# 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 $1,2,3$) $$4=\left(2-\dfrac{1}{2}\right)\left(3-\dfrac{1}{3}\right)$$

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.