# If $\beta=0.{a_1}^{k}{a_2}^{k}{a_3}^{k}\cdots\in\mathbb Q$, then $\alpha=0.a_1a_2a_3\cdots\in\mathbb Q$?

Question : For every even $$k\ge 4$$, is the following $$(\star)$$ true?

\begin{align}\text{If \beta=0.{a_1}^{k}{a_2}^{k}{a_3}^{k}\cdots\in\mathbb Q, then \alpha=0.a_1a_2a_3\cdots\in\mathbb Q.}\qquad(\star)\end{align}

Here, $$a_i$$ is the $$i$$-th decimal place of $$\alpha$$.

Example : For $$k=2,\alpha=0.12345,$$ we have $$\beta=0.1491625.$$

Motivation : I've been able to prove that the answer is NO for $$k=2$$, and that the answer is YES for every odd $$k\ge 3$$. The answer seems YES for any $$k\ge 3$$, but I'm facing difficulty for treating every even $$k$$ in general. Can anyone help?

Remark : One of the counterexamples for $$k=2$$ is $$\alpha=0.237237723777237777\cdots$$. This question has been asked previously on math.SE without receiving any answers.

• Maybe you can say a little more about the motivation for this question? -- E.g. why you think this is interesting, or what you could use it for. – Stefan Kohl Dec 7 '13 at 16:57
• @StefanKohl: Well, I thought $(\star)$ must be true for any $k\ge 2$, so I was a bit surprised to get a counterexample for $k=2$. The main reason why I think this is interesting might be that even though the conjecture must be true, proving that is not very easy. I love this kind of questions. Also, I don't know what we could use it for. In my opinion, we don't always need it though there may be anything useful that I don't notice. – mathlove Dec 7 '13 at 17:16
• Can you describe your proof for odd $k$? – Rodrigo Dec 7 '13 at 21:07
• @user43383: If $k$ is odd, then the one's digit of $1^k ,2^k ,\cdots,9^k$ are different from each other. Using this fact will enable you to prove what you desire. – mathlove Dec 8 '13 at 5:09
• Note that it is base dependent. In base $3$, we still have the property that the last digit is preserved when the number is taken to an odd power, but yet it fails for the even powers $2$ and $8$. For $2$, notice that $2^2=11_3$, and for $8$, $2^8=100111_3$. – Eric Naslund Dec 10 '13 at 23:24