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By Matiyasevich theorem, there is no algorithm to decide whether a given Diophantine equation $P(x_1,\dots, x_n)=0$ has a solution in positive integers. As suggested in What is the smallest unsolved Diophantine equation?, let us arrange the equations by $H(P)=\sum |a_i| 2^{d_i}$, where $a_i$ are coefficients on the monomials of $P$ and $d_i$ are their degrees. Then consider all equations in order of $H$, and try to decide the existence of a solution in positive integers. See Can you solve the listed smallest open Diophantine equations? for a related study for all solutions (positive or negative).

After equation Positive integers such that $(x+y)(xy-1)=z^2+1$ has been solved by Denis Shatrov, I was able to solve many other equations by similar methods, including all equations of size $H\leq 25$, and almost all equations of size $H=26$. The only remaining open are:

(a) Equation $$ (x+1)yz-y-z=x^3-2. $$ It implies that $x^3-2+z$ is divisible by $y$. Write $z=ty-x^3+2$ for integer $t$, substitute in the equation, and obtain $$ t((x+1)y-1) = (x+1)(x^3-2)+1 = x^4+x^3-2x-1. $$ So, the question is whether $x^4+x^3-2x-1$ has (for some integer $x\geq 2$) a positive divisor equal to $-1$ modulo $x+1$.

(b) Equation $x^3-xy^2+y+2z^2=0$. Update: This equation has no positive integer solutions as remarked by Denis Shatrov in a comment.

(c) Equations $$ y(x^3-z^2)=z \quad \text{and} \quad x^2y^2+x=z^3 $$ In the first equation, $z=yt$, where $t=x^3-z^2=x^3-(yt)^2$. Up to the names of the variables, this is the second equation. From the second, $x(xy^2+1)=z^3$, which is possible only if $x=u^3$ and $xy^2+1=v^3$, or $u^3y^2=v^3-1$. Integers of the form $u^3y^2$ are called powerful number, and the question reduces to the existence of positive integer $v$ such that $v^3-1$ is powerful.

(d) Equation $$ y(x^3-z^2)=x $$ We have $x=yt$ for $t=x^3-z^2=(yt)^3-z^2$, or $t(t^2y^3-1)=z^2$, hence $t=u^2$ and $(u^2)^2y^3-1=v^2$, or $u^4y^3=v^2+1$.

The question is, for each of these equations, whether it has a solution is positive integers. Equations (c) and (d) look difficult, but equation (a) looks doable.

kodlu
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Bogdan Grechuk
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    We generally frown on having more than one question in a question. Which will you accept if four different users solve your four equations? – Gerry Myerson Aug 22 '23 at 09:59
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    I have 4 open equations of the same size, and I though it is more convenient to have it all in one place rather than creating 4 questions. When I answer a question, I care about sharing knowledge not about whether "accept" button will be pressed. But I am not sure about other users. If formal "accept" is important for them, then this is indeed an issue. – Bogdan Grechuk Aug 22 '23 at 10:19
  • @BogdanGrechuk Did you test numerically that each of these $4$ has no solutions? – mathworker21 Aug 22 '23 at 12:33
  • Yes. As you may guess, there are hundreds of thousands of equations of size $H\leq 26$. My computer program excludes all families of equations that are easy to solve (e.g. linear ones), all equations that have no solutions for obvious reasons (like no solutions modulo some m), and then of course excludes all equations for which it can find a positive integer solution. Then it outputs a list of equations that it cannot exclude and I work with them manually. Equations with small solutions are therefore excluded and not go to the short list. – Bogdan Grechuk Aug 22 '23 at 12:51
  • For this reason, almost all equations in the short list happens to have no positive integer solutions. The only exception so far is the equation $x^3+y^3=z^3+2$ for which my program did not find any positive integer solutions, but there is a solution $1214928^3+3480205^3=3528875^3+2$ discovered in (Heath-Brown, D. R., Walter M. Lioen, and H. J. J. Te Riele. "On solving the Diophantine equation $x^3+y^3+z^3=k$ on a vector computer." Mathematics of computation 61.203 (1993): 235-244.) – Bogdan Grechuk Aug 22 '23 at 12:59
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    @BogdanGrechuk Thank you. And thanks for your work and your questions! – mathworker21 Aug 22 '23 at 13:27
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    The equation (b) is unsolvable, because $y(xy - 1) = x^3 + 2z^2$ and Jacobi symbol $\left(\frac{-2x}{xy - 1} \right)$ is equal to $-1$ (note that modulo 8 we see that $x, y$ are both even). – Denis Shatrov Aug 22 '23 at 13:38
  • Thank you! I had a feeling that I am missing something obvious with (b). Equation (a) looks more interesting (but should be doable), while equations (c) and (d) look difficult. – Bogdan Grechuk Aug 22 '23 at 14:57
  • Fixed the exponent of 2 in $H(P)$ please undo if I misinterpreted – kodlu Aug 22 '23 at 16:09

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For (a), we can introduce $s:=x+1$ and pose the question as $(sy-1)\mid (s^4 - 3s^3 + 3s^2 - 3s + 1)$. Equivalently, we have that $$\frac{s^2 + y^2 - 3s - 3y + 3}{sy-1}$$ is an integer. This problem is amenable to Vieta jumping, which implies that there are no positive solutions.

Max Alekseyev
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  • Thank you! I suspected that this equation is doable, but the solution turned out to be easier than I expected. Now we have (up to equivalence) just two open equations of size $H\leq 26$: (c) and (d), and both of them look difficult. – Bogdan Grechuk Aug 22 '23 at 19:17