You are Here: Home >< Maths

# Sum of arctangents Watch

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

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

Clues please because I'm not sure where I'm going with this one.
2. (Original post by Mr M)
I'm trying to find positive rational solutions for the following equation:

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 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.
3. (Original post by Farhan.Hanif93)
Have you gotten to the stage where you need to solve 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.
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.
4. (Original post by Mr M)
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.
I've gotten down to this:

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...
5. (Original post by Farhan.Hanif93)
I've gotten down to this:

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...
Yes, how inconvenient that the number of rationals is not finite (and small)!
6. (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.
7. (Original post by Farhan.Hanif93)
Have you gotten to the stage where you need to solve 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.
That simplifies to

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.
8. (Original post by jasbirsingh)
That simplifies to

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 than what you have up there.
9. (Original post by Farhan.Hanif93)
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 than what you have up there.
Oops sorry I didn't read the question properly.
10. (Original post by Farhan.Hanif93)
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 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.

11. 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).
12. (Original post by 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).
I believe that Mr M asked for the positive rational solutions for x and y.
EDIT: is a suggestion of yours that works.
13. 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).
14. (Original post by 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).
Thanks. That has moved me forward anyway.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 28, 2010
Today on TSR

### Oxford interview invitations

When can you expect yours?

### Official Cambridge interview invite list

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.