3
$\begingroup$

Do there exist integers $x,y,z$ such that $$ x^2+y^2-z^2 = xyz -2 \quad ? $$

Why this is interesting? First, this equation arose in an answer to the previous Mathoverflow question What is the smallest unsolved diophantine equation? but was not asked explicitly as a separate question. The context is that, in a well-defined sense for the notion of "smallness", the equation above is the "smallest" open Diophantine equation.

Second, this equation is one of the simplest non-trivial representative of the family of equations $ax^2+by^2+cz^2=dxyz+e$, which generalises a well-known Markoff equation $x^2+y^2+z^2=3xyz$. The well-known methods for the former (Vieta jumping) has been extended to the general case if $a,b,c$ are all natural numbers and are divisors of $d$ (see, for example, Fine, Benjamin, et al. "On the Generalized Hurwitz Equation and the Baragar–Umeda Equation." Results in Mathematics 69.1-2 (2016): 69-92). The question seems to be much more challenging when $a,b,c$ have different signs. The simplest case with different signs is $a=b=d=1$ and $c=-1$, which leads to the family of equations $x^2+y^2-z^2=xyz+e$. The equation above is the first non-trivial example from this family.

$\endgroup$
0
9
$\begingroup$

There is no solution.

Fix a solution $(x,y,z)$ with $|x|+|y|+|z|$ minimal. We will show a contradiction.

We can't have $xyz=0$ as we would then obtain one of the unsolvable equations $x^2+y^2= -2$, $x^2-z^2=-2$, $y^2-z^2=-2$.

If $xyz>0$, then by swapping the signs of two of $x,y,z$ if necessary we can assume $x,y,z>0$, and switching $x$ and $y$ we can assume $x \geq y$. We have the Vieta jump $x \to yz-x$, so if this is minimal we have $x \leq yz/2$. Since $f(x)=x^2+y^2-z^2 - xyz + 2 $ is convex and vanishes at $x$, we must $f(y) \geq 0$ or $f(yz/2) \geq 0$.

But $$f(y)= (2-z)y^2 -z^2 + 2$$ so $f(y) \leq 0$ imply $z<2$ and $z=1$ gives the impossible $x^2+y^2-xy=-1$ and $$f(yz/2) = y^2 -z^2 - y^2 z^2/4 +1= (1-z^2/4)y^2 - z^2 +1$$ which again is nonnegative only if $z<2$ which is impossible.

If$xyz<0$, then by swapping the signs if necessary we can assume $x,y,z<0$. We have the Vieta jump $z \to -xy-z$, so if this is minimal we have $z \geq -xy/2$. We have $g(z)=z^2 + xyz - x^2-y^2-2$ is convex and vanishes at $z$, we must have $g(0) \geq 0$ or $g(-xy/2) \geq 0$.

But $g(0) = -x^2 - y^2 - 2 <0$ and $g(-xy/2) = - x^2 y^2/4 - x^2 - y^2 -2 <0$.

So neither case is possible.

$\endgroup$
1
  • 3
    $\begingroup$ This is matter of taste, but I prefer testing Vieta jumps by product (rather than sum) Vieta theorem. Here we see that if $|x|=\max(|x|,|y|,|z|)$, then the second root $\tilde{x}$ w.r.t. to $x$ equals $\tilde{x}=(y^2+2-z^2)/x$ and $|\tilde{x}|<|x|$, we may perform a jump. If $|z|=\max(|x|,|y|,|z|)$, by changing signs assume that $x,y>0$, the second root w.r.t. $z$ satisfies $\tilde{z}z=-x^2-y^2-2<0$. If $z<0$, then $\tilde{z}=-xy-z<-z$ and so $|\tilde{z}|=\tilde{z}<|z|$, do a jump. Finally if $z>\max(x,y)$, then $z^2+xyz\ge (x+1)^2+y^2>x^2+y^2+2$. $\endgroup$ – Fedor Petrov May 17 at 13:39

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.