TITLE: Prove that $n^{2003}+n+1$ is composite for every $n\in \mathbb{N} \backslash\{1\}$
QUESTION [11 upvotes]: Prove that $n^{2003}+n+1$  is composite for every $n\in \mathbb{N} \backslash\{1\}$.
I tried with expanding $n^{2003}+1$, but I got nothing pretty not useful. I also couldn't get any improvement, let alone contradiction for assuming $n^{2003}+n+1=pq$ where $p,q\not= 1$. How should I do this and are there general tips on how to approach these problems, what to think about?

REPLY [10 votes]: 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.