Results 1 to 10 of 10

- March 2nd 2009, 02:18 AM #1

- Joined
- Dec 2008
- Posts
- 9

## x^3 = y^2 + 1

Hey guys, I was wondering do any of you have any insight into the problem mentioned in the title. Here it is again for good measure:

with

I reckon there's only 1 sol'n, i.e. x = 3 and y = 5

However, despite all my attempts to prove this, nothing seems to work. Any thoughts?

- March 2nd 2009, 02:52 AM #2

- March 2nd 2009, 09:02 AM #3

- March 2nd 2009, 12:58 PM #4
Catalan's conjecture states that and are the only consecutive natural numbers that are both powers. This result was first proved by Mihăilescu in 2002 (more than 150 years after Catalan first stated it). It follows as a special case that the equation has no solutions in natural numbers. Of course, there may be a much simpler way to prove this special case of the Catalan problem.

- March 2nd 2009, 02:57 PM #5

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- March 8th 2009, 04:43 AM #6

- Joined
- Dec 2008
- Posts
- 9

## x^3 = y^2 + 2

Here's the correct question - sorry:

with .

I reckon there's only 1 sol'n, i.e. x = 3 and y = 5

However, despite all my attempts to prove this, nothing seems to work. Any thoughts? I've tried my usual bag o' tricks already...

- March 8th 2009, 02:45 PM #7

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

I think I read in a book a long time ago that Euler spend seven years proving this result with elementary methods. However, if you know higher arithmetic, in particular then you can prove this result more easily. The steps are here.

- March 9th 2009, 02:17 AM #8

- Joined
- Dec 2008
- Posts
- 9

- March 10th 2009, 05:05 AM #9
According to…

http://mathworld.wolfram.com/NaturalNumber.html

… it is controversial if 0 is included or not in the ‘natural numbers’. If we accept that 0 is a 'natural number', then is solution of the equation …

Regards

- March 10th 2009, 09:27 PM #10

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10