<s>[INST] Prove that $n^5+n^4+1$ is composite for $n>1.$ 
Prove that $f(n)=n^5+n^4+1$ is composite for $n>1, n\in\mathbb{N}$.

This problem appeared on a local mathematics competition, however it looks like there is no simple method to solve it. I tried multiplying it by $n+1$ or $n-1$ and then tried factorizing it, but it was too tough for me.
Any help will be appreciated.
 [/INST] If $\omega$ is a (complex) root of $x^2+x+1$, then $\omega^3 - 1 = (\omega-1)(\omega^2+\omega+1) = (\omega-1)\cdot 0 = 0$, and hence $\omega^3 = 1$.
Observe then that $\omega^5 = \omega^3 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2$ and $\omega^4 = \omega^3 \cdot \omega^1 = 1 \cdot \omega = \omega$, so that $\omega^5+\omega^4 +1 = \omega^2 +\omega + 1 = 0$, hence $\omega$ is a root of $x^5+x^4+1$. Hence both roots of $x^2+x+1$ are roots of $x^5+x^4+1$ and thus $(x^2 + x + 1) \mid (x^5+x^4+1)$.
Note that by a similar argument we can show that $x^2+x+1$ is a factor of $x^8+x^7 + 1$ or even of $x^{101}+x^4+1$. All that matters is that the exponents give all the different residues modulo $3$. 
This problem appears as E$13$ in the Number Theory Chapter of Arthur Engel's Problem Solving Strategies. 
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