Prove that $n^{2003}+n+1$ is composite for every $n\in \mathbb{N} \backslash\{1\}$ [duplicate]

Let $w=e^{i2\pi/3}$. It's easy to see that $w$ and $w^2$ are all the roots of $x^2+x+1$ and roots of $x^{2003}+x+1$, therefore $x^2+x+1|x^{2003}+x+1$. So we have That $x^{2003}+x+1=(x^2+x+1)P(x)$, where $P(x)$ is some polynomial with integer coefficients. For $x\ge 2$, $x^{2003}+x+1$ is much bigger than $x^2+x+1$ so $P(x)$ is some integer greater than $2$ from which the conclusion follows.

Hint $\rm\ f = x^{3n+2}+x+1\ = \ x^2\:(\color{#0a0}{x^{3n}-1})\, +\, x^2+x+1\ $

therefore: $\,\ \rm\ x^2+x+1\ |\ x^3-1\ |\ \color{#0a0}{x^{3n}-1}\:\Rightarrow\: x^2+x+1\ |\ f$

Or $\rm\bmod\, x^2+x+1\!:\,\ \color{#c00}{x^3\equiv 1}\ \Rightarrow\ x^{3n+2}+x+1\equiv (\color{#c00}{x^3})^n x^2 + x + 1 \equiv x^2+x+1\equiv 0$

Remark $ $ Generally $\,\rm x^2+x+1\mid x^I + x^J + x^K\ $ if $\rm \ \{I, J, K\}\equiv \{0,1,2\}\pmod{\! 3}\,$ by here, which is a special case of the method of simpler multiples.