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Re: Demostration

On 14 Oct 2006, at 16:07, Miguel wrote:

> How canI to demostrate than the equation y^2=x^3+9 has 10 integer
> solutions?

It depends what you mean. It is easy enough to find 10 solutions. In  
fact, Mathematica's FindInstance can manage 8:

Rest[NestList[First[{x, y} /. FindInstance[y^2 == x^3 + 9 && x > First 
        {x, y}, Integers]] & , {-3, 0}, 4]]

{{-2, -1}, {0, 3}, {3, -6}, {6, 15}}

That's only 4, but of course for each solution {x,y}, the pair {x,-y}  
is also a solution, hence we get 8 solutions:

Sort[Join[l, l /. {x_Integer, y_Integer} -> {x, -y}]]

{{-2, -1}, {-2, 1}, {0, -3}, {0, 3}, {3, -6}, {3, 6}, {6, -15}, {6, 15}}

FindInstance can't find any more solutions:

FindInstance[y^2 == x^3 + 9 && x > 6, {x, y}, Integers]
FindInstance::"nsmet":"The methods available to FindInstance are  
insufficient to find the requested instances or prove they do not exist.
FindInstance[y^2 == x^3 + 9 && x > 6, {x, y}, Integers]

so we have to resort to brute force search:

Catch[Do[If[Sqrt[n^3 + 9.] == Floor[Sqrt[n^3 + 9.]], Throw[n],  
Continue[]], {n, 7, 1000}]]

Hence {40,253} and {40,-253} give us two more solutions.

I do not think it is possible to use Mathematica to prove that these  
are all the solutions. The reason is that while there is a general  
theorem whcih states that the Diophantine equation y^2==f[x] has at  
most a finite number of solutions if f[x] is a polynomial of degree  
 >=3, with integer coefficients and with distinct zeros, no method is  
known for determining the solutions or the number of solutions except  
in special cases. Since you have stated that in this case there are  
precisely 10 solutions, I assume this must be one of them, and there  
is some way to prove it which is not known to me (this is not my area  
and I do not follow recent development in it). But in any case, even  
if a way to prove this is known in this case, no such general  
algorithm exists and therefore it can't be known to Mathematica.

Andrzej Kozlowski
Tokyo, Japan

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