Generalization of a Diophantine Equation Problem

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I've been working a lot with Pythagorean triples and sums of squares that are themselves squares, specifically interlocking sums (where one square is part of two or more sums). As part of my work I needed to find Diophantine solutions to the expression found here: Triplets of distinct integers > 1 that return integer values. There are none. I have, however, generalized that problem and am looking for Diophantine (positive integer) solutions to: $$k=\frac{\frac{A^2-\alpha}{A} + \frac{B^2-\alpha}{B}}{\frac{C^2-\alpha}{C}}$$ Where $k\geq2$; $A>B>C>1$ if $k>2$ and $A>C>B>1$ if $k=2$; and $\alpha=n*lcm(A,B,C)$ for some integer $n\geq1$.

Here is what I have thus far: Start by re-writing $k$ as $$A-\frac{\alpha}{A}+B-\frac{\alpha}{B}-k(C-\frac{\alpha}{C})=0$$

Now, let $w$ be the factor shared by $A,B,$ and $C$. That is, $w=gcf(A,B,C)$.

Let $x,y,z$ be the factors shared by $A,B;A,C;B,C$ (respectively) EXCLUDING $w$. That is, $x=\frac{gcf(A,B)}{w}$, $y=\frac{gcf(A,C)}{w}$, and $z=\frac{gcf(B,C)}{w}$ (Note that this leads to the result that $gcf(x,\frac{C}{w})=1$ and likewise for $y,B$ and $z,A$.)

Finally, let $a,b,c$ be the factors of $A,B,C$ (respectively) not shared by either of the other two numbers (unless $a,b,$ or $c=1$). That is, $a=\frac{A}{wxy}$, $b=\frac{B}{wxz}$, and $c=\frac{C}{wyz}$.

From this it follows that $a,b,c$ are mutually prime (no pair shares a factor greater than $1$) and similarly $x,y,z$ are all mutually prime. Furthermore, from $gcf(x,\frac{C}{w})=1$ we have $x,c$ being mutually prime, and by the same logic $y,b$ and $z,a$ are mutually prime pairs. Note that $w$ need not be mutually prime to any of those six numbers so long as the previous conditions are met.

We now have the following: $A=awxy$, $B=bwxz$, $C=cwyz$, $\alpha=nabcwxyz$.

Substituting into the equation above we have: $$awxy-nbcz+bwxz-nacy-k(cwyz-nabx)=0$$

At this point I'm at a loss on how to proceed. I tried factoring and solving for $k$ and got: $$k=\frac{(ay+bz)(nc-wx)}{nabx-cwyz}$$ (switching the subtraction around as well) but I don't know where to go from here. Any help would be greatly appreciated.