# Real numbers whose digits are the even digits of their squares

The Context

The origin of my question is my own answer to this question, where the continuity of the function $$f: [0,1) \to [0,1)$$ that only preserves the odd digits of its input value is analyzed.

In this context, I will slightly modify the definition given there, as follows.

The function I am considering now is $$f: [0,+\infty) \to [0,+\infty)$$ that preserves only the even digits of the input. So if $$x$$ has decimal expansion

$$x = \sum_{k=-\infty}^{+\infty}a_k 10^k,$$

then

$$f(x)=\sum_{k=-\infty}^{+\infty}a_{2k}10^{k}.$$

So for example

$$f(5\textbf{8}7\textbf{4}1\textbf{2}.7\textbf{8}0\textbf{3}4\textbf{0}5\textbf{1})=842.8301.$$

With a very similar approach to the one given here, it can be shown that $$f$$ is continuous almost everywhere (the only exception being the numbers whose least significant digit occupies an odd position), and right-continuous everywhere.

Edit. I am assuming, as in the original question, to adopt, in case of ambiguity, the finite version of the number's decimal expansion.

Note the self-similarity

$$f\left(10^{2k} x\right)=10^k f(x), \ \ \forall k\in \Bbb Z.$$

Below an approximate plot of the function $$f$$ in the range $$[0,1]\times[0,1]$$.

Red dots represent points actually belonging to the graph of $$f$$. The yellow line represents the graph of $$s(x) = \sqrt x.$$ By self-similarity, a scaling of $$10^{2k}$$ of the $$x$$-axis and of $$10^{-k}$$ of the $$y$$-axis, for any $$k\in \Bbb Z$$, would give an exact replica of the given plot.

Introductory Observation

Aside from the trivial intersections between $$f$$ and $$s$$, that is all the points with coordinates

$$\left(10^{2k},10^k\right),$$

there are many other interesting intersections, such as (limiting ourselves to the range shown in the picture)

$$(0.25,0.5),$$ $$(0.36,0.6),$$ $$(0.0121,0.11),$$ and 'trikiest' ones, such as $$(0.5776,0.76),$$ or even $$(0.35295481,0.5941).$$

The Question

Is there any non-terminating decimal (or even irrational) $$x$$ such that $$y=f(x) = s(x),$$ that is, is there any non-terminating decimal $$y$$ whose square contains - in the even positioned digits - the digits of the original number $$y$$?

Edit. I emphasize again that no infinite sequences of $$9$$'s are allowed, since we are adopting the finite decimal expansion version of the number, if this ambuiguity arises.

• Perhaps I'm not understanding the question, but why aren't 5, 760, and 5941 solutions? Sep 17, 2019 at 15:40
• @rogerl you are correct, they are solutions. I only gave some examples of terminating decimals in the range of the picture. Infinite other solutions can be found from them, by self-similarity.
– dfnu
Sep 17, 2019 at 15:42

The answer is on the affirmative.

Consider the sequence $$\begin{eqnarray} \alpha_0 &=& 1,\\ \alpha_1 &=& 10005,\\ \alpha_2 &=& 1000505,\\ \alpha_3 &=& 10005050005,\\ \alpha_4 &=& 1000505000500000005,\\ \dots & &, \end{eqnarray}$$ where, for $$n>1$$, $$\alpha_{n+1}$$ is obtained from $$\alpha_n$$ by appending a sequence of $$2^{n-1}-1$$ zeros and then a $$5$$: $$\alpha_{n+1} = \alpha_n \cdot 10^{2^{n-1}}+5.$$

Let us first show, by induction, that $$f\left(\alpha_n^2\right) = \alpha_n.\tag{1}\label{eq1}$$ Suppose that \eqref{eq1} holds true for a given $$n$$. Then $$\begin{eqnarray} f\left(\alpha_{n+1}^2\right) &=& f\left(\left(\alpha_n \cdot 10^{2^{n-1}}+5\right)^2\right)=\\ &=& f\left(\alpha_n^2\cdot 10^{2^n}+\alpha_n\cdot 10^{2^{n-1}+1}+25\right). \end{eqnarray}$$ Note that

1. By induction and self-similarity of $$f$$, $$f\left(\alpha^2_n \cdot 10^{2^n}\right) = \alpha_n \cdot 10^{2^{n-1}};$$
2. For $$n\geq 3$$, the addition of the second term $$\alpha_n\cdot 10^{2^{n-1}+1}$$ does not modify any digits of the first term and has $$0$$'s in every even position;
3. The first and last term of the sum never interfere.

As a consequence

$$\begin{eqnarray} f\left(\alpha_{n+1}^2\right) &=& f\left(\alpha_n^2\cdot 10^{2^n} + 25\right) =\\ &=&\alpha_n \cdot 10^{2^{n-1}} + 5=\\ &=& \alpha_{n+1}. \end{eqnarray}$$

Consider now the sequence $$\begin{eqnarray} \beta_0 &=& 1,\\ \beta_1 &=& 1.005,\\ \beta_2 &=& 1.00505,\\ \vdots && \vdots\\ \beta_n &=& \alpha_n\cdot 10^{-2^n-2}. \end{eqnarray}$$ The sequence $$(\beta_n)$$ is monotonic and upper bounded, and thus convergent in $$\Bbb R$$, and so is the sequence $$(\beta_n^2)$$.

Let

$$(\beta_n^2) \to \xi.$$

Clearly $$\xi$$ is a non-terminating decimal. Thus, as shown in the answer to this question, $$f(x)$$ is continuous in $$\xi$$, and so is $$h(x) = f(x) - \sqrt x.$$

We therefore must have

$$\left(h\left(\beta^2_n\right)\right)\to h(\xi).$$

Since, by self-similarity of $$h$$, for each $$n$$ $$h\left(\beta_n^2\right) = 0,$$ it must be $$h(\xi) = 0,$$ that is $$f(\xi) = \sqrt \xi.$$

A little update

• The same reasoning applies to the sequences $$(5.0005, 5.000505, 5.0005050005,\dots)$$, and $$(6.0005, 6.000505, 6.0005050005,\dots)$$.
• In the above mentioned sequences any digit $$5$$ can be replaced by a $$0$$, obtaining thus a dense set of intersection points on the right neighborhoods of $$1$$, $$5$$, and $$6$$.