In 2018, Zidane asked What is the smallest unsolved Diophantine equation? The suggested way to measure size is substitute 2 instead of all variables, absolute values instead of all coefficients, and evaluate. For example, the size $H$ of the equation $y^2-x^3+3=0$ is $H=2^2+2^3+3=15$.

Below we will investigate only the solvability question: does a given equation has any integer solutions or not?

Selected trivial equations. The smallest equation is $0=0$ of size $H=0$. If we ignore equations with no variables, the smallest equation is $x=0$ of size $H=2$, while the smallest equations with no integer solutions are $x^2+1=0$ and $2x+1=0$ of size $H=5$. These equations have no real solutions and no solutions modulo $2$, respectively. The smallest equation which has real solutions and solutions modulo every integer but still no integer solutions is $y(x^2+2)=1$ of size $H=13$.

Well-known equations. The smallest not completely trivial equation is $y^2=x^3-3$ of size $H=15$. But this is an example of Mordell equation $y^2=x^3+k$ which has been solved for all small $k$, and there is a general algorithm which solves it for any $k$. Below we will ignore all equations which belong to a well-known family of effectively solvable equations.

Selected solved equations.

  • The smallest equation neither completely trivial nor well-known is $ y(x^2-y)=z^2+1$ of size $H=17$. As noted by Victor Ostrik, it has no solutions because all odd prime factors of $z^2+1$ are $1$ modulo $4$.

  • The smallest equation not solvable by this method is $ x^2 + y^2 - z^2 = xyz - 2 $ of size $H=22$. This has been solved by Will Sawin and Fedor Petrov On Markoff-type diophantine equation by Vieta jumping technique.

  • The smallest equation that required a new idea was $y(x^3-y)=z^2+2$ of size $H=26$. This one was solved by Will Sawin and Servaes by rewriting it as $(2y - x^3)^2 + (2z)^2 = (x^2-2)(x^4 + 2 x^2 + 4)$, see this comment for details.

  • Equation $ y^2-xyz+z^2=x^3-5 $ of size $H=29$ has been solved in the arxiv preprint Fruit Diophantine Equation (arXiv:2108.02640) after being popularized in this blog post.

  • Equation $ x(x^2+y^2+1)=z^3-z+1 $ of size $H=29$ has solution $x=4280795$, $y=4360815$, $z=5427173$, found by Andrew Booker. This is the smallest equation for which the smallest known solution has $\min(|x|,|y|,|z|)>10^6$.

  • Equation $ y^2-x^3y+z^3+3=0 $ of size $H=31$ has been the smallest open equation for over two years, and then has been solved by Denis Shatrov, see the answer.

  • Equation $ 3-y+x^2 y+y^2+x y z-2 z^2 = 0 $ of size $H=33$ has been the smallest open cubic equation for some time, and then has been solved by Denis Shatrov, see here.

  • Equation $ x^3 + y^3 + z^3 + xyz = 5 $ with $H=37$ has been listed here as the smallest open symmetric equation, but then I found solution $x=-3028982$, $y=-3786648$, $z=3480565$, see the answer for details how it was found.

Smallest open equations. The current smallest open equations are the equations $$ y^3+xy+x^4+4=0, $$ $$ y^3+xy+x^4+x+2=0, $$ $$ y^3+y=x^4+x+4 $$ and $$ y^3-y=x^4-2x-2 $$ of size $H=32$.

One may also study equations of special type. For example, the current smallest open equations in two variables are the ones listed above. The current smallest open cubic equation is $$ (x+1)y^2-xz^2=x^3+2x+2 $$ of size $H=34$, the current smallest open symmetric equation is $$ x^3+x+y^3+y+z^3+z = x y z + 1 $$ of size $H=39$, while the current smallest open 3-monomial equation is $$ x^3y^2 = z^3 + 6 $$ of size $H=46$.

The shortest open equations. I was told that it would be interesting to order equations by a more "natural" measure of size than $H$. Define the length of a polynomial $P$ consisting of $k$ monomials of degrees $d_1,\dots,d_k$ and integer coefficients $a_1,...,a_k$ as $l(P)=\sum_{i=1}^k\log_2|a_i|+\sum_{i=1}^k d_i$. This is an approximation for the number of symbols used to write down $P$ if we write the coefficients in binary, do not use the power symbol, and do not count the operations symbols. Note that $2^{l(P)}=\prod_{i=1}^k\left(a_i2^{d_i}\right)$ while $H(P)=\sum_{i=1}^k\left(a_i2^{d_i}\right)$. If we order equations by $l$ instead of $H$, then the current "shortest" open equations are $$ y^2-x^3y+z^4+1=0 $$ $$ 2 y^3 + x y + x^4 + 1 = 0 $$ and $$ x^3 y^2 = z^4+2 $$ of length $l=10$.

For each of the listed equations, the question is whether they have any integer solutions, or at least a finite algorithm that can decide this in principle.

The paper Diophantine equations: a systematic approach devoted to this project is available online: (arXiv:2108.08705). Paper last updated 13.04.2022.

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

Of course, in addition to the solvability question one may ask many other questions, e.g. whether the solution set is finite or infinite, how to describe all integer solutions, etc. Please see here for the current list of all smallest open equations in this project.

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    $\begingroup$ Concerning $x^3y^2=z^3+6$, once upon a time it was conjectured that no two powerful numbers differed by $6$. But then Narkiewicz found $7^325^2=463^2+6$. $\endgroup$ Jul 31, 2021 at 12:49
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    $\begingroup$ Unless I made an error, the first two equations and the symmetric one define smooth cubic surfaces. It would be interesting to calculate the Brauer group of the surfaces and determine whether there is a Brauer-Manin obstruction to solutions. If not, I suspect one can use techniques as in Heath-Brown's paper "The density of zeros of forms for which weak approximation fails" to guess the asymptotics for the number of solutions and make predictions about when one should be found. $\endgroup$
    – Will Sawin
    Jul 31, 2021 at 13:51
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    $\begingroup$ @Gottfried, since there haven't been any answers posted at all, I'd say it's a little premature to tag the question with big-list. $\endgroup$ Aug 16, 2021 at 13:13
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    $\begingroup$ If you look at the submission history, you will see that the paper has been submitted on the 2nd of August. So, equation (51) has been posed as an open problem in my paper submitted 2nd of August and then has been resolved in a later paper submitted 5th of August. It is a question to arxiv moderators why they needed 18 days to check that my paper is of sufficiently good quality to be accepted to the arXiv. $\endgroup$ Aug 20, 2021 at 12:37
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    $\begingroup$ @BogdanGrechuk A wordpress blog would work reasonably well: each equation could have its own blog post for discussion, as well as various tags ("H=46", "3-monomial", etc.) that one could use for navigation, and one could have additional pages that organize the equations in various ways (either updated by hand or by some automatic script). Alternatively one could look at databases such as the L-function and modular form database lmfdb.org for inspiration. $\endgroup$
    – Terry Tao
    Sep 16, 2021 at 14:52

3 Answers 3


The equation $$ x^3 + y^3 + z^3 + xyz = 5 $$ is solvable in integers. For example, take $$ x=-3028982, \quad y=-3786648, \quad z=3480565. $$ Verification is straightforward, but I would like to add more details how this solution has been found. If we just try $x$ and $y$ in order and then solve for $z$, then there are $10^{12}$ pairs $(x,y)$ even up to a million, hence finding the solution above was out of reach, at least for my computer.

Instead, I have noticed that linear change of variables $$ y \to y - z, \quad x \to x + 3 y $$ reduces the equation to $$ -5 + x^3 + 9 x^2 y + 27 x y^2 + 28 y^3 + x y z - x z^2 = 0 $$ from which we can see that $28y^3-5$ is divisible by $x$. Hence, we may choose any $y$, then choose $x$ among the divisors of $28y^3-5$ and solve the resulting quadratic equation in $z$. This allows to find the above solution in just a few hours on standard PC.

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    $\begingroup$ Nice, +1! If I may ask, how did you notice that particular change of variables (or did you just try several simple linear changes)? $\endgroup$ Feb 17, 2022 at 18:51
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    $\begingroup$ I made substitution $y \to y-z$ to cancel $z^3$ and made the equation quadratic in $z$. The term near $z^2$ in this quadratic equation was $(3y-x)$, which suggested the second substitution. $\endgroup$ Feb 17, 2022 at 19:13
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    $\begingroup$ I have added new smallest open symmetric equation to the question: $x^3+x+y^3+y+z^3+z = x y z + 1$ with $H=39$. A similar transformation applies, which allows me in 24 hours check for solutions up to $|y+z|\leq 5,000,000$. No solutions found. $\endgroup$ Feb 19, 2022 at 8:31
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    $\begingroup$ After the new smallest open equation with $H=39$, the next symmetric equation the program returned has been $x^3+y^3+z^3-xyz=11$ with $H=43$. I used similar transformation, then did a lot of optimizations in the code, and then run it for a week. After a few days, the search returned the solution $x=-106179597, y=-530403428, z=567068662$. This is the first equation in this project such that in the smallest known solution all variables exceed $10^8$ in absolute value. In fact, I think such solutions are the largest ones that can be found by a direct search on one standard PC. $\endgroup$ Mar 9, 2022 at 9:48

The equation $$y^2 - x^3y + z^3 + 3 = 0$$ has no solutions. Let $t = x^3 - 2y$. Then $$x^6 - 4z^3 = t^2 + 12$$ Modulo 9 analysis shows that $t$ is not divisible by 3. Modulo 32 analysis shows that $t$ is odd or is divisible by 4. Consider this equation as an equation in the ring $\mathbb{Z}[\omega]$, $\omega = \frac{-1 + i\sqrt{3}}{2}$. All the facts about Eisenstein integers used in the proof are from the Wikipedia article in the "Eisenstein integers" section. Proofs of all statements are contained, for example, in the book Ireland, Rosen "A classical introduction to modern number theory". Let $\pi \in \mathbb{Z}[\omega]$ be a prime divisor of $t^2 + 12$. Modulo $\pi^{\nu_{\pi}(t^2 + 12)}$ analysis shows that, either $3 \mid \nu_{\pi}(t^2 + 12)$ or $2$ is a cubic residue modulo $\pi$. The idea of solution is to use cubic reciprocity: $\left(\frac{\alpha}{\beta}\right)_3 = \left(\frac{\beta}{\alpha}\right)_3$ for any primary numbers $\alpha$, $\beta$. Primary number is the number of the form $(3n \pm 1) + (3m)\omega$. We need to consider 2 cases

Case 1. $t \equiv 1 \pmod{2}$. We can assume that $t \equiv 2 \pmod{3}$, because $-t$ is a solution whenever $t$ is a solution. $$t^2 + 12 = (t + 2\sqrt{3}i)(t - 2\sqrt{3}i) = (t + 2 + 4\omega)(t - 2 - 4\omega)$$ \begin{equation*} \begin{split} \left(\frac{2}{t + 2 + 4\omega}\right)_3 = \left(\frac{2}{\omega(t + 2 + 4\omega)}\right)_3 = \left(\frac{2}{-4 + (t - 2)\omega}\right)_3 = \\ = \left(\frac{-4 + (t - 2)\omega}{2}\right)_3 = \left(\frac{\omega}{2}\right)_3 = \omega \end{split} \end{equation*} Hence, $t + 2 + 4\omega$ has prime divisor $\pi$ for which 2 is a cubic nonresidue and $\nu_{\pi}(t + 2 + 4\omega) \ne 0 \pmod{3}$. This contradicts previous statement that $t^2 + 12$ cannot have such divisors.

Case 2. $t = 4r$. We can assume that $r \equiv 2 \pmod{3}$, because $-r$ is a solution whenever $r$ is a solution. $$t^2 + 12 = 4(2r + i\sqrt{3})(2r - i\sqrt{3}) = 4(2r + 1 + 2\omega)(2r - 1 - 2\omega)$$ \begin{equation*} \begin{split} \left(\frac{2}{2r + 1 + 2\omega}\right)_3 = \left(\frac{2}{\omega(2r + 1 + 2\omega)}\right)_3 = \left(\frac{2}{-2 + (2r - 1)\omega}\right)_3 = \\ = \left(\frac{-2 + (2r - 1)\omega}{2}\right)_3 = \left(\frac{-\omega}{2}\right)_3 = \omega \end{split} \end{equation*} Hence, $2r + 1 + 2\omega$ has prime divisor $\pi$ for which 2 is a cubic nonresidue and $\nu_{\pi}(2r + 1 + 2\omega) \ne 0 \pmod{3}$. This contradicts previous statement that $t^2 + 12$ cannot have such divisors.


I am very happy to see that the equation $$ y^2-x^3y+z^3+3 = 0, $$ which has been the smallest open equation for over two years, has now been solved by Denis Shatrov! Here, I would like to present an alternative argument which (while being longer and not self-contained because it uses some basic facts from the theory of quadratic forms and class group) works entirely over the integers. This proof is joint with Denis, all main ideas are his, all mistakes (if any) are mine.

We will need the following well-known facts.

Fact 1: A prime $p$ is a factor of $f(x)=x^3-2$ for some integer $x$ if and only if $p=3$, or $p \equiv 2(\text{mod}\, 3)$, or $p$ can be represented as $p=u^2+27v^2$ for some integers $u,v$.

Fact 2: Let $F$ be the principal form of discriminant $D$. Let $m$ be an odd positive integer with prime factorization $m=\prod_{i=1}^k p_i^{\alpha_i}$. Assume that, for each $i=1,\dots, k$ either (i) $p_i$ can be represented by $F$, or (ii) $\alpha_i$ is divisible by the class number $h(D)$. Then $m$ can be represented by $F$ but cannot be represented by any other (inequivalent) form $F'$ of the same discriminant.

Fact 3: For $D=-108$, the class group $H(-108)$ has three elements: the identity elements consisting on the forms equivalent to $u^2+27v^2$, and two other elements consisting on the forms equivalent to $4u^2\pm 2uv+7v^2$. In particular, $h(-108)=3$.

Fact 1 is a well-known characterization of primes $p$ modulo which $2$ is a cubic residue. Fact 2 is a direct corollary from, for example, Theorem 4.26 in [1]. Fact 3 is from standard tables of class groups of small discriminants, see, for example, pp 383-386 in [2].

A compbination of Fact 2 and Fact 3 implies the following lemma, conjectured by Denis Shatrov.

Lemma: Assume that $t$ is an integer such that $t \not \equiv 0(\text{mod} \, 3)$ and $t \not \equiv 2(\text{mod} \, 4)$, and let $t^2+12 = \prod_{i=1}^k p_i^{\alpha_i}$ be the prime factorization of $t^2+12$. Then there exists $i$ such that $\alpha_i$ is not divisible by $3$, and $p_i$ is an odd prime not representable as $u^2+27v^2$ for integers $u,v$.

Proof: Let $m=t^2+12$ if $t$ is odd, and $m=(t^2+12)/4$ if $t$ is divisible by $4$. Then $m$ is an odd positive integer, not divisible by $3$. If, by contradiction, in prime factorization $m=\prod p_i^{\alpha_i}$ we have, for all $i$, either $\alpha_i$ is divisible by $3$, or $p_i$ is of the form $u^2+27v^2$ for integers $u,v$, then Fact 2 applied with $F=u^2+27v^2$ and $D=-108$ would imply that $m$ cannot be represented by any other (inequivalent) form $F'$ of the same discriminant. In other words, $m$ is not of the form $4u^2+2uv+7v^2$ for integers $u,v$.

Let us now prove that $m$ is always of this form. Indeed, if $t$ is odd, then either $t=6r+1$ or $t=6r-1$ for some integer $r$. In the first case, $$ m=(6r+1)^2+12 = 4(-2r+1)^2+2(-2r+1)(2r+1)+7(2r+1)^2, $$ that is, $m$ is of the form $4u^2+2uv+7v^2$ for $u=-2r+1$ and $v=2r+1$. Similarly, in the second case, $$ m=(6r-1)^2+12 = 4(-2r-1)^2+2(-2r-1)(2r-1)+7(2r-1)^2. $$ If $t$ is even, then it is divisible by $4$ but not by $3$, and must therefore be equal to either $4(3r+1)$ or $4(3r-1)$. In the first case, $$ m = (t^2+12)/4 = 4(3r+1)^2+3 = 4(-2r)^2+2(-2r)(2r+1)+7(2r+1)^2, $$ while in the second case $$ m = (t^2+12)/4 = 4(3r-1)^2+3 = 4(-2r)^2+2(-2r)(2r-1)+7(2r-1)^2. $$ This finishes the proof of the lemma.

Now we are ready to prove that our equation, or, equivalently, equation $$ x^6 - 4z^3 = t^2+12 $$ has no integer solutions. Modulo $9$ analysis of this equation shows that $t$ is not divisible by $3$. Further, if $t$ is even, then so is $x$, and, by writing $x=2u$, $t=2v$ and cancelling $4$, we obtain $16u^6-z^3=v^2+3$. Modulo $8$ analysis of this equation shows that $v$ cannot be odd, hence $t \not \equiv 2(\text{mod} \, 4)$. Thus, by the Lemma, in the prime factorization $t^2+12 = \prod_{i=1}^k p_i^{\alpha_i}$ there exists $i$ such that $\alpha_i$ is not divisible by $3$, and $p_i$ is odd and not of the form $u^2+27v^2$ for integers $u,v$. Because $t$ is not divisible by $3$, $p_i>3$. Because $p_i$ is a divisor of $t^2+12$, this implies that $\left(\frac{-3}{p_i}\right)=1$, or $p_i \equiv 1(\text{mod}\, 3)$. Thus, by Fact 1, $p_i$ is not a divisor of $X^3-2$ for any integer $X$.

Now, write $x=p_i^\beta x_1$ and $z=p_i^\gamma z_1$, where $\text{gcd}(x_1,p_i)=\text{gcd}(z_1,p_i)=1$. Then $$ t^2+12 = x^6 - 4z^3 = p_i^{6\beta} x_1^6 - 4 p_i^{3\gamma} z_1^3. $$ If $6\beta > 3\gamma$, then $\alpha_i = 3\gamma$. Similarly, if $6\beta < 3\gamma$, then $\alpha_i = 6\beta$, in both cases a contradiction with the fact that $\alpha_i$ is not divisible by $3$. Hence, we must have $6\beta = 3\gamma < \alpha_i$, thus $x_1^6-4z_1^3$ is divisible by $p_i$. Now, $\text{gcd}(x_1,p_i)=1$ implies the existence of integer $\overline{x}_1$ such that $x_1\overline{x}_1 \equiv 1(\text{mod}\, p_i)$. Then, modulo $p_i$, $$ 0 = -2 \overline{x}_1^6 (x_1^6-4z_1^3) = -2(x_1\overline{x}_1)^6+(2\overline{x}_1^2z_1)^3 = -2+X^3, $$ where $X=2\overline{x}_1^2z_1$. This contradiction finishes the proof.

[1] Buell, Duncan A., Binary quadratic forms: classical theory and modern computations, Springer-Verlag, New York,1989, https://doi.org/10.1007/978-1-4612-4542-1

[2] Rose, H. E., A course in number theory, The Clarendon Press, Oxford University Press, New York, 1994


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