Let $a,b ∈N$ with $\gcd(a,b) = 1$. Find the number of ordered pairs $(a,b)$ such that: $\frac {a}{ b} + \frac {201b}{10201a} ∈ N$. 
Let $a,b \in \mathbb{N}$ two coprime numbers, so $\gcd(a,b) = 1$. 
  Find the number of ordered pairs $(a,b)$ such that: $\frac {a}{ b} + \frac {201b}{10201a} \in \mathbb{N}$.

I got $10201ab | (10201a^2 +201b^2)$. How to proceed further?
Edit: As pointed out in the comment, I noticed $101^2=10201$. Therefore it can also be written as: $101^2ab|101(101a^2+2b^2)-b^2$

Right.. so $\frac {201b^2}{10201a} = kb -a$ (for some natural $k$) is an integer.
But $a$ and $b$ are relatively prime so $a|201 = 3*67$.
So 4 cases: $a=1, 3, 67,$ or $201$.
In each case if we let $a'=\frac {201}3$  we get $\frac ab + \frac {a'b}{101^2} = k$ and $a + \frac {a'b^2}{101^2} = kb$ and as $\gcd(201, 101) = 1$ we have $\gcd(201,a)=1$ and $b = 101m$ is a multiple of $101$.
So we have $a + a'm^2 = 101km$ which means $m|a$.
If $m=a$ we get $a + 201*a = 202a = 101ka$ and $k= 2$.
And we get 
$\frac {1,3,67,201}{101,303,6767,20301} + \frac {201*(101,303,6767,20301)}{10201*(1,3,67,201)}= 2$.
The only things to check are if $m|a$ but $m < a$. But if $m < a$ then $a'm\le 67$ and $\frac am+a'm = 101k$ will never be possible.
