prove the least postive integer $4$ can define the form $(a-b)(c-d)$ Below the conjecture look is true, but how to prove?

Let $x,y,z,w\in Q^{+}$,and such $xyzw=1$,if postive integer $n=(x-y)(z-w)$,show that:$$n_{\min}=4?$$
  because By Now I found this example(other words,I can't find any example is $1,2,3$)
  $$4=\left(2-\dfrac{1}{2}\right)\left(3-\dfrac{1}{3}\right)$$

 A: So you have
$$
\left\{ \begin{gathered}
  0 < \text{integer}\,n \hfill \\
  \left( {x - y} \right)\left( {z - w} \right) = n \hfill \\
  x\,y\,z\,w = 1 \hfill \\ 
\end{gathered}  \right.
$$
which translates into
$$
\left\{ \begin{gathered}
  n = \left( {x - y} \right)\left( {z - w} \right) = \frac{{\left( {x - y} \right)\left( {z - w} \right)}}
{{x\,y\,z\,w}} = \left( {\frac{1}
{y} - \frac{1}
{x}} \right)\left( {\frac{1}
{w} - \frac{1}
{z}} \right) \hfill \\
  x\,y\,z\,w = 1 \hfill \\
  \frac{1}
{x}\frac{1}
{y}\frac{1}
{z}\frac{1}
{w} = 1 \hfill \\ 
\end{gathered}  \right.
$$
Now, apart from the trivial solution $(1,1,1,1)$ which would give $n=0$,
the inversion symmetry that appears above tells you that for $1<x,y$, you can have either
$$
n = \left( {x - \frac{1}
{y}} \right)\left( {y - \frac{1}
{x}} \right) = \frac{{\left( {xy - 1} \right)^2 }}
{{xy}}
$$
or:
$$
n = \left( {x - \frac{1}
{x}} \right)\left( {y - \frac{1}
{y}} \right) = \frac{{\left( {x^2  - 1} \right)\left( {y^2  - 1} \right)}}
{{xy}} = \frac{{\left( {x + 1} \right)\left( {x - 1} \right)\left( {y + 1} \right)\left( {y - 1} \right)}}
{{xy}}
$$
and the conclusion follows easily, just because $n$ is increasing with $x$ and $y$, and there is just to find the first couple $(x,y)$ that renders one of the above expression integral, i.e. $(2,3)$ with the 2nd expression.
