0
$\begingroup$

In order to obtain infinite integer non trivial solutions of the equation $X^4-DY^4=Z^4$ (all numbers natural) we do the following. We set $X=(r_1●p+p)$, $Y=p$, $Z=(r_1●p)$, $D=(4r_1^3+6r_1^2+4r_1+1)$, where $r_1$ is one rational solution of a particular cubic equation.

Let’s have the equation $(f●a+a)^4-(4f^3+6f^2+4f+1)●a^4=(f●a)^4$.

In the quantity $4f^3+6f^2+4f+1$ we set $f=k-1/2$ and we obtain the cubic $4k^3+k$. We equate this cubic with any integer $±m$ and we obtain $4k^3+k=±m$. Let’s say $r_1=b/p$ is one rational solution of this cubic equation. We set the value $b/p$ at $X, Y, Z, D$ and we obtain non trivial solutions of the above equation.

Let’s have $4k^3+k=15$ so $r_1=3/2$ and $5^4-34●2^4=3^4$. It is easily shown that the cubic $4k^3+k=±m$ has infinite rational solutions.

I tried to find a technique to obtain infinite solutions of the equation $X^5-DY^5=Z^5$ by applying the method of binomial separation but I could not find one. Does anyone know if such a technique exists?

$\endgroup$
2
$\begingroup$

If you fix $D\ne0$, the equation $X^4-DY^4=Z^4$ has only finitely many solutions $(X,Y,Z)\in\mathbb{Z}^3$ satisfying $\gcd(X,Y,Z)=1$ by Faltings's theorem, since the curve has genus 3. So it would help if you specified that you're treating $X,Y,Z,D$ as four variables. (The use of $D$ suggests that it is fixed.) Also, at least on my computer, there's an operation that's appearing as a filled in circle, so I have no idea what that means. (Maybe's it's multiplication?)

Anyway, here's an idea (although not a solution): Let's consider the general equation $X^n-WY^n=Z^n$ and look for solutions in integers $(X,Y,Z,W)$ with (say) $\gcd(X,Y,Z)=1$. My suggestion would be to embed this into $\mathbb{P}^2\times\mathbb{P}^2$ by homogenizing. Assuming it's nonsingular (you can check), it's easy enough to compute its canonical bundle. Then you can probably use Vojta's conjecture to make a good guess for the value of $N$ such that for all $n\ge N$, the solutions to the equation are not Zariski dense. (This actually might be interesting to work out, since you're allowing rational solutions in the first $\mathbb{P}^2$, but taking integer solutions for the second $\mathbb{P}^2$.) Anyway, that should give at least an idea whether there are likely to be a lot of solutions when $n=5$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The filled circle does mean multiplication. Thanks for your suggestions. I was hoping for a technique that yields infinite solutions, not just a lot. $\endgroup$ – Vassilis Parassidis Jan 26 '14 at 1:23
  • $\begingroup$ But the variety $X^n - W Y^n = Z^n$ is clearly rational because the equation can be solved for $W$. We can even make $\gcd(X,Y,Z) = 1$ by fixing $Y \neq 0$ and (for example) taking $X$ coprime to $Y$ and $Z = X + Y^n$. $\endgroup$ – Noam D. Elkies Jan 26 '14 at 3:18
  • $\begingroup$ If we set $k=a/2$ where $a$ an integer, then in the equation $4k^3+k=±m$ $m$ is always an integer and so the equation $X^4 –DY^4=Z^4$ is solvable. $\endgroup$ – Vassilis Parassidis Jan 26 '14 at 3:52
  • $\begingroup$ @NoamD.Elkies Good point, the 3-fold is rational. So certainly the solutions in $\mathbb{Q}^4$ are dense. And by modifying your example to, say, $Z=X+cY^n$ with $c\in\mathbb{Z}$, you'll get a Zariski dense set of solutions $(W,[X,Y,Z])$ that are Zariski dense in $\mathbb{Z}\times\mathbb{P}^2(\mathbb{Q})$. $\endgroup$ – Joe Silverman Jan 26 '14 at 12:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.