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

Question

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: $101^2ab|101(101a^2+2b^2)-b^2$

Answer 1

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.

Answer 2

$\frac {a}{ b} + \frac {201b}{10201a} =n$

$10201=101^2$

$\frac {10201a^2+201b^2}{ 10201ab} =n$

$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 $

$d=200,\;n=2$

$c=200*101b$

$a=\frac {10201nb \pm 200*101b}{ 2*10201}=b \pm \frac {100b}{ 101}$

$\Rightarrow b=101$

$a=(1,201),\;(a,b)= {(1,101), (201,101)}$

Answer 3

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.