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In general, it is not clear What does it mean to solve an equation? in integers. In this question, let us assume that an equation $$ P(x_1,\dots,x_n)=0 $$ is solved if we have proved that its integer solution set $S \subset {\mathbb Z}^n$ can be represented as the finite union $S=S_1 \cup \dots \cup S_m$, where each $S_i$ is either a polynomial family or a family defined by recurrence relations. Here, $S \subset {\mathbb Z}^n$ is a polynomial family if there exists polynomials $P_1,\dots,P_n$ in some variables $u_1,\dots,u_k$ and integer coefficients such that $(x_1,\dots,x_n) \in S$ if and only if there exists integers $u_1,\dots,u_k$ such that $x_i=P_i(u_1,\dots,u_k)$ for $i=1,\dots,n$.

Following Zidane What is the smallest unsolved Diophantine equation? , let us define size $H$ of the equation $P=0$ as a result of substitution 2 instead of all variables, absolute values instead of all coefficients, and evaluating.

All equations with $H \leq 8$ are easy to solve. However, simple-looking equation $xy-zt=1$ with $H=9$ has been open for decades. In 2010, Vaserstein1 https://annals.math.princeton.edu/wp-content/uploads/annals-v171-n2-p07-p.pdf proved that

  • The solution set to $xy-zt=1$ is a polynomial family with $46$ parameters.

As a corollary of this, Vaserstein solved the following families of equations

  • $xy-zt=D$ for any integer $D$;
  • $yz=x^2+D$ for any integer $D$;
  • $x_1x_2+x_3x_4+Q(x_5,\dots,x_n)=D$ for quadratic form $Q$ and integer $D$.

In addition, the following equations/families has been solved:

I was also able to solve all equations with $H=13$ except of the following ones. $$ x^3 + 1 = yz $$ $$ x^2y=z^2 \pm 1 $$ $$ x^2y=tz+1 $$ $$ x^2 + y^2 \pm 1 = zt $$ $$ x^2 \pm 1 = yzt $$ $$ x_1x_2x_3+x_4x_5=1 $$

For each of the listed equations, the question is to find all integer solutions. Specifically, check whether the set of all integer solutions is a finite union of polynomial families and/or families defined by recurrence relations. You do not need to write the resulting families explicitly, because, as example $xy-zt=1$ indicates, they may be quite complicated.

See here Can you solve the listed smallest open Diophantine equations? for a version of this question where we only want to check whether any integer solution exists, and here On the smallest open Diophantine equations: beyond Hilbert's 10 problem for a version where we also check whether the solution set is finite or infinite (and find all solutions if there are finitely many).

1Vaserstein, Leonid, Polynomial parametrization for the solutions of Diophantine equations and arithmetic groups, Ann. Math. (2) 171, No. 2, 979-1009 (2010). ZBL1221.11082, JSTOR.

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The equation $x^2+1 = yzt$ has a parametric solution as follows.

We factorize $x^2+1 = yzt$ in $\mathbb Z[i].$

Let $(Y,Z,T)=(a+bi,c+di,e+fi)$ then

$x+i = YZT = (acf+ead+ebc-bdf)i+ace-fad-fbc-bde.$

Hence we get a parametric solution $x = acf+ead+ebc-bdf$ if $ace-fad-fbc-bde =1.$

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    $\begingroup$ We are not allowed to use the "if" condition, otherwise there is a trivial answer $x=a$, $y=b$, $z=c$, $t=d$ if $a^2+1=bcd$. We can only use "if" condition if we know how to solve the equation under if statement. So we have reduced the problem of writing parametric solutions of $x^2+1=yzt$ to the problem of writing parametric solutions of $ace-fad-fbc-bde=1$. This equation does not look any easier for me than the original one. $\endgroup$
    – Bogdan Grechuk
    Commented Jan 20, 2022 at 7:42
  • $\begingroup$ This condition is necessary to obtain all integer solutions, just as we did for $yz=x^2+x+1$. $\endgroup$
    – Tomita
    Commented Jan 20, 2022 at 9:53
  • $\begingroup$ With $yz=x^2+x+1$ we used the condition $ab-cd=1$ which we were allowed to use only because this equation has been solved in the Vaserstein paper. He proved that $ab-cd=1$ if and only if $a=P(u)$, $b=Q(u)$, $c=R(u)$, $d=T(u)$ where $P,Q,S,T$ are some polynomials and $u$ is a vector of variables without any conditions. $\endgroup$
    – Bogdan Grechuk
    Commented Jan 20, 2022 at 14:05
  • $\begingroup$ Then is we have $x=ad+bc+bd$, $y=a^2+ab+b^2$, $z=c^2+cd+d^2$, then $x=P(u)T(u)+Q(u)R(u)+Q(u)T(u)$, $y=P^2(u)+P(u)Q(u)+Q^2(u)$, $z=S^2(u)+S(u)T(u)+T^2(u)$ gives the final answer without any conditions on parameters $u$. This is not available for the condition $ace-fad-fbc-bde=1$. In general, we may use any condition defined by an equation that we have already solved (or that is solved in the literature). $\endgroup$
    – Bogdan Grechuk
    Commented Jan 20, 2022 at 14:16
  • $\begingroup$ Okay, I hope it will be parameterized. $\endgroup$
    – Tomita
    Commented Jan 21, 2022 at 0:56

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