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?


1 Answer 1


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 $n=5$.

  • $\begingroup$ The filled circle does mean multiplication. Thanks for your suggestions. I was hoping for a technique that yields infinite solutions, not just a lot. $\endgroup$ Jan 26, 2014 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 $Z = X + Y^n$. $\endgroup$ Jan 26, 2014 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. $\endgroup$ Jan 26, 2014 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 $\mathbb{Z}\times\mathbb{P}^2(\mathbb{Q})$. $\endgroup$ Jan 26, 2014 at 12:15

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