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.

  • $\begingroup$ Since you are asking for integer solutions, yes, this is a diophantine equation: $A + B + C = ABC$. $\endgroup$ – hardmath Aug 1 '14 at 13:39
  • $\begingroup$ I have asked a more basic, follow-up question at math.stackexchange.com/questions/884695/… $\endgroup$ – Doubt Aug 1 '14 at 15:34
  • 1
    $\begingroup$ @Doubt sorry it was arctan not tan $\endgroup$ – abkds Aug 1 '14 at 16:48
  • $\begingroup$ @Zoro - it still does not seem to work. Did you mean $\arctan(1/A) = \arctan(C) - \arctan(1/B)$? $\endgroup$ – Doubt Aug 1 '14 at 20:32
  • $\begingroup$ $axy+bx+cy=d$ is equivalent $a^2xy+abx+acy+bc=ad+bc$ or $(ax+c)(ay+b)=ad+bc$ This can be used to your case $A=7$ $\endgroup$ – Bumblebee Aug 4 '14 at 9:48

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 $C$.

  • $\begingroup$ Dude I just asked for a hint you gave the complete solution $\endgroup$ – abkds Aug 1 '14 at 13:42
  • $\begingroup$ @Zoro deal with it you nub, it's a good solution $\endgroup$ – John Fernley Aug 1 '14 at 13:44
  • $\begingroup$ @Zoro The issue with problems like this is that I don't see any good hint, once I hint that $ABC \leq 3C$, that's basically the solution. $\endgroup$ – N. S. Aug 1 '14 at 13:53
  • $\begingroup$ yeah you are right . Thanks a lot :)! $\endgroup$ – abkds Aug 1 '14 at 13:54
  • $\begingroup$ @Zoro P.S. This is a pretty standard technique for Diophantine equations in multiple variables, where both sides are positive and one side has degree (much) lower than the other side. $\endgroup$ – N. S. Aug 1 '14 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.