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.

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    $\begingroup$ What exactly is your question? Are you envisaging using MO to record ongoing progress on this open-ended project? If so, I don't think MO is the right venue for that. $\endgroup$ Dec 17, 2021 at 18:29
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    $\begingroup$ The question, for each of the listed equations, is whether the solution set is finite or infinite (and if finite, list the solutions). $\endgroup$ Dec 17, 2021 at 19:35
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    $\begingroup$ I have solved hundreds other equations with $H \leq 22$ but cannot solve these ones. The equations look nice and I hope mathoverflow users will enjoy trying to solve them. $\endgroup$ Dec 17, 2021 at 20:00
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    $\begingroup$ For the first two, thinking about random $x \approx N^{1/3}, y \approx N^{1/2}, z\approx N^{1/4}$ suggests there should be infinitely many solutions on $1/2 + 1/3 +1/4>1$. For the last two, $x \approx N^{1/3}, y \approx N^{1/2}, z\approx N^{1/6}$ suggests there may be infinitely many but it should be harder to show as $1/2 + 1/3+1/6=1$. $\endgroup$
    – Will Sawin
    Dec 18, 2021 at 14:36
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    $\begingroup$ One of the integer solution of the equation $xyz=x^3+y^2-2$ is $(x,y,z)=(-5629441, -14347779969589, 2548702)$ with $|x|,|z|<10^{7}$. $\endgroup$
    – Tomita
    Dec 19, 2021 at 6:11

1 Answer 1


This is a partial solution. The equation


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]
  • $\begingroup$ Thank you! I have now moved this equation to the "solved" section! One small question - how you have found the expressions of $x$ and $z$ in terms of $n$? Just computer search for quadratic expressions with small coefficients? $\endgroup$ Dec 20, 2021 at 12:13
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    $\begingroup$ Yes, since this equation has many small solutions, I thought it would be reduced to quadratic diophantine equation problem. $\endgroup$
    – Tomita
    Dec 20, 2021 at 13:11
  • $\begingroup$ Actually, the same method solves another equation, $xyz=x^3+y^2-2$. We need $D=x^2z^2-4x^3+8$ to be a perfect square. Select $x=6n^2+1$ and $z=6n$, then $D=4(6n^2-1)^2(3n^2+1)$. It is left to note that there are infinitely many $n$ such that $3n^2+1$ is a perfect square. I will now move this equation to the "solved" section as well! So, there are currently two equations left with $H\leq 22$. $\endgroup$ Dec 20, 2021 at 20:09
  • $\begingroup$ @BogdanGrechuk which? $\endgroup$
    – EGME
    Dec 20, 2021 at 21:43
  • $\begingroup$ $y(z^2-y)=x^3-2$ and $xyz=x^3+y^2+2$, as stated in the question. $\endgroup$ Dec 20, 2021 at 21:50

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