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: $101^2ab|101(101a^2+2b^2)-b^2$
 A: $\frac {a}{ b} + \frac {201b}{10201a} =n$
$10201=101^2$
$\frac {10201a^2+201b^2}{ 10201ab} =n$
$10201a^2-10201nba+201b^2=0 $ 
$Let\; solve\; for\; a\; this\; quadratic\; equasion$
$ D=101^4n^2b^2-4*101^2*201b^2=101^2*b^2(101^2n^2-4*201)=c^2, с \in \mathbb{N} $
$\frac {c}{ 101b}=d, d\in \mathbb{N} $
$ 101^2n^2-4*201=d^2 $
$ 101^2n^2-d^2=4*201=1*2^2*3*67 $
$ (101n-d)(101n+d)=4*201=1*2^2*3*67 $
$ RHS's\; factors\; have\; the\; same\; parity,\;since\; their\; difference\; must\; be\; even $
$ Hence,\;101n-d=2,\;101n+d=402  $
$d=200,\;n=2$
$c=200*101b$
$a=\frac {10201nb \pm 200*101b}{ 2*10201}=b \pm \frac {100b}{ 101}$
$\Rightarrow b=101$
$a=(1,201),\;(a,b)= {(1,101), (201,101)}$
A: Hints From there $10201ab | (10201a^2 +201b^2)$ you get that $10201a | 201b^2 = (10201a^2 +201b^2)- (10201a)a $, and because $a$ and $b$ are coprime you get $a | 201$ (and also $10201|b^2$). Using the same idea you can get $b|10201$. This gives you only finitely many cases to check. 
A: 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.
