It is not difficult to see (a,b)=(k,1) for all k. If a,b>1 then a=b2*a so b2* =1 .hence only solution set is (k,1).

Sorry i made a typo. the numbers are coprime

It is well known that $\frac{a}{b}$ have finite decimal representation iff $b=2^x5^y$, where $x,y\geq 0$. If $x=y$ then $\frac{a}{10^x}=10^x.a$ i.e. $a=10^x.a\times 10^x$. Since $a$ is an integer we must have $a>10^{x-1}$ but since $(a,10)=1$ then we have $a<10^x$ therefore $a=10^{2x}+a$ which is impossible. If $x>y$ then $5^{x-y}a=2^x5^y.a\times 10^x$ and since $(a,2)=1$ we must have $5^{x-y}a=2^{2x}5^{x+y}+a$ i.e. $a(5^{x-y}-1)=2^{2x}5^{x+y}$ and hence $a=5^{x+y}$ and $5^{x-y}-1=4^x$ so $x=1$ and $y=0$. If $x<y$ then $a(2^{y-x}-1)=5^{2y}2^{x+y}$ and hence $5^{2y}=2^{y-x}-1$ which is impossible. Only solution is $(a,b)=(5,2)$.

dan23 wrote: Find all coprime natural numbers $a$ & $b$ such that: $\frac{a}{b}=b.a$ (note that if $a=92$ and $b=13$ then $b.a=13.92$) Let $p$ be the number of digits of $a$ and the equation is $\frac ab=b+10^{-p}a$ with $10^p>a\ge 10^{p-1}$ $\iff$ $a(10^p-b)=b^210^p$ with $10^p>a\ge 10^{p-1}$ Since $a,b$ are coprime, we get $a|10^p$ and so $a\in\{10^{p-1},2.10^{p-1},5.10^{p-1}\}$ If $a=10^{p-1}$, we get $10b^2+b-10^p=0$ and so $10|b$ and so $p=1$ (else $a,b$ are not coprime) which is not a solution. If $a=2.10^{p-1}$, we get $10b^2+2b-2.10^p=0$ and so $5|b$ and so $p=1$ (else $a,b$ are not coprime) which is not a solution. If $a=5.10^{p-1}$, we get $10b^2+5b-5.10^p=0$ and so $2|b$ and so $p=1$ (else $a,b$ are not coprime) and so $b=2$ and $a=5$ which indeed is a solution Hence the unique answer : $\boxed{(a,b)=(5,2)}$ which gives $\frac 52=2.5$ * edited : too late **

anyone know how to solve this equation for gcd(a,b)>1?