Giving integer images (bis) Prove the statement below (the restriction $x < y < z$  is to avoid apparent uncertainties but the property is valid for all $x, y, z$ really). 
$$F(x,y,z) = \frac{(y+z)x^n}{(z-x)(y-x)} +\frac{(z+x)y^n}{(z-y)(x-y)}+\frac{(x+y)z^n}{(x-z)(y-z)} \in \mathbb Z.$$

This is straightforward if one exploits the innate symmetry in the expression. Namely
$G = -(x\!-\!y)(y\!-\!z)(z\!-\!y)F = f(x,\color{#c00}y,z)\!+\!f(\color{#c00}y,z,x)\!+\!f(z,x,\color{#c00}y)\, $ for $\,f(x,y,z) = x^n(y^2\!-\!z^2)$
$\smash[b]{\require{cancel}\ \ \ {\rm mod}\,\ x\!-\!y\!:\ \ \color{#c00}{y\equiv x}\ \Rightarrow\ G\equiv\!\! \cancel{f(x,\color{#c00}x,z)}\!\! +\!\!\!\!\!\!\!\!\! \underbrace{f(\color{#c00}x,\color{#0a0}{z,x})}_{\Large \cancel{-f(x,\color{#0a0}{x,z})}\quad\ \  }\!\!\!\!\!\!\!\!+\!\underbrace{f(z,x,\color{#c00}x)}_{\large 0}\equiv 0\ \Rightarrow\ x\!-y\mid G}$
$\ \ \ {\rm by}\ \ f(x,\color{#0a0}{y,z})\, =\, -f(x,\color{#0a0}{z,y})\,$ 
