solution of Diofantine equation

Question

Find number of non-negative integer solutions of equation: $x+y^2+z=x^2z+y$.

I have tried to rearrange it like $x^2z-y^2=x-y+z$, but I don't have an idea what to do in the next step. Thanks for any advice.

Answer 1

Let,

$$x+y^2+z = x^2z+y\tag1$$

As Lucian points out, the solvable $x$ (so far) are,

$$x = 4, 40, 58, 94, 130, 148, 238, 310, 400,\dots$$

For example, looking at $x=40$, one has to find positive integers to solve,

$$y = \tfrac{1}{2}\left(\sqrt{6396z-159}\right)$$

The first few are,

$$z = \color{blue}{28}, \color{red}{70}, \color{green}{1000}, \color{b}{1204}, \color{blue}{2050}, \color{red}{2338}, \color{green}{5128}, \color{b}{5578},\color{blue}{7270}, \color{red}{7804}, \color{green}{12454}, \color{b}{13150},\dots$$

where it seems every fourth solution is in a polynomial progression. After inspecting $x=40$ and others, one can make a general conclusion.

Theorem. "If $y_0, z_0$ is a solution to $(1)$, then,

$$y = y_0 + (x^2 - 1) k$$

$$z = z_0 + (2y_0 - 1) k + (x^2 - 1) k^2$$

is also a solution for any $k$, thus yielding an infinite family."

Example, $x = 40$:

1. First with $y_0,\, z_0 = 212,\,28$: $$y= 212 + 1599 k = 212, 1811, 3410,\dots\\ z=28 + 423 k + 1599 k^2 = 28, 2050, 7270,\dots$$
2. Second with $y_0,\, z_0 = 335,\,70$: $$y= 335 + 1599 k = 335, 1934, 3533,\dots\\ z=70 + 669 k + 1599 k^2 = 70, 2338, 7804,\dots$$

and so on for four infinite families.

Note: I haven't checked if each solvable $x$ yields multiple quadratic families that yield distinct $z$.