# Numbers of the kind $0.aaa\ldots =\frac{1}{aaa\ldots a}$

What are the integers $$a$$ such that $$0.\underbrace{aaa\ldots}_{\infty\text{ times}} =\frac{1}{\underbrace{aaa\ldots}_{k\text{ times}}}$$

eg. $$0.333\ldots=1/3\\$$ while $$0.1616\ldots\ne1/16\ne1/1616\ne1/\underbrace{161616\ldots}_{\text{any }k}$$

I was able to reduce $$a$$ to $$a=\frac{10^d-1}{\sqrt{10^{k\,d}-1}}$$
where $$d$$ is the number of digits in $$a$$ in base $$10$$. If this is correct, is $$k=1,d=1$$ the only solution for integral $$a$$?

• @player3236 Careful: $a$ is not assumed to be a single digit. Commented Feb 8, 2021 at 3:34
• For base $10$, $10^{kd}-1$ is a square only for $kd = 1$ since $10^{kd} - 1 \equiv 3 \pmod 4$ otherwise. Hence $a$ is rational only for $k=d=1$. Commented Feb 8, 2021 at 3:45
• @player3236 That's a complete answer; I think you should write that up. Commented Feb 8, 2021 at 3:57

I think a complete answer should cover all bases (no pun intended). We will have:

$$a = \frac {b^d-1}{\sqrt{b^{kd}-1}}$$

for any base $$b$$. However for $$k\ge 2$$:

$$\sqrt{b^{kd}-1}\ge \sqrt{b^{2d}-1} = \sqrt{(b^d-1)(b^d+1)} \ge b^d-1$$

forcing $$a<1$$. Hence we must have $$k=1$$, with $$a = \sqrt{b^d-1}$$, or $$b^d - a^2 = 1$$.

By Catalan's conjecture (or Mihăilescu's theorem, or if you are interested, this $$1+a^2 = b^d$$ is a special case proven by V. A. Lebesgue using Gaussian integers), if $$d \ge 2$$ there are no solutions. This forces $$d = 1$$, and we have only the uninteresting case:

$$1+a^2 = b$$

so our choice of $$b$$ must be one more than some square. For base $$10$$, $$10 = 1+3^2$$, so $$0.\dot3 = \dfrac13$$.

Let $$a$$ have $$d$$ digits in base $$b$$. Then \begin{align} a\ge b^{d-1}=1{\underbrace{00..0}_{d-1\text{ times}}}{}_{b}\\ \end{align} Therefore \begin{align} \underbrace{aaa\ldots a}_{k\text{ times}}&\ge1000\ldots0100\ldots0100\ldots0_b\\ &>1000\ldots0000\ldots0000\ldots0_b\\ &=b^{k\, d -1} \end{align} So \begin{align} \frac{1}{aaa\ldots a}& If $$k\,d>1$$,then \begin{align} \frac{1}{aaa\ldots a}&
where $$a_1a_2a_3\ldots a_d$$ is written showing its digits in base $$b$$. If $$k\,d>2$$, there is a non-zero number of $$0$$s after the decimal in the RHS of eqn. $$1$$ implying $$a_1=0$$. If $$k\,d=2$$, then the LHS $$<0.1$$ which again implies $$a_1=0$$. In either case, this contradicts the assumption that $$a$$ has $$d$$ digits in base $$b$$. Therefore, for postive integral $$k,d$$ $$k\,d=1\implies k =1,d=1$$ So the required numbers are of the form \begin{align} 0.a_1a_1a_1\ldots&=\frac{1}{a_1}\\ a_1.a_1...&=b/a_1\\ a_1+\frac{1}{a_1}&=\frac{b}{a_1}\\ a_1&=+ \sqrt{b-1}=a \end{align}

same as @player3236's result above. For e.g. $$0.\bar{2}_5=\left(\frac{1}{2}\right)_{5}$$. Note that $$b>2$$ since $$a=0,1$$ don't satisfy the condition in the question.