<s>[INST] Integer solutions for $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xyz}=1$ Is there a beautiful solution for this equation over the integers?  
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xyz}=1$$
 [/INST] $$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}+\frac{1}{xyz}\color{red}{=1+1=2}$$
Rewriting $1+\frac{1}{x}=\frac{x+1}{x}$ and mutliplying both sides by $xyz$ (assumed nonzero), we get $$(x+1)(y+1)(z+1)=2xyz$$
Now, $x,y,z$ are all integers, so this gives a lot of conditions.  For example, if we take $z=5$, we have $$(x+1)(y+1)6=10xy$$
or $$3(x+1)(y+1)=5xy$$
Hence, either $x$ or $y$ is a multiple of $3$.  Also, either $x+1$ or $y+1$ is a multiple of $5$.  Assume that we have $x=3k, y+1=5m$, so we rewrite as
$$3(3k+1)5m=5(3k)(5m-1)$$
Cancelling, we get $$(3k+1)m=k(5m-1)$$
If $m=1$, then $3k+1=4k$, so $k=1$, which gives $(x,y,z)=(3,4,5)$.
If $m=2$, then $6k+2=9k$, so $k=2/3$, which gives $(x,y,z)=(2,9,5)$.
If $m=3$, then $9k+3=14k$, so $k=3/5$, which does not give an integer solution.
Etc, etc, we can find as many solutions as we want by trial and error.
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