For every $(a,b)$

Number Theory Level 4

What is the minimum value of $n$ , such that for each ordered pair of integers $(a,b)$ with $ab\neq 0$ , the number of ordered pair of integers $(c,d)$ with the following two properties doesn't exceed $n$ .

Property-I : $\frac{a}{b} = \frac{c}{d}$

Property-II : $c$ and $d$ are coprime.

4 3 2 1

1 solution

Muhammad Rasel Parvej
Dec 4, 2016

The answer is $\boxed{2}$ .

For a positive fraction $\frac{a}{b}$ , we have two choices for such $\frac{c}{d}$ :

$c=\frac{|a|}{gcd(a,b)}$ and $d=\frac{|b|}{gcd(a,b)}$

OR

$c=\frac{-|a|}{gcd(a,b)}$ and $d=\frac{-|b|}{gcd(a,b)}$ .

Example :

$\frac{a}{b}=\frac{2}{4}$ has two such $\frac{c}{d}$ , namely, $\frac{1}{2}$ and $\frac{-1}{-2}$ . (Although they're equal in magnitude, we're counting them as different ordered pairs of integers $(c,d)$ .)
$\frac{a}{b}=\frac{-3}{-6}$ has two such $\frac{c}{d}$ , namely, $\frac{1}{3}$ and $\frac{-1}{-3}$ .

For a negative fraction $\frac{a}{b}$ , we have again two choices for such $\frac{c}{d}$

$c=\frac{-|a|}{gcd(a,b)}$ and $d=\frac{|b|}{gcd(a,b)}$

OR

$c=\frac{|a|}{gcd(a,b)}$ and $d=\frac{-|b|}{gcd(a,b)}$ .

Example :

$\frac{a}{b}=-\frac{3}{9}$ has two such $\frac{c}{d}$ , namely, $\frac{-1}{3}$ and $\frac{1}{-3}$ .

Note : $gcd(a,b)$ is positive.

For every ( a , b ) (a,b) ( a , b )

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For every $(a,b)$