An equation of the form A + B + C = ABC [duplicate]

So I was on a SPOJ spree until I came across this question . The question says $$\tan(\frac{1}{A}) = \tan(\frac{1}{B}) + \tan(\frac{1}{C})$$

where we have to find the $\min(B+C)$ for a fix $A$ where $A,B$ and $C$ all are positive integers. After some rearrangement I got $A+B+C = ABC$ . I have no clue how to solve such an equation for positive integers . I just tried some value of $A$ as in I tried $7$ which gives $7BC = 7+B+C$ but by trial and error ( for finding positive integer solutions ) it doesn't seem any $B$ and $C$ will satisfy the equation . Any hints on how to proceed ?

PS : I don't have much knowledge but is this a diophantine equation.

This is pretty standard. Since the equation is symmetric in $A,B,C$, it suffices to find all solutions with $A \leq B \leq C$ and then by permutations you get all of them.

Then $$ABC = A+B+C \leq C+C+C =3C$$ which implies $AB \leq 3$.

Now, since $A \leq B$, there are only three possibilities such that $A B \leq 3$. In each of them $ABC=A+B+C$ becomes a simple equation in $C$.