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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!

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  • $\begingroup$ Just to inform, this question for $gcd(a,b)=1$ was in the Iran NMO second round exam, question 1 :) $\endgroup$ – CODE Jun 29 '14 at 11:16
  • $\begingroup$ From $10^na^2+ba-10^nb=0$, it's not hard to show that $b$ must be of the form $2^r5^sk^2$ with $(k,10)=1$. This rules out numbers like $b=2600$ and $b=6500$. $\endgroup$ – Barry Cipra Apr 3 at 12:21
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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).

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    $\begingroup$ I really want to know if there are solutions when gcd$(a,b)>1$. $\endgroup$ – Next Dec 31 '13 at 12:56

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