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). 
$$F(x,y,z) = \frac{(y+z)x^n}{(z-x)(y-x)} +\frac{(z+x)y^n}{(z-y)(x-y)}+\frac{(x+y)z^n}{(x-z)(y-z)} \in \mathbb Z.$$

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).
