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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?

Lazar Ljubenović
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2 Answers2

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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.

Zero
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    Can you give some insight on how you found this. (Interesting approach) – Kirthi Raman Mar 10 '12 at 22:16
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    @KirthiRaman I noticed that $2003\equiv 2\pmod{3}$ and that the polynomial $n^{2003}+n+1$ have only 1's and 0's as coefficients. This approach can be generalized to polynomials of the form $x^m+x^{m-1}+...+x+1$, for example $1+x+x^2+x^3|x^{23}+x^6+x+1$ – Zero Mar 10 '12 at 22:21
  • @Kirthi You can find a couple algebraic derivations in my answer. – Bill Dubuque Mar 10 '12 at 23:38
  • Thanks to both Bill Dubuque and Diego S. (I am catching up with what I have forgotten all these years) – Kirthi Raman Mar 11 '12 at 01:05
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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.

Bill Dubuque
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