" argument shows that conversely for any $$N$$ one can find integers $$n_1,\ldots,n_{D-1} \in [-N,N]$$, not all zero, such that $$d(\sum_{j=1}^{D-1} n_j x^j, {\bf Z}) \ll N^{1-D}$$.
• I see, pretty neat, thanks! "Not new" but for me it is dropped from the sky... Do you know f by any chance such arguments extend even further to infinite sums? i.e. I'm looking for $x_j$ such that for integers $n_j$, the distance from $\sum_j x_j n_j$ to $\mathbb{Z}$ is bounded from below by the inverse value of some norm of $\|(n_j)\|$ on the space of integer sequences... – kaleidoscop Nov 29 '19 at 21:04
• Seems the best $C$ is $1/(6x) = 0.13228342\ldots$. For large $k$, if $l+mx+nx^2 = (x-1)^k$ then its product with $(|m|+|n|)^2$ depends on the phase of $(e^{2\pi i/3}x - 1)^k$, which can be made arbitrarily close to the optimum. As with $\sqrt2$ for $d(nx,{\bf Z})$, the choice $x = \root3\of 2$ here is the most familiar but not quite the best; taking for $x$ the real root of $x^3-x+1$, or the root $2\cos(2\pi/7)$ of $x^3+x^2-2x-1$, should yield somewhat larger $C$. To be sure the exact value of $C$ is partly an artifact of the choice of quadratic factor: [cont'd] – Noam D. Elkies Dec 9 '19 at 21:34
• changing $(|m|+|n|)^2$ to say $m^2+n^2$ or $\max(|m|,|n|)^2$ would change both the liminf and (probably) the sequence of algebraic units that attain it. But the liminf would remain finite and positive. – Noam D. Elkies Dec 9 '19 at 21:36
• Very cool. Thank you. For $x = \root 3 \of 2$ I could only get down to .132283780 before running out of patience. Two more digits and an ISC lookup and I may have got there. – O. S. Dawg Dec 10 '19 at 5:11