Find $xy+yz+xz$ given that $x+y+z=xyz$ for $x,y,z \in \mathbb{Z}^{+}$

Question

I have to calculate $xy+yz+xz$ using $x+y+z=xyz$ where $x,y,z \in \mathbb{Z}^{+} $.

I tried to solve the equation $x+y+z=xyz$ and got the triples without any method. By observing the equation I found the triples $(x,y,z)=(1,2,3),(2,1,3),(3,2,1),(2,3,1)$. All triples above showed that $xy+yz+xz=11$.

I also found that $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1$ and tried to apply some egyptian fractions methods but failed.

Answer 1

Hint: As you have tried, you may solve directly $1/a+1/b+1/c=1$ for $a=xy, b=yz, c=xz$ over positive integers.

The solutions are $(2,3,6), (2,4,4), (3,3,3)$ and permutations of these (see What are the integer solutions of 1/x+1/y+1/z=1?)

Then you can solve for $x,y,z$ and find that only $1,2,3$ satisfies your equation.

Answer 2

You say you found solutions without any method, so I assume that you are looking for a method to confirm what you found. Here is a method:

WLOG, we can assume that $x\le y \le z$ (i.e., the unknowns can be arranged according to their size without affecting the solution), which means that $x+y+z \le 3z$. This implies $xyz \le 3z$, which in turn implies $xy\le 3$. Hence, $x,y=1,1\text{ or }1,2 \text{ or }1,3$. It is a simple matter to test these three options.

$1+1+z=1\cdot 1\cdot z \Rightarrow 2+z=z$. This has no solutions in the positive integers.

$1+2+z=1\cdot 2\cdot z \Rightarrow 3+z=2z \Rightarrow z=3$

$1+3+z=1\cdot 3\cdot z \Rightarrow 4+z=3z \Rightarrow z=2$. This solution is identical to the previous case except for order.

So all solutions are found among $x,y,z=1,2,3$ and permutations thereof.