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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.

Arctic Char
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abkds
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  • Since you are asking for integer solutions, yes, this is a diophantine equation: $A + B + C = ABC$. – hardmath Aug 01 '14 at 13:39
  • I have asked a more basic, follow-up question at http://math.stackexchange.com/questions/884695/from-tan1-a-tan1-b-tan1-c-to-a-b-c-abc – Doubt Aug 01 '14 at 15:34
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    @Doubt sorry it was arctan not tan – abkds Aug 01 '14 at 16:48
  • @Zoro - it still does not seem to work. Did you mean $\arctan(1/A) = \arctan(C) - \arctan(1/B)$? – Doubt Aug 01 '14 at 20:32
  • $axy+bx+cy=d$ is equivalent $a^2xy+abx+acy+bc=ad+bc$ or $(ax+c)(ay+b)=ad+bc$ This can be used to your case $A=7$ – Bumblebee Aug 04 '14 at 09:48

1 Answers1

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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 $C$.

N. S.
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