Hi everyone !
Solve in N˛ this diophantine equation :
$\displaystyle y^2=x^3-5x$
Good luck !
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
$\displaystyle y^2=x^3-5x$ since y˛ is positif and x is negatif x˛-5 must be negatif and we get
if x=-2 we get y˛=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 !
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.
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:
$\displaystyle y=\sqrt{(2n+1)^3-5(2n+1)}=\sqrt{4(2n+1)(n^2+n-1)}=2\sqrt{(2n+1)(n^2+n-1)}$
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 )