$\displaystyle a^2 + 2 = b^3 $

a=5 and b=3

Prove that a=5 and b=3 are the only solutions.

Any ideas?

Printable View

- Nov 9th 2006, 10:22 AMRuichanProve this is the only solution
$\displaystyle a^2 + 2 = b^3 $

a=5 and b=3

Prove that a=5 and b=3 are the only solutions.

Any ideas? - Nov 9th 2006, 11:09 AMCaptainBlack
- Nov 9th 2006, 12:53 PMRuichan
- Nov 9th 2006, 01:42 PMCaptainBlack
- Nov 9th 2006, 03:23 PMRuichan
Actuarial Math but i don't think it's related.

Other stuffs he sort of provided are:

$\displaystyle (a^2 -25) = b^3 - 27$

$\displaystyle

(a + 5)(a - 5) = (b-3)(b^2+3b+9)

$

a (subscript k) +5 = $\displaystyle k^2 +10k $

1.(0,37) = 37

2.(37,98) = 61

3. (98,189) = 91

4. (189,316) = 127

2.-1. = 24

3.-2. = 30

4.-3. = 36

He did say that he proved it long time ago, nobody else, at least those he knows could prove it.

I actually have no idea what's he talking about though. - Nov 9th 2006, 06:10 PMThePerfectHacker
In the 17th Century my Favorite mathemation Pierre de Fermat wrote that 26 is the only number contained in between a square and a cube.

That leads to the equation,

$\displaystyle x^2+2=y^3$.

Fermat never wrote down the proof (like usual) and many people failed to prove it (including me :mad: ) but about 100 years later Leonard Euler proved it. Though (I think) it he spend 7 years on this problem! (Just goes to show the Fermat was the master of Diophantine Equations). I myself have no idea how to show that.

I happen to know that $\displaystyle x^3+1=y^2$ also has no solutions accept $\displaystyle x=2,y=3$ which is (special case of Catalan Conjecure--- now proven). But was shown to be true also by Euler I think and again all texts on number theory omit the prove because it is way too long (though clever). However, I think there is a simpler proof due to Gauss based on the Quadradic fields and complex factorization of the cube which leads to an impossibility. But I do not think Gauss' method works here because the expression $\displaystyle x^3+2$ is not factorable while $\displaystyle x^3+1=(x+1)(x^2-x+1)$ which is the entire argument. Finally there is certainly a way to show this using the theory of Elliptic curve because this leads to a the square root of a cubic polynomial, but sadly I am an noob :( so I cannot know of one.

By the way, I really do not think your professor proved it using Elementary number theory. I can promise you that this problem is a problem only for the gods. I think he just lied. - Nov 9th 2006, 09:00 PMtopsquark
Crap. I was really hoping TPH would come through on this one (as he likes these) and I've now spent several hours working on it. All I can say is that the solution must be of the form:

a = 210x + 5, b = 210y + 3

or

a = 210x + 5, b = 210y + 33

(x and y integers, not necessarily positive)

Obviously the first pair contains the (presumably only) solution and I have yet to find any solutions using the second. Unfortunately the method I am using will never generate only one solution, so it's a dead end (though an enjoyable piece of work.)

-Dan - Nov 9th 2006, 09:04 PMQuick
- Nov 9th 2006, 09:06 PMRuichan
I don't think he's lying though. His solution was published in reader's digest or whatever magazine. I wasn't paying attention to what he was saying coz I was practically sleeping with my eyes open by 8.30pm last night.

Guess I just need to ask him next week on which magazine that solution was published in. However, it was published almost 2 decades ago though.

He didn't proved the theory, just proved that a = 5 and b = 3 are the only solutions. - Nov 10th 2006, 05:17 AMtopsquark
- Nov 10th 2006, 05:20 AMQuick
- Nov 10th 2006, 05:22 AMtopsquark
I don't know, maybe someone will gain some inspiration from what I was trying and succeed where I failed. I was trying an inductive type proof that a and b could only be expressed as:

$\displaystyle a = \left (2 \prod_{i = 1}^n(2i+1) \right )m + 5$

$\displaystyle b = \left (2 \prod_{i = 1}^n(2i+1) \right )q + 3$

where m and q are integers.

If this is true and you can do an induction, then we can show that n is without bound. If this is true then the only solution is for m, q = 0. That is, a = 5, b = 3. But I couldn't manage to prove the inductive step. (I kept on finding other forms that I couldn't prove were incorrect.)

-Dan - Nov 10th 2006, 06:18 AMThePerfectHacker
I would love to see it.

---

Topsquark just drop it. You are not going to solve it. I spend hours on these diophantine equations including this one. Trust me when I say that they are not easy, they might seem easy. Especially when you introduce cubes. I have seen proofs of several diophantine equations for example Fermat's last theorem for n=3, by Euler, it is huge, just pages and pages and pages all using manipulations divisibility arguments. They also involve beyond belive algebraic manipulations and the most unusual substitution. Which is why Euler solved most of them, becuase he was one of the few that was able to do this.

Solving diophantine equations in this type is called an "elementary" solution. My problem of the week used such an argument. There are also other non-elementary methods, one became famous in 19th century using field theory. And today there is a another one using elliptic curves. I think your professor solve it using the two latter methods, not the classical, but it would be nice if you can post it here. I would like to see how it is approach using modern math. - Nov 16th 2006, 01:04 AMRuichan
Mathematics or Mathematical Monthly. It's from Mathematics Association of America, Journal form. That issue was published between 1980-1981.

Prof doesn't remember how to prove it, HEHEHE, he's looking for the hard copy of his proof. He forgot where he place his hardcopy.

That's why he's asking if we know how to prove it.

He said he only remembered that he used the modular formula or whatever.

I'm too lazy to go search for through the last 20 years of that issue which his proof was accredited. - Nov 16th 2006, 06:22 AMThePerfectHacker