Hi everyone !

Solve in N² this diophantine equation :

$\displaystyle y^2=x^3-5x$

Good luck !

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- Jan 8th 2011, 01:34 PMTaraskDiophantine equation .
Hi everyone !

Solve in N² this diophantine equation :

$\displaystyle y^2=x^3-5x$

Good luck ! - Jan 8th 2011, 02:01 PMAlso sprach Zarathustra
- Jan 9th 2011, 02:08 AMTarask
Hi !

I actually do have a solution using graphs , equation of tangents , but it is too long and of course not beautiful at all !

I wanted a more "algebraic" solution , and i started this way :

if x=0 then y=0

suppose that x and y are different from 0

If x<0

$\displaystyle y^2=x^3-5x$ http://latex.codecogs.com/gif.latex?...20y^2=x(x^2-5) since y² is positif and x is negatif x²-5 must be negatif and we get http://latex.codecogs.com/gif.latex?...%20\right%20\}

if x=-2 we get y²=2 which is impossible , and if x=-1 we get y=2 or y=-2

so (-1,2) and (-1,-2) are solutions.

We now have to treat the case of x>0 , and here is where i got stuck ! - Jan 9th 2011, 02:53 AMtonio
- Jan 9th 2011, 02:55 AMAlso sprach Zarathustra
- Jan 9th 2011, 03:09 AMTarask
Hi again !

thanks for paying attention to my exercice !

Here is what Wolfram gave me http://www4b.wolframalpha.com/Calcul...=3&w=200&h=206

it also gave me that the only integer solutions are the ones i found , really weird !!!! I think we should use the gcd ! - Jan 9th 2011, 03:12 AMPaulRS
For $\displaystyle x,y > 0$ :

We have that $\displaystyle x\cdot (x^2 - 5)$ is a square, now of course $\displaystyle \text{gcd}(x,x^2-5)$ divides $\displaystyle x\cdot x + (-1)\cdot (x^2 - 5) = 5$ thus $\displaystyle \text{gcd}(x,x^2-5)$ is either 1 or 5 (in case that $\displaystyle 5|x$ )

So we separate in the 2 cases:

**Case 1:**$\displaystyle \text{gcd}(x,x^2-5) = 1$.

Here it immediately follows that both $\displaystyle x$ and $\displaystyle x^2-5$ should be squares for their product to be a square.

However, note that $\displaystyle x^2-5$ if $\displaystyle x$ is big enough (since the "distance" between 2 consecutive squares increases) it can't be an square since $\displaystyle x^2$ is already an square. How big is big enough? note that for $\displaystyle x>3$ it will be impossible, and in fact you can see we get no solutions for this case.

**Case 2:**$\displaystyle \text{gcd}(x,x^2-5) = 5$.

Here $\displaystyle x = 5\cdot k$ thus $\displaystyle y^2 = 5^2\cdot k \cdot (5\cdot k^2 - 1) $

Thus $\displaystyle \left(\displaystyle\frac{y}{5}\right)^2 = k \cdot (5\cdot k^2 - 1)$ but now of course $\displaystyle k$ and $\displaystyle 5\cdot k^2 - 1$ are coprime and so they both must be squares.

If $\displaystyle k=1$ we have a solution ( $\displaystyle x = 5$ ; $\displaystyle y=10$ ).

$\displaystyle k = a^2$ then we want $\displaystyle 5\cdot a^4 -1 = b^2$

mmmm... stuck. - Jan 9th 2011, 03:30 AMAlso sprach Zarathustra
An Idea...

We need to prove that there are only solutions of the form: (a,2a) {and (a,-2a)}.

...and if we prove that we can make a conclusion that the only solution are: (0,0),(-1,2),(-1,-2),(5,10),(5,-10).

{by solving: a^3-4a^2-5a=0} - Jan 9th 2011, 03:34 AMTarask
- Jan 9th 2011, 03:44 AMAlso sprach Zarathustra
By showing that $\displaystyle \sqrt{2n(4n^2-5)}$ is not an integer we will conclude that $\displaystyle a$ is must to be odd.

e.i:

...hence(after proving the above) $\displaystyle a$ is odd, say $\displaystyle a=2n+1$ it easy to show that $\displaystyle y$ as integer must be even:

$\displaystyle y=\sqrt{(2n+1)^3-5(2n+1)}=\sqrt{4(2n+1)(n^2+n-1)}=2\sqrt{(2n+1)(n^2+n-1)}$ - Jan 9th 2011, 03:53 AMTarask
- Jan 9th 2011, 04:09 AMAlso sprach Zarathustra
- Jan 9th 2011, 04:18 AMTarask
- Jan 9th 2011, 04:26 AMAlso sprach Zarathustra
- Jan 9th 2011, 04:34 AMTarask
Your french is good apparently , but for me , i don't know a single word of Hebrew :p

Here is the file http://www.animath.fr/IMG/pdf/cours-arith1.pdf see page 68-.. (i hope that the mods and admins don't mind )