Diophantine equation .

**Tarask** · January 8th 2011, 01:34 PM

Hi everyone !
Solve in N² this diophantine equation :
$y^2=x^3-5x$
Good luck !

**Also sprach Zarathustra** · January 8th 2011, 02:01 PM

Originally Posted by Tarask

Hi everyone !
Solve in N² this diophantine equation :
$y^2=x^3-5x$
Good luck !

Try to draw this function...

You will see that you need to check only values of x between -2 and 2.

**Tarask** · January 9th 2011, 02:08 AM

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
$y^2=x^3-5x$ $\Leftrightarrow y^2=x(x^2-5)$ since y² is positif and x is negatif x²-5 must be negatif and we get $x\in \left \{-2;-1 \right \}$
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 !

**tonio** · January 9th 2011, 02:53 AM

Originally Posted by Also sprach Zarathustra

Try to draw this function...

You will see that you need to check only values of x between -2 and 2.

I can't see why: $(5,10)\,,\,(5,-10)$ are solutions, say...

Tonio

**Also sprach Zarathustra** · January 9th 2011, 02:55 AM

Originally Posted by tonio

I can't see why: $(5,10)\,,\,(5,-10)$ are solutions, say...

Tonio

Me too...

**Tarask** · January 9th 2011, 03:09 AM

Hi again !
thanks for paying attention to my exercice !
Here is what Wolfram gave me

it also gave me that the only integer solutions are the ones i found , really weird !!!! I think we should use the gcd !

**PaulRS** · January 9th 2011, 03:12 AM

For $x,y > 0$ :

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

So we separate in the 2 cases:

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

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

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

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

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

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

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

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

mmmm... stuck.

**Also sprach Zarathustra** · January 9th 2011, 03:30 AM

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}

**Tarask** · January 9th 2011, 03:34 AM

Originally Posted by Also 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}

Using parity may be ?

**Also sprach Zarathustra** · January 9th 2011, 03:44 AM

By showing that $\sqrt{2n(4n^2-5)}$ is not an integer we will conclude that $a$ is must to be odd.

e.i:
...hence(after proving the above) $a$ is odd, say $a=2n+1$ it easy to show that $y$ as integer must be even:
$y=\sqrt{(2n+1)^3-5(2n+1)}=\sqrt{4(2n+1)(n^2+n-1)}=2\sqrt{(2n+1)(n^2+n-1)}$

**Tarask** · January 9th 2011, 03:53 AM

Originally Posted by Also sprach Zarathustra

By showing that $\sqrt{2n(4n^2-5)}$ is not an integer we will conclude that $a$ is must to be odd.

Well if we consider the functions f such that $f(x)=8x^3-10x$ and g such that $g(x)=x^2$ we find that their graphs never meet

if i'm not wrong of course ....

**Also sprach Zarathustra** · January 9th 2011, 04:09 AM

Originally Posted by Tarask

Well if we consider the functions f such that $f(x)=8x^3-10x$ and g such that $g(x)=x^2$ we find that their graphs never meet

if i'm not wrong of course ....

Interesting!

**Tarask** · January 9th 2011, 04:18 AM

Originally Posted by Also sprach Zarathustra

Interesting!

Thanks !
By the way , as i said , i have a solution using graphs , it is similar to what we did !
It's in french , if you are interested in it , i'll publish the file so that you can take a look at it

**Also sprach Zarathustra** · January 9th 2011, 04:26 AM

Originally Posted by Tarask

Thanks !
By the way , as i said , i have a solution using graphs , it is similar to what we did !
It's in french , if you are interested in it , i'll publish the file so that you can take a look at it

I will be glad to look at it.
My French is like your Hebrew...

Merci mon ami.

**Tarask** · January 9th 2011, 04:34 AM

Originally Posted by Also sprach Zarathustra

I will be glad to look at it.
My French is like your Hebrew...

Merci mon ami.

Your french is good apparently , but for me , i don't know a single word of Hebrew

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 )

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