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- Thread starter jacks
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- Thread starter
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- Jan 26, 2012

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There are details you may need to fill in yourself but:A polynomial $f(x)$ has Integer Coefficients such that $f(0)$ and $f(1)$ are both odd numbers. prove that $f(x) = 0$ has no Integer solution

\(f(0)\) odd implies that the constant term is odd

then \(f(1)\) odd implies that there are an even number of odd coeficients of the non constant terms.

So if \(x \in \mathbb{Z}\) then \(f(x)\) is odd and so cannot be a root of \(f(x)\)

CB