Let $f(x)$ be a polynomial with integer coefficients. Assume that 3 divides the value of $f(n)$ for each integer $n$. prove that when $f(x)$ is divided by $x^3-x $ , the remainder is of the form $3r(x)$, where $r(x)$ is a polynomial with integer coefficients.
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its supposed to x^3 -x... I made a typo – May 09 '14 at 14:29
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click on "edited ...ago" and you will see what is the difference with your original post, put your maths between $$. – Math137 May 09 '14 at 14:31
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Let $\pi : \mathbb{Z}[x] \to \mathbb{Z}_3[x]$ be the natural quotient map, and write $\overline{f} := \pi(f)$. By your hypothesis, $\overline{f}$ has 3 roots. Hence $$ \overline{f} = \overline{ax(x-1)(x+1)} = \overline{a(x^3-x)} $$ Hence $$ f(x) - a(x^3-x) \in \ker(\pi) = 3\mathbb{Z}[x] $$ which is what you want.
Prahlad Vaidyanathan
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