$x,y,z >3$ olur.
$x\leq y\leq z$ olsun. Bu takdirde $\frac{1}{z} \leq \frac{1}{y} \leq \frac{1}{x}$ ve $\frac{1}{3} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \leq \frac{3}{x}$, $x \leq 9$ yani $4\leq x \leq 9$ elde edilir.
$x=4$ ise $\frac{1}{3} = \frac{1}{4} + \frac{1}{y} + \frac{1}{z} $, $\frac{1}{12} = \frac{1}{y} + \frac{1}{z} \leq \frac{2}{y}$, $y\leq 24$
$\frac{1}{z} = \frac{y-12}{12y}$, $z=\frac{12y}{y-12} = 12+\frac{144}{y-12}$
$13 \leq y \leq 24$
$(x=4,y=13, z=156), (x=4,y=14, z=84), (x=4,y=15, z=60), (x=4,y=16, z=48) (x=4,y=18, z=36), (x=4,y=20, z=30),(x=4,y=21, z=28),(x=4,y=24, z=24)$
$x=5$ ise $\frac{1}{3} = \frac{1}{5} + \frac{1}{y} + \frac{1}{z} $, $\frac{2}{15} = \frac{1}{y} + \frac{1}{z} \leq \frac{2}{y}$, $y \leq 15$
$\frac{1}{z} = \frac{2y-15}{15y}$, $z=\frac{15y}{2y-15} = 7+\frac{y}{2y-15}$
$8 \leq y \leq 15$.
$(x=5,y=8, z=120), (x=5,y=9, z=45), (x=5,y=10, z=30), (x=5,y=12, z=20),(x=5,y=15, z=15)$
$x=6$ ise $\frac{1}{3} = \frac{1}{6} + \frac{1}{y} + \frac{1}{z} $, $\frac{1}{6} = \frac{1}{y} + \frac{1}{z} \leq \frac{2}{y}$
$y \leq 12$
$\frac{1}{z} = \frac{y-6}{6y}$, $z=\frac{6y}{y-6} = 6+\frac{36}{y-6}$
$7 \leq y \leq 12$.
$(x=6,y=7, z=42), (x=6,y=8, z=24),(x=6,y=9, z=18),(x=6,y=10, z=15),(x=6,y=12, z=12)$
$x=7$ ise $\frac{1}{3} = \frac{1}{7} + \frac{1}{y} + \frac{1}{z} $, $\frac{4}{21} = \frac{1}{y} + \frac{1}{z} \leq \frac{2}{y}$
$y \leq 10$
$\frac{1}{z} = \frac{4y-21}{21y}$, $z=\frac{21y}{4y-21} = 5+\frac{y+105}{4y-21}$
$7 \leq y \leq 10$.
$(x=7,y=7, z=21)$
$x=8$ ise $\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$
$\frac{1}{z} = \frac{5y-24}{24y}$, $z=\frac{24y}{5y-24} = 4+\frac{4y+96}{5y-24}$
$8 \leq y \leq 9$.
$(x=8,y=8, z=12)$
$x=9$ ise $\frac{1}{3} = \frac{1}{9} + \frac{1}{y} + \frac{1}{z} $, $\frac{2}{9} = \frac{1}{y} + \frac{1}{z} \leq \frac{2}{y}$
$y=9$
$(x=9,y=9, z=9)$
$21$ tane çözüm vardır.