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Umm. This comes from Diophantine quartic equation in four variables and will finish the most important part if it can be done.

Four positive integers $w,x,y,z.$ One equation and two inequalities $$ wxyz = (w+x+y+z)^2, $$ $$ w \geq x \geq y \geq z \geq 1, $$ $$ xyz \geq 2(w+x+y+z). $$

I am hoping for an upper bound. Since i made $w$ biggest, it would be an UPPER BOUND on $w.$ For example, I am running a computer program to find all such quadruples with $w \leq 1000.$

Sample question: is it true that $w \leq 1000?$

This is the method of Hurwitz 1907. I have a pdf. His techniques are almost right for this problem, to the point where i am already convinced that the answer to the question by hardmath comes out the exact same way.

EDIT, these imply easily that $$\color{green}{ x+y+z \geq w}. $$ Could be useful.

EDIT: almost forgot, these are what I believe to be all such quadruples:

w  x  y  z     xyz  2(w+x+y+z)
4  4  4  4      64     32
6  6  3  3      52     36
8  5  5  2      50     40
10  10  9  1    90     60
12  6  4  2     48     48   
15  10  3  2    60     60   
18  9  8  1     72     72   
21  14  6  1    84     84   
30  24  5  1   120    120   
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  • $\begingroup$ Useful indeed! I'd just finished adding to my CW post under my Question that the inequality $w \gt x+y+z$ is necessary and sufficient for the Diophantine solutions to be realized by (nondegenerate) quadrilaterals. Four of your nine fundamental solutions satisfy this, and the other five have $w = x+y+z$. This last condition causes one side of the "quadrilateral" to have zero length, i.e. to degenerate into a triangle. $\endgroup$ – hardmath Jan 10 '14 at 11:52
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    $\begingroup$ @hardmath, I'm not sure i've gotten across the Markov Tree aspect of this. The quadruple (30,24,5,1), in the first layer, gives new solutions (54,30,5,1); (605,30,24,1); (3481, 30, 24,5). Each of these gives three new solutions, making a second layer, and so on forever. By solutions, i now mean $(w+x+y+z)^2 = wxyz.$ $\endgroup$ – Will Jagy Jan 10 '14 at 17:46
  • $\begingroup$ If the growth of these "trees" of quadruples is similar to the two families that @TitoPiezas originally described, then there will be only finitely many (for each tree) that satisfy $w \lt x+y+z$ (taking $w$ to be the largest entry for convenience). So infinitely many (positive) solutions of the equation, but only finitely many that give rise to cyclic quadrilaterals per the motivating Question. $\endgroup$ – hardmath Jan 10 '14 at 17:52
  • $\begingroup$ @hardmath, I see, you have the inequality the correct way in this recent comment, incorrect in the first one. I cannot say I noticed this additional condition in answering. I will post a few layers of one of the trees, maybe at your question. But you should consider making a new question including: max(w,x,y,z) < sum of the others $\endgroup$ – Will Jagy Jan 10 '14 at 18:48
  • $\begingroup$ Yes, I did get it backwards in that first comment. I had not ferreted out the condition when I posted my Question, so I was quite surprised to see it appear in yours (above: "these imply easily...") just after I posted an Edit to my CW post (derivation of the equation). There probably is enough "loose threads" to entertain another Question. If I've understood it correctly, you have (B) of the original Answer in hand now, and this probably yields an approach on your conjecture (A). I'm scurrying to catch up! $\endgroup$ – hardmath Jan 10 '14 at 20:20
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Indeed, one can get an upper bound and resolve this equation in general. The cheapest way is to denote $x+y+z=\alpha w$ and note that $\alpha\le 3.$ Plugging this into our equations we end up with:

$$xyz=(\alpha+1)^2w$$ and $$xyz\ge 2(\alpha+1)w.$$ Combining latter, we get $\alpha\ge 1.$ Recall that $x+y+z=\alpha w,$ so $$\frac{xyz}{x+y+z}=\frac{(\alpha+1)^2}{\alpha}\le 3+\frac{1}{3}+2=\frac{16}{3},$$ when $1\le \alpha\le 3.$ So we have the bound $xyz\le \frac{16}{3}(x+y+z)\le \frac{16}{3}(x+x+x)=16x.$ So $yz\le 16.$ One can then proceed and find all solutions.

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  • $\begingroup$ Very good. Beginning to catch up. Your $\alpha \geq 1$ and $w \geq x$ give $yz \geq 4.$ If $yz \geq 6$ then the part about $16/3$ gives $x \leq 16(y+z)/(3yz - 16).$ Oh, and $w \leq x+y+z$ in any case, that is your $\alpha \geq 1$ again. So the special cases are $(y,z)= (5,1), (4,1),(2,2).$ $\endgroup$ – Will Jagy Jan 10 '14 at 2:18
  • $\begingroup$ From $w \geq x$ and your $xyz \geq 2 (1 + \alpha) w$ and $\alpha \geq 1,$ IF $yz = 4$ THEN $w=x$ AND $\alpha = 1.$ So both $x = w$ and $x+y+z = w.$ This contradicts the assumption that $yz = 4.$ Still need to finish $(y,z) = (5,1),$ which does give one quadruple. $\endgroup$ – Will Jagy Jan 10 '14 at 2:32
  • $\begingroup$ When $y=5,z=1$ then $5x \geq 2(1+\alpha) w \geq 2(1+\alpha) x,$ or $5 \geq 2 (1 + \alpha)$ and $\alpha \leq 3/2.$ Then $5x/(x+6) \leq 25/6$ and $x \leq 30.$ $\endgroup$ – Will Jagy Jan 10 '14 at 2:55

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