Question

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

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}{x y}}\right)=\tan ^{-1}\left(\frac{1}{7}\right)$
$\Rightarrow \frac{x+y}{x y-1}=\frac{1}{7} \Rightarrow 7 x+7 y=x y-1 \Rightarrow(7-y) x=-7 y-1 $
$\Rightarrow x=\frac{7 y+1}{y-7}=\frac{7(y-7+7)+1}{y-7} \Rightarrow x=7+\frac{50}{y-7}$
Here, $y=8,9,12,17,32,57$ are satisfy. ]