Iranian Mathematical Olympiad

Let $k$ be the number of digits of $a$. Therefore, $10^{k-1}\le a<10^k$. We have $\frac{a}{b}=b.a=b+\frac{a}{10^k}$ and consequently $\frac{a-b^2}{b}=\frac{a}{10^k}$. Since $(a,b)=1$, $(a-b^2,b)=1$; therefore, $\frac{a-b^2}{b}$ is the irreducible form of $\frac{a}{10^k}$. Thus, for some natural number $s$, $sb=10^k$ and $s(a-b^2)=a$.

Now, $a-b^2|a$ and $(a-b^2,a)=1$. Hence, $a-b^2=\pm 1$, but $\frac{a-b^2}{b}=\frac{a}{10^k}>0$, so $a-b^2=1$. Therefore, $a=b^2+1$ and $\frac{1}{b}=\frac{a}{10^k}$.

Since $a\ge 10^{k-1}$, we have $\frac{1}{b}=\frac{a}{10^k}\ge \frac{1}{10}$ which implies $b\le 10$. On the other hand, $ab=b(b^2+1)=10^k$, so prime factors of $b$ and $b^2+1$ are only 2 or 5. Since $b\le 10$, we have $b\in \{1,2,4,5,8,10\}$. A simple calculation shows that only for $b=2$, $b(b^2+1)$ is a power of 10. For this case, $a=5$ and $k=1$. Therefore, $\frac{5}{2}=2.5$, this is the only solution.