Easy to get the polynomial $f(x)$: $f(x)=P(x)(x^2-x+1)(x+2)(x+1)x(x-1)$ for any $x \in \mathbb{R}$

Scorpion.k48 wrote: Easy to get the polynomial $f(x)$: $f(x)=P(x)(x^2-x+1)(x+2)(x+1)x(x-1)$ for any $x \in \mathbb{R}$ I think the main part is to prove that $(n^2-n+1)(n+2)(n+1)n(n-1)$ has at least five distinct prime divisors for all $n\geq 8$.

Scorpion.k48 wrote: Easy to get the polynomial $f(x)$: $f(x)=P(x)(x^2-x+1)(x+2)(x+1)x(x-1)$ for any $x \in \mathbb{R}$ Why $f(x)=P(x)(x^2-x+1)(x+2)(x+1)x(x-1)$ for any $x \in \mathbb{R}$[/quote]

Haismit wrote: Why $f(x)=P(x)(x^2-x+1)(x+2)(x+1)x(x-1)$ for any $x \in \mathbb{R}$ Note that $x^3-2x^2+2x-1=(x-1)(x^2-x+1)$ and $x^3+4x^2+4x+3=(x+3)(x^2+x+1)$. So we have $(x+3)(x^2+x+1)f(x)=(x-1)(x^2-x+1)f(x+1)$. Put $x=1$ to obtain $f(1)=0$. Put $x=-3$ to obtain $f(-2)=0$. So $f(x)=(x-1)(x+2)g(x)$ for some polynomial $g(x)$. Plugging this into the equation and cancelling, you have $(x+2)(x^2+x+1)g(x)=(x^2-x+1)xg(x+1)$. Put $x=0$ to obtain $g(0)=0$. Put $x=-2$ to obtain $g(-1)=0$. So $g(x)=x(x+1)h(x)$ for some polynomial $h(x)$. Plugging this into the equation and cancelling, we have $(x^2+x+1)h(x)=(x^2-x+1)h(x+1)$. Since the roots of $x^2-x+1$ and $x^2+x+1$ are disjoint, we must have $h(x)=(x^2-x+1)q(x)$ for some polynomial $q(x)$. Plugging this into the equation and cancelling, we have $q(x)=q(x+1)$ and hence $q$ is constant. So $f(x)=c(x-1)x(x+1)(x+2)(x^2-x+1)$ for some constant $c \in \mathbb{Z} \backslash \{0\}$.