Sum of arctangents

Original post by Mr M

I'm trying to find positive rational solutions for the following equation:

\tan^{-1}x + \tan^{-1}y + \tan^{-1}(\frac{6}{xy}) = \pi

Obviously any pair of 1, 2, 3 are solutions.

Clues please because I'm not sure where I'm going with this one.

Have you gotten to the stage where you need to solve

x^2y+xy^2-6xy+6=0

by taking the tan of both sides from the start? There's symmetry in there, perhaps that helps. You could rearrange this to be in the form of a quadratic in x and then consider the formula to get an answer in terms of y and repeat this for a quadratic in y (and you'll find that these two equations are the same, just with the x and y's swapped around). Although, I'm not sure how I would finish off the problem.
EDIT: New idea, divide through the above equation by 6xy. May leave you with something much easier to deal with.
EDIT2: Drawing the graph may also be useful.

(edited 13 years ago)

Reply 2

Original post by Farhan.Hanif93

Have you gotten to the stage where you need to solve

x^2y+xy^2-6xy+6=0

I did get to there and tried to look at what I needed under the square root so that the answers would be rational. I couldn't get that to work. The answers do lie between 0.9358 and 3.3055 but that doesn't really help either.

Reply 3

Original post by Mr M

I've gotten down to this:

x+y+\frac{6}{xy}=6

I'm not sure how to get any further. If you do figure it out, could I ask you to post your solution up on here as I think I'd be interested to see it? Would have been so much easier if x and y were just integers...

Reply 4

Original post by Farhan.Hanif93

I've gotten down to this:

x+y+\frac{6}{xy}=6

Yes, how inconvenient that the number of rationals is not finite (and small)!

Reply 5

Original post by Mr M

Yes, how inconvenient that the number of rationals is not finite (and small)!

Indeed, I think I'll have to leave this for someone with a few more brain cells than me to round off.

Reply 6

jasbirsingh

Original post by Farhan.Hanif93

Have you gotten to the stage where you need to solve

x^2y+xy^2-6xy+6=0

That simplifies to

x^2 + (y-6)x + \frac{6}{y} = 0

After that wouldn't you just put some numbers to produce a pair of solutions i.e putting y = 2 gives x = 1.

which is a valid pair of solution.

Reply 7

Original post by jasbirsingh

That simplifies to

x^2 + (y-6)x + \frac{6}{y} = 0

After that wouldn't you just put some numbers to produce a pair of solutions i.e putting y = 2 gives x = 1.

which is a valid pair of solution.

Yes but how would you find ALL of the valid rational solutions? That's not quite so easy. I found all the integer solutions instantly. Finding all of the rational solutions is far more difficult than that.
It appears to be easier to work with

x+y+\frac{6}{xy}=6

than what you have up there.

Reply 8

jasbirsingh

Original post by Farhan.Hanif93

x+y+\frac{6}{xy}=6

than what you have up there.

Oops sorry I didn't read the question properly.

Reply 9

Original post by Farhan.Hanif93

x+y+\frac{6}{xy}=6

than what you have up there.

Actually, I haven't been asked to find ALL of them. One more will do but I can't even manage that.

:sad:

Reply 10

DFranklin

By computer search:
When y = 8, you get x^2 + 2x + 3/4 = 0, so x = -1/2 or x=-3/2 both work.

Also y = -3/2, y = 25/21, y= 49/15, y = 54/35, -361/68, 867/76. (Haven't verified whether any of these work, although y = -3/2 is a symmetry pair to the y=8 case so it will).

Reply 11

Original post by DFranklin

I believe that Mr M asked for the positive rational solutions for x and y.
EDIT:

(x,y)=(\frac{49}{15},\frac{25}{21})

is a suggestion of yours that works.

(edited 13 years ago)

Reply 12

DFranklin

Guessing somewhat about which solutions will 'pair', I suspect you'll find y=49/15, x = 54/35 will be such a solution.

(Haven't time to check right now).

Reply 13