Question

If $x$ and $y$ are positive integers satisfying ${\tan }^{-1}\left(\frac{1}{x}\right)+{\tan }^{-1}\left(\frac{1}{y}\right)={\tan }^{-1}\left(\frac{1}{7}\right),$ then the number of ordered pairs of $(x,y)$ is

Answer 1

${\tan }^{-1}\left(\frac{1}{x}\right)+{\tan }^{-1}\left(\frac{1}{y}\right)={\tan }^{-1}\left(\frac{1}{7}\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{\frac{1}{x}+\frac{1}{y}}{1-\frac{1}{xy}}\right)={\tan }^{-1}\left(\frac{1}{7}\right)$
$\Rightarrow \frac{x+y}{xy-1}=\frac{1}{7}$
$\Rightarrow 7x+7y=xy-1$
$\Rightarrow x(y-7)=7y+1$
$\Rightarrow x=\frac{7y+1}{y-7}=\frac{7(y-7+7)+1}{y-7}$
$\Rightarrow x=7+\frac{50}{y-7}$
Since $x$ is a positive integer, $y$ can take the values $8,9,12,17,32,57.$
Hence, number of ordered pairs of $(x,y)$ is $6.$