Solution Of Diophantine Equations 
Find all integer solutions for the equation:
  $$(x+y)(y+z)(z+x)=txyz$$
  such that gcd$(x, y)=1$, gcd$(y, z)=1$ and gcd$(z, x)=1$

Now we can write our equation as
$$(\frac{x+y}{y})(\frac{y+z}{z})(\frac{x+z}{x})=t$$
This gives $$(1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=t$$
Now as $t$ is an integer and gcd$(x, y)=1$, gcd$(y, z)=1$ and gcd$(z, x)=1$ so the only possibility is $$x=y=z=1$$ giving a solution as $$(x, y, z, t)=(1, 1, 1, 8)$$
Am I Right?
 A: Romania TST, 1995. This is the cleanest way of doing it:
Note, $x+y\equiv y\pmod{x}$, and $x+z\equiv z\pmod{x}$. Since $(x,yz)=1$, it follows that, $x\mid y+z$, and thus, $x\mid x+y+z$ (this is similar to above). From symmetry, $y\mid x+y+z$ and $z\mid x+y+z$ are also obtained. Now, since $x,y,z$ are pairwise coprime, we have $xyz\mid x+y+z$. Thus,
$$
\frac{x+y+z}{xyz} = \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz} \in\mathbb{Z}^+.
$$
From here, it is not hard to conclude (all three equal to one, exactly two one, only one is one, and the last case, all three $\geqslant 2$, in which case the object above is less than $1$). 
A: Hint:
Say $x,y,z$ are positive. We have $x|y+z$ and $y|x+z$ and $z|x+y$. We can assume $x\geq y\geq z$. Then $y+z= kx$ where $k=1$ or $k=2$.
If k=1we have $(2y+z)(2z+y)=tyz$ and so $y|2z^2$ so $y=1$ or $y=2$.
If k=2 we have $(3y+z)(3z+y) =2tyz$ so $y| 3z^2$ so $y=1$ or $y=3$...
A: The number of variables can be reduced so your equation can be written as:
$$a+\frac {1}{a}+b+\frac{1}{b}+ab+\frac{1}{ab}=t-2$$
