An arctan problem including a diophantine equation

Question

This is a follow-up question to An equation of the form A + B + C = ABC . I totally messed up with making the equation from the question specification . Actually the question was $$ \arctan(\frac{1}{A}) = \arctan(\frac{1}{B}) + \arctan(\frac{1}{C})$$

So the equation after rearranging becomes $$\frac{1}{A} = \frac{B+C}{BC-1}.$$ Now we have to find the $\min(B+C)$ where A is given to be a fixed positive integer and $B$ and $C$ and are some positive integers which satisfy the equation. I tried some values which satisfy the equation but I don't have enough mathematical background on solving diophantine equations.

PS : Sorry for the mistake in the previous post

Answer 1

Write your equation as $$B = \dfrac{AC+1}{C-A}$$ Let $C = A + x$, so this becomes $$ B =A + \dfrac{A^2+1}{x}$$ Thus $x$ must divide $A^2+1$, and $B + C = 2A + x + \dfrac{A^2+1}{x}$. You'll want a divisor of $A^2+1$ that is closest to $\sqrt{A^2+1}$ (on one side or the other).

Answer 2

$$\begin{align} \frac1A={B+C\over BC-1} &\iff BC-AB-AC-1=0 \\&\iff (B-A)(C-A)=BC-BA-BC+A^2=A^2+1. \end{align}$$ Thus as Robert Isreal pointed out, to minimize $B+C$, you have to find a divisor of $A^2+1$ that is closest to $\sqrt{A^2+1}$, and let it be $B-A$.