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). 
$$F(x,y,z) = \frac{(y+z)x^n}{(z-x)(y-x)} +\frac{(z+x)y^n}{(z-y)(x-y)}+\frac{(x+y)z^n}{(x-z)(y-z)} \in \mathbb Z.$$
 A: If $n = 1$, then the whole expression is equal to $-1$. If $n=2$, it is equal to $0$. Let us therefore assume that $n > 2$.
We have:
$$F(x,y,z) = \frac{y^nz^2 - y^2z^n + x^n(y-z)(y+z) + x^2(z^n-y^n)}{(x-y)(x-z)(y-z)} = \frac{f(x,y,z)}{(x-y)(x-z)(y-z)}.$$
What we want to know is if the numerator is divisible by $x-y$, $x-z$ and $y-z$. Let's check (for example) $y-z$.
It is clear that $y-z$ divides the second term.
Finally, observe that
$$y^nz^2 - y^2z^n + x^2(z^n-y^n) = y^2z^2(y^{n-2}-z^{n-2}) - x^2(y^n-z^n)$$
and
$$a^k-b^k = (a-b)(a^{k-1} + a^{k-2}b + \dots + b^{k-1}).$$
We apply this with $a = y$ and $b = z$ after noticing that if $a,b \in \mathbb Z$, then $a+b$ and $ab$ is an integer (so the expression in brackets is indeed integer).
A: 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 $\,f(x,y,z) = x^n(y^2\!-\!z^2)$
$\smash[b]{\require{cancel}\ \ \ {\rm mod}\,\ x\!-\!y\!:\ \ \color{#c00}{y\equiv x}\ \Rightarrow\ G\equiv\!\! \cancel{f(x,\color{#c00}x,z)}\!\! +\!\!\!\!\!\!\!\!\! \underbrace{f(\color{#c00}x,\color{#0a0}{z,x})}_{\Large \cancel{-f(x,\color{#0a0}{x,z})}\quad\ \  }\!\!\!\!\!\!\!\!+\!\underbrace{f(z,x,\color{#c00}x)}_{\large 0}\equiv 0\ \Rightarrow\ x\!-y\mid G}$
$\ \ \ {\rm by}\ \ f(x,\color{#0a0}{y,z})\, =\, -f(x,\color{#0a0}{z,y})\,$ 
