$x$, $y$, $z$ are positive integers such that $x + y \mid xy$ and $y + z \mid yz$. Prove that $\gcd(x, y, z) > 1$. 
$x$, $y$, $z$ are positive integers such that $x + y \mid xy$ and $y + z \mid yz$. Prove that $\gcd(x, y, z) > 1$.

Let $\gcd(x, y) = m$ and $\gcd(y, z) = n$, we have that $$\gcd(m, n) > 1 \implies \gcd(x, y, z) > 1$$
Suppose that $\gcd(m, n) = 1$ and $$x = mx', y = mny', z = nz' (\gcd(x', y', z') = 1)$$
We have that $$\left. \begin{align} \frac{xy}{x + y} = \frac{mx' \cdot mny'}{mx' + mny'} = \frac{mnx'y'}{x' + ny'}\\ \frac{yz}{y + z} = \frac{mny' \cdot nz'}{mny' + nz'} = \frac{mny'z'}{my' + z'} \end{align} \right\} \in \mathbb N \implies \frac{mnx'y'z'}{\gcd(x' + ny', my' + z')} \in \mathbb N$$
Then I was stuck.

If $x+y\mid xy$, then we have $xy=k(x+y)$ for some positive integer $k$.
Rewrite it as $(x-k)(y-k)=k^2$. Let $d=\gcd(x-k,y-k)$. We then have $x-k= a^2d, y-k=b^2d,k=dab$ for some coprime positive integers $a,b$. That is, $x=a(a+b)d$ and $y=b(a+b)d$.
Similarly, we have $y=p(p+q)s$ and $z=q(p+q)s$ for some positive integers $p,q,s$ with $p,q$ coprime.
Suppose $\gcd(x,y,z)=1$. Since $a+b$ divides $x,y$, it must be prime to $z$, and therefore prime to $(p+q)s$. But it divides $y=p(p+q)s$, so it must divide $p$. In particular, we have $p\geq a+b$.
Similarly, $p+q$ must divide $b$, and hence $b\geq p+ q$.
This is a contradiction.
