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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, Vaserstein https://annals.math.princeton.edu/wp-content/uploads/annals-v171-n2-p07-s.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).

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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. $\endgroup$ Jan 20 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
    Jan 20 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. $\endgroup$ Jan 20 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). $\endgroup$ Jan 20 at 14:16
  • $\begingroup$ Okay, I hope it will be parameterized. $\endgroup$
    – Tomita
    Jan 21 at 0:56

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