Number theory: find $a, b$ such that $\frac{a}{b} = b.a$ in a general base $\mathcal{B}\neq 10$

Question

I was playing with numbers and thinking about this "coincidence"

$$\frac{5}{2} = 2.5$$

That is, for positive $a$ and $b$ we have

$$\frac{a}{b} = b.a$$

And those questions came into my mind:

1. Could we find all such integers pair $a, b$? (clearly in base $10$)

And due to the fact that we work in base $10$, a more general problem popped up, that is, to write our numbers in a base $\mathcal{B}\neq 10$ and thence look for triplets $(\mathcal{B}, a, b)$ such that

$$\frac{a}{b} = b.a$$

When $a, b$ are written in base $\mathcal{B}$.

2. Could we find a general formula that will produce infinitely many such integer triples?

I am not really into number theory, except for few little questions, so this is more a sort of "I am asking to you experts in the field" question.

If for some reason this problem is unclear or wrong or impossible, just tell me!

Thank you!

Answer 1

In the base $10$ we have $$\frac{a}{b}=\frac{10b+a}{10}$$ which gives $$a=\frac{10b^2}{10-b}$$ and $$\frac{10b^2}{10-b}\leq9$$ gives $$b\leq\frac{\sqrt{3681}-9}{20},$$ which gives $$b\le2.$$ Since $b=1$ is not valid, we obtain $b=2$ and $a=5$ only.

Answer 2

$$a=b\left(b+\dfrac a{10}\right)$$

$$\iff10a=b(10b+a)\iff a=\dfrac{10b^2}{10-b}=\dfrac{10(10-c)^2}c$$ setting $10-b=c$

$\implies 0<c<10$

$$a=10c-200+\dfrac{1000}c\implies c|1000$$

Answer 3

My idea would be to do $\frac{a}{b}=b+\frac{a}{10}$ and this leads to $a=\frac{10b^2}{(10-b)}$. I think this may be a possible solution for case 1. Actually I do not have an idea for $\mathcal{B}\neq 10$.