# Solution needed 4 a very good problem!

1. Sep 29, 2005

### rahul963

Solve x^3 - y^2 = 2006 in integers.

2. Sep 29, 2005

### Zurtex

What have you tried so far? What sort of mathematics have you learnt to solve this problem?

3. Sep 29, 2005

### rahul963

number theory problem? do u knowsolution r not?

4. Sep 29, 2005

### rahul963

zurtex? do u hve ny idea?

5. Sep 29, 2005

### Zurtex

Well yes I do know a solution and I would have a guess it's the only solution, but I only know this solution by running the problem through mathematica, I was wondering what sort of number theory you know so I might be able to help you rather than just give you an answer.

6. Sep 29, 2005

### Trepidation

$$x^3 - y^2 = 2006$$

$$y^2 = x^3 - 2006$$

$$y = \sqrt{x^3 - 2006}$$

So then, using substitution...

$$x^3 - \sqrt{x^3 - 2006} = 2006$$

I'm sure this isn't nearly sophisticated enough, and I've probably gotten into a rut of some sort (I know I don't know how to continue ><)... It's what I tried, though-- and now I'm curious as to how you would continue, or if I'm completely wrong... If this is digressing from the topic, I apologize-- just ignore me if that is the case...

Last edited: Sep 29, 2005
7. Sep 29, 2005

### Icebreaker

You can't do that. You have a single equation with two variables which you substituted back into your equation. It will give you something in along the lines of 0 = 0 or x = x. And it does:

$$x^3 - x^3 + 2006 = 2006$$

8. Sep 29, 2005

### Jeff Ford

I don't know the theory behind it, but it only took a few minutes of guess and check to solve.

9. Sep 29, 2005

### David

Is this a math olympiad problem by any chance?