Math Help - x^3 = y^2 + 1

1. 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:

$x^3 = y^2 + 1$ with $x,y \in \mathbb{N}$

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?

2. Originally Posted by Myrc

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?
Hello

Something is wrong with your solution or question

3. $3, 5$ doesn't work, as $3^3=27$ and $5^2+1=26$.

If this equation will ever be satisfied by $x,y \in \mathbb{N}$, one of them must be even and the other must be odd.

That's all I can figure out, although you may want to subtract the 1 from both sides and factor $x^3-1$ into $(x-1)(x^2+x+1)$ to see where that gets you.

4. Catalan's conjecture states that $8=2^3$ and $9=3^2$ 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 $x^3=y^2+1$ has no solutions in natural numbers. Of course, there may be a much simpler way to prove this special case of the Catalan problem.

5. Originally Posted by Myrc
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:

$x^3 = y^2 + 1$ with $x,y \in \mathbb{N}$

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 want to add something to what Opalg said. There are easier ways to show that $x^3 = y^2+1$ has no other solutions however those approaches require an understanding of higher arithmetic. Thus, there is no quick and easy way to answer this question.

6. x^3 = y^2 + 2

Here's the correct question - sorry:

$x^3 = y^2 + {\color{red}2}$ with $x,y \in \mathbb{N}$.

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...

7. Originally Posted by Myrc
Here's the correct question - sorry:

$x^3 = y^2 + 2$ with $x,y \in \mathbb{N}$.

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...
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 $\mathbb{Z}[\sqrt{-2}]$ then you can prove this result more easily. The steps are here.

8. Awesome, thanks for that, finally I can rest in peace.
Although, I would love to see how Euler did it with elementary N.T., it would have to be a lot more complicated than the ring method though...

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 $x=1,y=0$ is solution of the equation $x^3 = y^2 + 1$

Regards

10. Originally Posted by chisigma
it is controversial if 0 is included or not in the ‘natural numbers’. If we accept that 0 is a 'natural number', [/FONT][/COLOR]then $x=1,y=0$ is solution of the equation $x^3 = y^2 + 1$
There is nothing controversial. Some people just define it different ways.