Consider following diophantine equation in $\mathbb Z[x,y,z]$ in three integer variables $x,y,z$

$$x^2+L(y,z)x+L_1(y)L_2(z)=0$$ where $L(y,z)$ is a non-homogeneous linear polynomial in $y,z$ and $L_1(y),L_2(z)$ are linear non-homogeneous in $y,z$ respectively.

In general such an equation is difficult to solve if $x,y,z$ are independent.

However suppose that there are univariate linear non-homogeneous polynomials $L'(y)$ and $L''(z)$ such that if $(x^*,y^*,z^*)$ is a solution to the three variable diophantine equation then

1. $GCD(L'(y^*),L''(z^*)=1$

2. There exists $a,b\in\mathbb Z_{>1}$ with $a|L'(y^*)$ and $b|L''(z^*)$ such that $ab=x^*$

3. With $\overline x^*=\frac{L'(y^*)}{a}\frac{L''(z^*)}{b}$ we have $(\overline x^*,y^*,z^*)$ also a solution

holds always. Then the unknown $x$ depends on $y,z$ and so in principle we should be able to eliminate $x$ and make this as a two variable quadratic diophantine equation.

1. One cannot say there is no small degree algebraic relation between $x$ and $y,z$ since we have already provided one by the quadratic equation. Unless we provide a different relation we cannot use elimination theory. However there is arithmetic relation. There is some hope. Is there a way to at least in principle reduce the problem to solving two unknowns through some functions coming from arithmetic?

2. What is the minimal degree of any other polynomial relation between $x,y,z$?

It is unclear if either of following two is possible at least when $L'(y)=L_1(y)$ and $L'(z)=L_2(z)$:

I. There is an explicit arithmetic function that relates $x$ and $y,z$.

II. There is any other low degree polynomial relation between $x$ and $y,z$. Note that by formulating the problem itself we have provided a quadratic relation between $x$ and $y,z$. So there is a polynomial relation. The problem is if there is another polynomial relation.

Is it possible to think in following way if solutions are considered in $\mathbb K$ where $\mathbb K$ is a ring? Over $\mathbb K=\mathbb Q$ or $\mathbb K=\mathbb C$ it seems the equation cuts a codimension $1$ object in $\mathbb P^3(\mathbb K)$. However if $\mathbb K=\mathbb Z$ then the equation somehow refers at most a codimension $2$ object with some symmetry in solutions? If so there should be a way to look at $2$ dimensional solutions over $\mathbb Z$ algebraically and if so any tools from arithmetic geometry applies here? Is there another way to think from algebraic or arithmetic geometry?

Consider following diophantine equation in $\mathbb Z[x,y,z]$ in three integer variables $x,y,z$

$$x^2+L(y,z)x+L_1(y)L_2(z)=0$$ where $L(y,z)$ is a non-homogeneous linear polynomial in $y,z$ and $L_1(y),L_2(z)$ are linear non-homogeneous in $y,z$ respectively.

In general such an equation is difficult to solve if $x,y,z$ are independent.

However suppose that there are univariate linear non-homogeneous polynomials $L'(y)$ and $L''(z)$ such that if $(x^*,y^*,z^*)$ is a solution to the three variable diophantine equation then

1. $GCD(L'(y^*),L''(z^*)=1$

2. There exists $a,b\in\mathbb Z_{>1}$ with $a|L'(y^*)$ and $b|L''(z^*)$ such that $ab=x^*$

3. With $\overline x^*=\frac{L'(y^*)}{a}\frac{L''(z^*)}{b}$ we have $(\overline x^*,y^*,z^*)$ also a solution

holds always. Then the unknown $x$ depends on $y,z$ and so in principle we should be able to eliminate $x$ and make this as a two variable quadratic diophantine equation.

1. One cannot say there is no small degree algebraic relation between $x$ and $y,z$ since we have already provided one by the quadratic equation. Unless we provide a different relation we cannot use elimination theory. However there is arithmetic relation. There is some hope. Is there a way to at least in principle reduce the problem to solving two unknowns through some functions coming from arithmetic?

2. What is the minimal degree of any other polynomial relation between $x,y,z$?

It is unclear if either of following two is possible at least when $L'(y)=L_1(y)$ and $L'(z)=L_2(z)$:

I. There is an explicit arithmetic function that relates $x$ and $y,z$.

II. There is any other low degree polynomial relation between $x$ and $y,z$. Note that by formulating the problem itself we have provided a quadratic relation between $x$ and $y,z$. So there is a polynomial relation. The problem is if there is another polynomial relation.

Is it possible to think in following way if solutions are considered in $\mathbb K$ where $\mathbb K$ is a ring? Over $\mathbb K=\mathbb Q$ or $\mathbb K=\mathbb C$ it seems the equation cuts a codimension $1$ object in $\mathbb P^3(\mathbb K)$. However if $\mathbb K=\mathbb Z$ then the equation somehow refers at most a codimension $2$ object with some symmetry in solutions? If so there should be a way to look at $2$ dimensional solutions over $\mathbb Z$ algebraically and if so any tools from arithmetic geometry applies here? Is there another way to think from algebraic or arithmetic geometry? I do not know if I made sense in these last points. However proceeding if it is dimension $2$ over $\mathbb P^{3}(\mathbb Z)$ then does it have any possibility of being looked at as an affine curve in two dimensions cut by an two variable polynomial or at least cover by such affine curves?

Consider following diophantine equation in $\mathbb Z[x,y,z]$ in three integer variables $x,y,z$

$$x^2+L(y,z)x+L_1(y)L_2(z)=0$$ where $L(y,z)$ is a non-homogeneous linear polynomial in $y,z$ and $L_1(y),L_2(z)$ are linear non-homogeneous in $y,z$ respectively.

In general such an equation is difficult to solve if $x,y,z$ are independent.

However suppose that there are univariate linear non-homogeneous polynomials $L'(y)$ and $L''(z)$ such that if $(x^*,y^*,z^*)$ is a solution to the three variable diophantine equation then

1. $GCD(L'(y^*),L''(z^*)=1$

2. There exists $a,b\in\mathbb Z_{>1}$ with $a|L'(y^*)$ and $b|L''(z^*)$ such that $ab=x^*$

3. With $\overline x^*=\frac{L'(y^*)}{a}\frac{L''(z^*)}{b}$ we have $(\overline x^*,y^*,z^*)$ also a solution

holds always. Then the unknown $x$ depends on $y,z$ and so in principle we should be able to eliminate $x$ and make this as a two variable quadratic diophantine equation.

1. One cannot say there is no small degree algebraic relation between $x$ and $y,z$ since we have already provided one by the quadratic equation. Unless we provide a different relation we cannot use elimination theory. However there is arithmetic relation. There is some hope. Is there a way to at least in principle reduce the problem to solving two unknowns through some functions coming from arithmetic?

2. What is the minimal degree of any other polynomial relation between $x,y,z$?

It is unclear if either of following two is possible at least when $L'(y)=L_1(y)$ and $L'(z)=L_2(z)$:

I. There is an explicit arithmetic function that relates $x$ and $y,z$.

II. There is any other low degree polynomial relation between $x$ and $y,z$. Note that by formulating the problem itself we have provided a quadratic relation between $x$ and $y,z$. So there is a polynomial relation. The problem is if there is another polynomial relation.

Is it possible to think in following way if solutions are considered in $\mathbb K$ where $\mathbb K$ is a ring? Over $\mathbb K=\mathbb Q$ or $\mathbb K=\mathbb C$ it seems the equation cuts a codimension $1$ object in $\mathbb P^3(\mathbb K)$. However if $\mathbb K=\mathbb Z$ then the equation somehow refers at most a codimension $2$ object with some symmetry in solutions? If so any tools from arithmetic geometry applies here? Is there another way to think of this from algebraic or arithmetic geometry?

Consider following diophantine equation in $\mathbb Z[x,y,z]$ in three integer variables $x,y,z$

$$x^2+L(y,z)x+L_1(y)L_2(z)=0$$ where $L(y,z)$ is a non-homogeneous linear polynomial in $y,z$ and $L_1(y),L_2(z)$ are linear non-homogeneous in $y,z$ respectively.

In general such an equation is difficult to solve if $x,y,z$ are independent.

However suppose that there are univariate linear non-homogeneous polynomials $L'(y)$ and $L''(z)$ such that if $(x^*,y^*,z^*)$ is a solution to the three variable diophantine equation then

1. $GCD(L'(y^*),L''(z^*)=1$

2. There exists $a,b\in\mathbb Z_{>1}$ with $a|L'(y^*)$ and $b|L''(z^*)$ such that $ab=x^*$

3. With $\overline x^*=\frac{L'(y^*)}{a}\frac{L''(z^*)}{b}$ we have $(\overline x^*,y^*,z^*)$ also a solution

holds always. Then the unknown $x$ depends on $y,z$ and so in principle we should be able to eliminate $x$ and make this as a two variable quadratic diophantine equation.

1. One cannot say there is no small degree algebraic relation between $x$ and $y,z$ since we have already provided one by the quadratic equation. Unless we provide a different relation we cannot use elimination theory. However there is arithmetic relation. There is some hope. Is there a way to at least in principle reduce the problem to solving two unknowns through some functions coming from arithmetic?

2. What is the minimal degree of any other polynomial relation between $x,y,z$?

It is unclear if either of following two is possible at least when $L'(y)=L_1(y)$ and $L'(z)=L_2(z)$:

I. There is an explicit arithmetic function that relates $x$ and $y,z$.

II. There is any other low degree polynomial relation between $x$ and $y,z$. Note that by formulating the problem itself we have provided a quadratic relation between $x$ and $y,z$. So there is a polynomial relation. The problem is if there is another polynomial relation.

Is it possible to think in following way if solutions are considered in $\mathbb K$ where $\mathbb K$ is a ring? Over $\mathbb K=\mathbb Q$ or $\mathbb K=\mathbb C$ it seems the equation cuts a codimension $1$ object in $\mathbb P^3(\mathbb K)$. However if $\mathbb K=\mathbb Z$ then the equation somehow refers at most a codimension $2$ object with some symmetry in solutions? If so there should be a way to look at $2$ dimensional solutions over $\mathbb Z$ algebraically and if so any tools from arithmetic geometry applies here? Is there another way to think from algebraic or arithmetic geometry?