Let $ f(x)=5x^9+6x^8+3x^6+8x^5+9x^3+6x^2+8x+3 $. Prove that $x^nf(x)+12$ is reducible in $\mathbb Z[x]$ for any positive integer $n$.
This problem is due to polish mathematician Andrzej Schinzel.
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1see this paper. – mathlove Feb 21 '15 at 09:40
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This is a typical application of covering congruences as in this answer. – Bill Dubuque Feb 28 '15 at 02:08
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Let $\omega=e^{\frac{\pi i}{6}}$ be a primitive twelfth root of unity. Since: $$ f(\omega) = f(i) = 12i,\qquad f(-1)=-12,\qquad f(\omega^2)=12\omega^4$$ for any $n\in\mathbb{N}$ we have that $x^n f(x)+12$ is divided by some polynomial among: $$ (x+1),\quad(x^2+1),\quad(x^2+x+1),\quad(x^2-x+1),\quad(x^4-x^2+1).$$
Jack D'Aurizio
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What do the roots of unity have anything to do with? This was done by modular arithmetics in the paper. – mathreadler Feb 21 '15 at 10:59
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1@mathreadler: just to be clear, to compute $p(i)$ is more or less the same as computing $p(x)\pmod{(x^2+1)}$. – Jack D'Aurizio Feb 21 '15 at 11:20
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