How many different pairs $(a,b)$ of $+ve$ integers are there such that $a \leqq b$ and $\frac{1}{a}+\frac{1}{b} =\frac{1}{9} ?$
I approached this problem by simplifying the equation in term of $b$ which is $b=\frac{9a}{a-9}$ and then substituting the value of $b$ in $a \leqq b$ and I got $a \leqq 18$ and similarly doing for $b$, I got $b \geqq 18$.
So from here I can say that one such pair is $(18,18)$ and the rest I found out by manually putting the value of $a$ less than $18$ till $1$ as $a \in [1,18]$ in the equation $\frac{1}{a}+\frac{1}{b} =\frac{1}{9} $ and finding $b$ and by doing so I found two another pairs i.e. $(12,36)$ and $(10,90)$.
But this approach is very tedious and takes a lot of time by putting the value in the equation and then finding the value of the another variable which satisfies the condition. Can someone please explain to me a better way to solve this question so that I do not have to find the pairs by manual putting and finding the value method?
Thanks in Advance!
Your equation is equivalent to $$(a-9)(b-9)=81=3^4,$$ and $a\le b$ implies $a-9=\sqrt{(a-9)(a-9)}\le\sqrt{(a-9)(b-9)}=\sqrt{81}=9$. So $a-9$ must be in $\{1,3^1,3^2\}$ (all divisors of $3^4$ not greater than $9$).
Obviously, we don't have to consider products of two negative divisors of $81$, since one of them would have to be $\le-9$, and thus $a$ or $b$ not positive.