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This is related to this post. First, we have,

Theorem: "If $w_0, z_0$ is a solution to,

$$1-4x-4(1-x^2)z = w^2\tag1$$

then,

$$w = w_0+2(x^2-1)n$$

$$z = z_0+w_0\,n+(x^2-1)n^2$$

is also a solution for any $n$."

Let us limit $w,x,z$ to positive integers. For example, the four smallest solutions when $x=40$ are,

$$z = F_1(n):= \color{brown}{28} + 423 n + 1599 n^2\\ z = F_2(n):= \color{brown}{70} + 669 n + 1599 n^2\\ z = F_3(n):= \color{brown}{1000} + 2529 n + 1599 n^2\\ z = F_4(n):= \color{brown}{1204} + 2775 n + 1599 n^2$$

as $F_i(0)$. The next four smallest are $F_i(1)$, and so on. The following,

$$x=4,40,58,94,130,148,238,310,\color{blue}{400}, 490, 544, 598, \color{blue}{670}, \color{blue}{904}, 1570,\dots$$

are solvable. The value $x=4$ has one family, the others have four, except those in blue which have eight.

Question: Is there a way, based on just knowing $x$, to predict the number of families? In other words, what is it about $x$ that determines this number?

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