# No two integers are equidistant from the square root of 2.

This is a rather simple consequence of the fact that $\sqrt{2}$ is irrational. Those who understood the proof or came up with one of their own won't be surprised at the following generalization:

No two rational numbers are equidistant from the square root of 2.

We may try to generalize even further. First we claimed that no two integers are equidistant from $\sqrt{2}.$ In other words, the set of all integers does not contain a pair of points equidistant from $\sqrt{2}.$ Next, we established that another, bigger set - the set $\mathbb{Q}$ of all rational numbers - has the same property, i.e., no two of its elements lie at the same distance from $\sqrt{2}.$ Can we augment the latter set still preserving the property in question? So another question arises naturally:

What is the biggest set such that no two of its points are equidistant from $\sqrt{2}.$

The first of the two question is simple, the second is less so (I have thoughts about the latter, but not a complete answer.) We can augment the set $\mathbb{Q}$ by, say, $\sqrt{3}$ and even consider the extension field $\mathbb{Q}[\sqrt{3}].$ On top of this, extension field $\mathbb{Q}[\sqrt{3}, \sqrt{5}]$ also has the desired property. Similarly to the theory of constructible numbers, we may even form an expanding sequence of extensions $\mathbb{Q}[\sqrt{m_{1}}, \sqrt{m_{2}}, \ldots , \sqrt{m_{n}}]$ each of which will suite our purpose as long as it does not contain $\sqrt{2}$ itself. (If one of these fields does contain $\sqrt{2},$ then, being a field, it of necessity also contains $0$ and $2\sqrt{2}$ that are equidistant from $\sqrt{2}.$ The converse is also true: if a field contains two numbers equidistant from $\sqrt{2}$ then it contains $\sqrt{2}.$ (Why?) We can construct such a sequence in many ways, e.g., by selecting distinct $m_{i}$ from the set of prime numbers.

Let's now switch the view. From the stand point of $\sqrt{2}$, all distances from it to the set $\mathbb{Q}$ of rational numbers are different. Are there any other points with the same property? Let $S$ denote the set of all such points. What can be said about it?

First, $S$ contains no rational points. Second, if a real number is equidistant from two distinct rational points, the number is necessarily the average of the two points, and is, therefore, rational. From here, $S$ consists of all irrational points.

The latter result has been generalized by Schoenberg [Cofman, p 187] to higher dimensions. In two dimensions, consider a grid $\mathbb{Q}^2$ of pairs of rational numbers. Schoenberg gave a characterization of the set $S$ of points in the plane each of which lies at different distances from the grid points. According to Schoenberg, $S$ consists of those points $(x_{1}, x_{2})$ that do not belong to any line that admits an equation

$a_{1}x_{1} + a_{2}x_{2} + a_{3} = 0,$

where $a_{1}, a_{2}, a_{3}$ are rational numbers not all equal to $0.$ As an example, point $(\sqrt{2},\sqrt{2})$ lies on the line $x_{1} - x_{2} = 0$ and, therefore, does not belong to $S.$ On the other hand, point $(\sqrt{2},\sqrt{3})$ does belong to $S.$

Stepping back from the grid $\mathbb{Q}^{2}$ to pairs of integal points, we may now answer a question posed by Hugo Steinhaus in 1957. W. Sierpinski gives a background for the question and a solution credited to André Schinzel. As a mater of fact Steinhaus published his question in a journal for the secondary school teachers.

Is it possible to construct, for any (non-negative) integer $n,$ a circle such that its interior contains exactly $n$ grid points?

Not only such circles exist for any $n$, the circles can be selected with the same center. Let, for example, point $(\sqrt{2}, \sqrt{3})$ be chosen as the center of the circles in question. (Sierpinski picks the center at $(\sqrt{2}, 1/3).$ I follow Steinhaus' choice in his out-of-print-but-avaialble-online problem book.) If the radius is sufficiently small, the circle contains no (integer) grid points. Let the radius grow. At some point, the circle will pass through a grid point for the first time. No other grid point may lie on the same circle. From that moment on and until the circle passes through a grid point for the second time, the circle will contain exactly $1$ grid point. Afterwards and until it passes through another grid point for the third time, it will contain only $2$ grid points, and so on. If $f(r)$ is the number of grid points inside the circle with radius $r,$ then $f(r)$ changes in jumps of $1$, never $2$ or more. And clearly it grows without bound as $r$ grows. Therefore, for any $n$ there is radius $r$ (not uniquely determined) such that the circle with radius $r$ contains exactly $n$ grid points.

Sierpinski extends the question to higher dimensions, $3,$ in particular:

Is it possible to construct, for any (non-negative) integer $n,$ a sphere such that its interior contains exactly $n$ points with integer coordinates?

The answer is in affirmative and, again, is based on a judicious selection of a point to serve as a common center of the sought spheres. Sierpinski chooses $(\sqrt{2},\sqrt{3},\sqrt{5}),$ but, as before, this is just an idomatic selection. The point serves to answer the question because for the expression $(u\sqrt{2}+v\sqrt{3}+w\sqrt{5})$ may be rational for rational $u,v,w$ only when all three numbers vanish, $u=v=w=0.$

Further on, Sierpinski establishes asymptotic behavior of $f(r)$ - the number of grid points in the enterior of a circle of radius $r.$. Whatever the center of the circle, the number of grid points within can be estimated with

$\displaystyle\pi(r-\frac{1}{\sqrt{2}})^{2}\le n\le\pi(r+\frac{1}{\sqrt{2}})^{2}.$

If $r_{n}$ is one of the radii for which the inequality holds then

$\displaystyle\mbox{lim}_{n\rightarrow\infty}\frac{r_n}{\sqrt{n}}=\frac{1}{\sqrt{\pi}}.$

**Note**: This is a very surprising fact proved by A. Schinzel (dimension 2) and T. Kulikowski (dimension above 2) that, for every $n,$ there exists a sphere that passes through exactly $n$ integer points.

### References

- J. Cofman,
*What To Solve?*, Oxford Science Publications, 1996. - W. Sierpinski,
__Sur Quelques Problemes Concernant les Points aux Coordonnées Entières__,*L'Enseignement Mathématique*, Volume 4 (Series 2), 1958, 25-31 (available online) - H. Steinhaus,
*One Hundred Problems in Elementary Mathematics*, Problems 23, 24.

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Copyright © 1996-2018 Alexander Bogomolny

### Simple consequence

Assume that for two different integers $m$ and $n,$ $|m - \sqrt{2}| = |n - \sqrt{2}|.$ Then, after squaring both sides of the equality and some simplifications, we get $2\sqrt{2}(n - m) = n^{2} - m^{2}.$ Or $2\sqrt{2} = n + m.$ Which implies rationality of $\sqrt{2}.$ A contradiction.

### Why?

Field is a set closed under arithmetic operations. A field always contains a zero $(0)$ and a unit $(1)$ elements. Since $1 + 1 = 2$ one may be prompted to assert that every field contains $2$ as well. But this is not true in general. E.g., the finite field $\mathbb{Z}_{2}$ that contains no other elements but $0$ and $1$, does not, in particular, contain $2.$ All extension fields of $\mathbb{Q}$ naturally contain $2.$ They, therefore, contain the average $(a + b)/2$ of any two elements.

Now, if two numbers $a$ and $b$ are equidistant from $\sqrt{2},$ then they lie on the opposite sides from $\sqrt{2}$ with the latter serving as their average.

### $S$ contains no rational points

A rational number $r$ is equidistant from two rational numbers: $0$ and $2r.$ Therefore, no rational number may belong to $S.$

### According to Schoenberg

Points $(x_{1}, x_{2})$ in the plane equidistant from the grid points $(p_{1},p_{2})$ and $(q_{1},q_{2})$ satisfy

$(x_{1} - p_{1})^{2} + (x_{2} - p_{2})^{2} = (x_{1} - q_{1})^{2} + (x_{2} - q_{2})^{2},$

or

$(2q_{1} - 2p_{1})x_{1} + (2q_{2} - 2p_{2})x_{2} + p_{1}^{2} - q_{1}^{2} + p_{2}^{2} - q_{2}^{2} = 0$

which is a linear equation with rational coefficients:

(*) | $a_{1}x_{1} + a_{2}x_{2} + a_{3} = 0.$ |

Conversely, let line $L$ be given by equation (*). Note that both $a_{1}$ and $a_{2}$ may not be $0;$ for, otherwise, $a_{3}$ will be $0$ as well in contradiction with the requirement that not all three coefficients are $0.$ Assume, e.g., that $a_{1}$ is not $0.$ Point $M(-a_{3}/a_{1},0)$ then lies on $L.$ By direct verification, points $A(-a_{3}/a_{1} + a_{1},a_{2})$ and $B(-a_{3}/a_{1} - a_{1}, -a_{2})$ lie on a line through $M$ perpendicular to $L.$ Since $M$ is equidistant from the grid points $A$ and $B$ so are all the points of $L$.

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Copyright © 1996-2018 Alexander Bogomolny