TITLE: Solve $\frac{b}{a}=a.b$ in decimal
QUESTION [10 upvotes]: Are there any equations of the form $\dfrac{b}{a}=a.b$ other than $\dfrac{5}{2}=2.5$?

Denote $n=\lfloor\log_{10}b\rfloor+1,$ then 
$$\frac{b}{a}=a.b=a+\frac{b}{10^n},\\
b=a^2+\frac{ab}{10^n},a^2<b<a^2+a<(a+1)^2, a=\lfloor\sqrt{b}\rfloor,10^n\mid ab.$$
I searched $b<10^5$ such that $10^n\mid ab$ and got a table:
$$\begin{array}{|c|c|c|}\hline
 b & a & d(b)=a^2+a b/10^n-b \\\hline
 5 & 2 & 0 \\
 75 & 8 & -5 \\
 100 & 10 & 1 \\
 400 & 20 & 8 \\
 500 & 22 & -5 \\
 640 & 25 & 1 \\
 900 & 30 & 27 \\
 2600 & 50 & -87 \\
 5000 & 70 & -65 \\
 6500 & 80 & -48 \\
 9375 & 96 & -69 \\
 10000 & 100 & 10 \\
 25625 & 160 & 16 \\
 31250 & 176 & -219 \\
 40000 & 200 & 80 \\
 62800 & 250 & -143 \\
 65625 & 256 & 79 \\
 76000 & 275 & -166 \\
 90000 & 300 & 270 \\\hline
\end{array}$$
$\frac{b}{a}=a.b$ iff $(b)=0$. For example, $d(640)=1\neq0$ and $\frac{640}{25}=25.6$.
Also we can search the solutions start with $a,$ then $n=\lfloor\log_{10}a \rfloor+1$ or $n=\lfloor\log_{10}a\rfloor+2$, $b=\frac{10^n a}{10^a-n}.$
I think someone has conducted research and gives a complete solution to this problem, because the problem is so interesting and natural, so I hope you can give me a reference or write an answer here, thanks in advance!

REPLY [7 votes]: I assume you require a fraction in shortest terms, i.e. $\gcd(a,b)=1$. 
To continue from $$ \frac ba= a+\frac b{10^n}\implies 10^nb=10^na^2+ab$$
we see that not only $$ 10^n\mid 10^n(b-a^2)=ab$$
but also
$$ a\mid a\cdot(10^na+b)=10^n b$$
and $$ b\mid b\cdot(10^n-a)=10^n a^2.$$
With $\gcd(a,b)=1$ this implies $a\mid 10^n$ and $b\mid 10^n$ and (with negatives excluded) ultimately $(a,b)=(2^n,5^n)$ or $(a,b)=(5^n,2^n)$ or $(a,b)=(10^n,1)$ or $(a,b)=(1,10^n)$. The last case is excluded because $b<10^n$ is required, the penultimate case is excluded because it leads to $\frac ba<1<a+\frac b{10^n}$.
So we want to solve 
$$ 10^n 5^n=10^n2^{2n}+2^n5^n\qquad\text{or}\qquad 10^n 2^n=10^n5^{2n}+5^n2^n,$$
or equivalently
$$ 5^n=4^{n}+1\qquad\text{or}\qquad 2^n=25^{n}+1.$$
The only solution is indeed $n=1$ for the left variant (and no solution on the right).