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Let $x,y,z$ be positive integers $x<y<z$ and satisfy $x+y+z = xyz$ , show that the only solution satisfying the above equation is $(x,y,z) = (1,2,3)$ .

My method was as follows : we have $x+y+z = xyz$ , rearranging a bit we have $1 = \frac{x+y+z}{xyz} = \frac{1}{xy} +\frac{1}{yz} + \frac{1}{zx}$, now as $x<y<z$ therefore we have $xy<yz$ and $xy<xz$ , or $\frac{1}{xz} < \frac{1}{xy} and \frac{1}{yz}<\frac{1}{xy}$ , so we have $\frac{1}{xy} +\frac{1}{yz} + \frac{1}{zx} = 1 < \frac{3}{xy}$ or we have yz < 3 , therefore only possible is yz=2 , hence we get $x=1,y=2$ , now $3+z = 2z$ or $z = 3$ , so we showed only solution is $(1,2,3)$

ProblemDestroyer
  • 967
  • The famous equation $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ in your case gives $(a,b,c)=(2,3,6)$, so that $(x,y,z)=(1,2,3)$. Or you can do it directly: $x\le y\le z$ implies that $xyz = x+y+z \le 3z \Rightarrow xy \le 3$

    Now taking the cases $(x,y) = (1,1), (1,2), (1,3)$ we see that $x=1, y=2, z=3$ is the only solution.

     – Dietrich Burde May 14 '22 at 17:03
  • Thanks :) @DietrichBurde – ProblemDestroyer May 14 '22 at 17:11

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