Hi everyone !

Solve in N² this diophantine equation :

Good luck !

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- Jan 8th 2011, 02:34 PMTaraskDiophantine equation .
Hi everyone !

Solve in N² this diophantine equation :

Good luck ! - Jan 8th 2011, 03:01 PMAlso sprach Zarathustra
- Jan 9th 2011, 03:08 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

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, 03:53 AMtonio
- Jan 9th 2011, 03:55 AMAlso sprach Zarathustra
- Jan 9th 2011, 04:09 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, 04:12 AMPaulRS
For :

We have that is a square, now of course divides thus is either 1 or 5 (in case that )

So we separate in the 2 cases:

**Case 1:**.

Here it immediately follows that both and should be squares for their product to be a square.

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

**Case 2:**.

Here thus

Thus but now of course and are coprime and so they both must be squares.

If we have a solution ( ; ).

then we want

mmmm... stuck. - Jan 9th 2011, 04:30 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, 04:34 AMTarask
- Jan 9th 2011, 04:44 AMAlso sprach Zarathustra
By showing that is not an integer we will conclude that is must to be odd.

e.i:

...hence(after proving the above) is odd, say it easy to show that as integer must be even:

- Jan 9th 2011, 04:53 AMTarask
- Jan 9th 2011, 05:09 AMAlso sprach Zarathustra
- Jan 9th 2011, 05:18 AMTarask
- Jan 9th 2011, 05:26 AMAlso sprach Zarathustra
- Jan 9th 2011, 05:34 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 )