On the smallest open Diophantine equations: beyond Hilbert's 10 problem

Question

In 2018, Zidane asked What is the smallest unsolved Diophantine equation? The suggested way to measure size of the equation is substitute 2 instead of all variables, absolute values instead of all coefficients, and evaluate.

In the Mathoverlow question Can you solve the listed smallest open Diophantine equations? I list the current smallest equations for which it is open whether there exists any integer solution at all (Hilbert's 10 problem).

However, there are some famous equations, like $x^3+y^3+z^3=3$ of size $H=2^3+2^3+2^3+3=27$, for which the Hilbert's 10 problem is trivial (in this example, $x=y=z=1$ is a solution), but the equation can hardly be classified as solved, because we do not even know whether the solution set is finite.

Here, I consider more general problem: for a given polynomial Diophantine equation, determine whether the solution set is finite, and if so, list all the solutions. This is a much better approximation of our intuition what does it mean to solve an equation, but still avoids a subtle issue what counts as an acceptable description of the solution set if it is infinite (see What does it mean to solve an equation? for some discussion of this).

Selected solved equations.

• The smallest equation that required a new idea turned out to be $y^2+z^2=x^3-2$ of size $H=18$, see Representing $x^3-2$ as a sum of two squares for the proof that it has infinitely many integer solutions.

• Equations $ y(z^2-y)=x^3+2 $ and and $ xyz=x^3+y^2-2 $ of size $H=22$ has been listed as open and then solved by Tomita, see the answer below.

Smallest open equations.

The current smallest open equations are the equations $$ y^2-yz^2+x^3-2=0 $$ and $$ xyz=x^3+y^2+2 $$ of size $H=22$. These are the only remaining open equations with $H \leq 22$.

One may also study equations of special types. For example, the current smallest open symmetric equation (that is, invariant under cyclic shift of the variables) is $$ x^2y+y^2z+z^2x=1 $$ of size $H=25$. The current smallest open equations in two variables are $$ y^3+y=x^4+x $$ and $$ y^3-y=x^4-x $$ of size $H=28$, the current smallest open 3-monomial equation is $$ x^3y^2=z^3+2 $$ of size $H=42$, while the current smallest open homogeneous equation is $$ x^4+x^3 y-y^4+y^3 z+z^4=0 $$ of size $H=80$, see Existence of rational points on some genus 3 curves.

For the listed equations, the Hilbert 10th problem is trivial, because there are some obvious small solutions. The question, for each of the listed equations, is whether the solution set is finite or infinite, and if finite, list the solutions.

The plan is to list new smallest open equations once these ones are solved. The solved equations will be moved to the "solved" section.

Answer 1

This is a partial solution. The equation

$$y(z^2-y)=x^3+2\tag{1}$$

has infinitely many integer solutions.

Since $y = 1/2z^2 \pm 1/2\sqrt{z^4-4x^3-8}$, the expression $z^4-4x^3-8$ must be a perfect square. On the other hand, substitute $x=-3n^2-2n-2$ and $z=3n+1$ to $z^4-4x^3-8$, then we get
$$z^4-4x^3-8 = (25+8n+12n^2)(1+2n+3n^2)^2,$$ where $n$ is arbitrary integer. Hence we must find integer solutions of $$v^2 = 25+8n+12n^2\tag{2}.$$ We know equation $(2)$ has infinitely many integer solutions (Gauss's theorem, Mordell's book p.57). Recursive solutions are given as follows.
\begin{align*}(v_0,n_0)&=(\pm 5,0),\\ (v_{k+1},n_{k+1}) &= (7v_k + 24n_k + 8,2v_n + 7n_k + 2).\end{align*}

             k      x     y     z
            [12],[-458, 10510, 37]  
            [12],[-458, -9141, 37]
            [172],[-89098, 26729099, 517] 
            [172],[-89098, -26461810, 517] 
            [2400],[-17284802, 71887487358, 7201]  
            [2400],[-17284802, -71835632957, 7201]
            [33432],[-3353162738, 194174947774195, 100297]  
            [33432],[-3353162738, -194164888285986, 100297]
            [465652],[-650496286618, 524648022526642094, 1396957]  
            [465652],[-650496286618, -524646071037782245, 1396957]
            [-8],[-178, 2654, 23]  
            [-8],[-178, -2125, 23]
            [-108],[-34778, 6538075, 323]  
            [-108],[-34778, -6433746, 323]
            [-1500],[-6747002, 17535455598, 4499]  
            [-1500],[-6747002, -17515214597, 4499]
            [-20888],[-1308883858, 47355417946259, 62663]  
            [-20888],[-1308883858, -47351491294690, 62663]
            [-290928],[-253916721698, 127949397813283390, 872783]  
            [-290928],[-253916721698, -127948636063118301, 872783]

Answer 2

Equation $$ xyz=x^3+y^2+2 \quad\quad (1) $$ has infinitely many integer solutions. This is a special case of the following theorem, that has been stated in 1952 by Mordell in [1] and has been proved in 2015 by Schinzel [2, Corollary 2].

Theorem: For any non-zero integers $a,b,c$ and positive integers $m,n$ satisfying $(m-1)(n-1)>1$, the congruence $$ ax^m + by^n + c \equiv 0 \,(\text{mod}\, xy); $$ has infinitely many solutions in integers $(x,y)$ such that $\text{gcd}(y,c)=1$.

Equation (1) corresponds to the case $(a,b,c,m,n)=(1,1,2,3,2)$ of this theorem. Let us illustrate the idea of the proof by applying it to equation (1). Modulo 4 analysis shows that $x,y$ are odd and therefore coprime. Let us denote $x_1=x$ and $x_2=y$. Then (1) implies that $$ x_1^3+2 \equiv 0\,(\text{mod}\,\,x_2) \quad \text{and} \quad x_2^2+2 \equiv 0\,(\text{mod}\,\,x_1). $$ Then $x_3=\frac{x_2^2+2}{x_1}$ is an odd integer coprime to $x_2$. Then, modulo $x_2$, $$ 0 \equiv x_3^3(x_1^3+2) = (x_2^2+2)^3 + 2x_3^3 \equiv 2(4+x_3^3). $$ Because $x_2$ is odd, this is equivalent to $$ x_3^3+4 \equiv 0\,(\text{mod}\,\,x_2) \quad \text{and} \quad x_2^2+2 \equiv 0\,(\text{mod}\,\,x_3). $$ Then $x_4=\frac{x_3^3+4}{x_2}$ is an odd integer coprime to $x_3$. Then, modulo $x_3$, $$ 0 \equiv x_4^2(x_2^2+2) = (x_3^3+4)^2 + 2x_4^2 \equiv 2(8+x_4^2). $$ Because $x_3$ is odd, this is equivalent to $$ x_3^3+4 \equiv 0\,(\text{mod}\,\,x_4) \quad \text{and} \quad x_4^2+8 \equiv 0\,(\text{mod}\,\,x_3). $$ Now select any positive integer $n$. By iterating this process $n$ times, we obtain an equivalent system of congruences $$ x_{2n-1}^3+A_n \equiv 0\,(\text{mod}\,\,x_{2n}) \quad \text{and} \quad x_{2n}^2+B_n \equiv 0\,(\text{mod}\,\,x_{2n-1}). $$ for some integers $A_n$ and $B_n$. Choose any solution to this system in odd integers, say $x_{2n-1}=x_{2n}=1$, and then, working backwards, compute $x_{2n-2}, \dots, x_2, x_1$. Then $(x,y)=(x_1,x_2)$ is a solution to (1). Because $n$ was an arbitrary integer, this implies that (1) has infinitely many integer solutions.

[1] Mordell, L. J., The congruence (a x^ 3 + b y^ 3 + c \equiv 0 \pmod {xy}), and integer solutions of cubic equations in three variables, Acta Math. 88, 77-83 (1952). ZBL0047.04104.

[2] Schinzel, A., On the congruence (f(x)+g(y)+c\equiv 0\pmod{xy}) (completion of Mordell’s proof), Acta Arith. 167, No. 4, 347-374 (2015). ZBL1371.11006.