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

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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.

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    $\begingroup$ How does the conclusion follow easily? $\endgroup$ – TonyK Oct 11 '16 at 14:48
  • $\begingroup$ @TonyK added the explanation, hope it is satisfactory, although not "orthodox". Do you know a more "analytic" approach? $\endgroup$ – G Cab Oct 11 '16 at 15:15
  • $\begingroup$ @TonyK, did not hear your point of view $\endgroup$ – G Cab Oct 13 '16 at 21:13

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