##x\lt y \lt z## positive integer numbers and if ##\dfrac 1x+\dfrac 1y+\dfrac 1z=\dfrac 13##

[littlemathquark]

TL;DR

İf $x\lt y \lt z$ positive integer numbers and if $\dfrac 1x+\dfrac 1y+\dfrac 1z=\dfrac 13$ then find the minumum and maximum value of sum of $x+y+z$

I can solve diophantine equation of $\dfrac 1x+\dfrac 1y+\dfrac 1z=\dfrac 13$. But do I have to find the individual values of x, y, and z to find the minimum and maximum value of the sum? Isn't there an algorithm that gives the smallest and biggest sum of denominator values?

 

[fresh_42]

Well, there is an algorithm. WA says that there are $15$ solutions, with $31$ being the minimum and $173$ the maximum. it looks as if they eliminated $z$ and solved the remaining two-dimensional lattice problem.

Observation: $3\,|\,xyz$ and looking at the results, we see that $3\,|\,z.$ Could be a point to start with.

 

[littlemathquark]

Well, there is an algorithm. WA says that there are $15$ solutions, with $31$ being the minimum and $173$ the maximum. it looks as if they eliminated $z$ and solved the remaining two-dimensional lattice problem.

Well, I didn't mean that. I was asking if there's an algorithm that can directly output the values for the minimum as $1/3 =1/6+1/10+1/15$ or for the maximum as $1/3=1/4+1/13+1/156$
İn fact I know algorithm $1/x=(1/x(x+1)) +(1/(x+1))$ and I think it's produce maximum denominators.

 

[fresh_42]

I didn't mean that. I was asking if there's an algorithm that can directly output the values for the minimum as $1/3 =1/6+1/10+1/15$ or for the maximum as $1/3=1/4+1/13+1/156$
İn fact I know algortihm $1/x=(1/x(x+1)) +(1/(x+1))$ and I think it's produce maximum denominators.

Yes, and I answered your question. There exists an algorithm simply because WA uses one. I don't know which one, but I suggested a reduction to a lattice problem. Look up lattice algorithms.

Here is a starting point:
https://sites.math.washington.edu/~rothvoss/lecturenotes/IntOpt-and-Lattices.pdf

 

[littlemathquark]

My solution like this: I solved for $x\ge y\ge z$

İt's clear $x,y,z >3$



Let $x\leq y\leq z$. Then $\frac{1}{z} \leq \frac{1}{y} \leq \frac{1}{x}$ and $\frac{1}{3} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \leq \frac{3}{x}$, $x \leq 9$ that is $4\leq x \leq 9$

İf $x=4$ then $\frac{1}{3} = \frac{1}{4} + \frac{1}{y} + \frac{1}{z}$, $\frac{1}{12} = \frac{1}{y} + \frac{1}{z} \leq \frac{2}{y}$, $y\leq 24$

$\frac{1}{z} = \frac{y-12}{12y}$, $z=\frac{12y}{y-12} = 12+\frac{144}{y-12}$

$13 \leq y \leq 24$

$(x=4,y=13, z=156), (x=4,y=14, z=84), (x=4,y=15, z=60), (x=4,y=16, z=48) (x=4,y=18, z=36), (x=4,y=20, z=30),(x=4,y=21, z=28),(x=4,y=24, z=24)$



if $x=5$ then $\frac{1}{3} = \frac{1}{5} + \frac{1}{y} + \frac{1}{z}$, $\frac{2}{15} = \frac{1}{y} + \frac{1}{z} \leq \frac{2}{y}$, $y \leq 15$

$\frac{1}{z} = \frac{2y-15}{15y}$, $z=\frac{15y}{2y-15} = 7+\frac{y}{2y-15}$

$8 \leq y \leq 15$.

$(x=5,y=8, z=120), (x=5,y=9, z=45), (x=5,y=10, z=30), (x=5,y=12, z=20),(x=5,y=15, z=15)$



if $x=6$ then $\frac{1}{3} = \frac{1}{6} + \frac{1}{y} + \frac{1}{z}$, $\frac{1}{6} = \frac{1}{y} + \frac{1}{z} \leq \frac{2}{y}$

$y \leq 12$

$\frac{1}{z} = \frac{y-6}{6y}$, $z=\frac{6y}{y-6} = 6+\frac{36}{y-6}$

$7 \leq y \leq 12$.

$(x=6,y=7, z=42), (x=6,y=8, z=24),(x=6,y=9, z=18),(x=6,y=10, z=15),(x=6,y=12, z=12)$



if $x=7$ then $\frac{1}{3} = \frac{1}{7} + \frac{1}{y} + \frac{1}{z}$, $\frac{4}{21} = \frac{1}{y} + \frac{1}{z} \leq \frac{2}{y}$

$y \leq 10$

$\frac{1}{z} = \frac{4y-21}{21y}$, $z=\frac{21y}{4y-21} = 5+\frac{y+105}{4y-21}$

$7 \leq y \leq 10$.

$(x=7,y=7, z=21)$



if $x=8$ then $\frac{1}{3} = \frac{1}{8} + \frac{1}{y} + \frac{1}{z}$, $\frac{5}{24} = \frac{1}{y} + \frac{1}{z} \leq \frac{2}{y}$

$y \leq 9$

$\frac{1}{z} = \frac{5y-24}{24y}$, $z=\frac{24y}{5y-24} = 4+\frac{4y+96}{5y-24}$

$8 \leq y \leq 9$.

$(x=8,y=8, z=12)$



if $x=9$ then $\frac{1}{3} = \frac{1}{9} + \frac{1}{y} + \frac{1}{z}$, $\frac{2}{9} = \frac{1}{y} + \frac{1}{z} \leq \frac{2}{y}$

$y=9$

$(x=9,y=9, z=9)$

There is $21$ solutions.

 

[fresh_42]

You said $0<x<y<z$, so there are fewer than 21 solutions.

 

[littlemathquark]

Yes, I solved
for $x\ge y\ge z\ge 0$