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1971 Canadian MO Problems/Problem 5

Revision as of 14:08, 19 January 2012 by Thepisong (talk | contribs) (Created page with "<math>p(0)=a_nx^{n}+a_{n-1}x^{n-1}+...a_1x+a_0=a_0</math> We know that p(0) is odd, so we know <math>a_0</math> is odd. By Vieta's this means that all the roots of polynomial p...")
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$p(0)=a_nx^{n}+a_{n-1}x^{n-1}+...a_1x+a_0=a_0$

We know that p(0) is odd, so we know $a_0$ is odd.

By Vieta's this means that all the roots of polynomial p(x) are also odd.

$p(1)=a_n(x-r_1)(x-r_2)...(x-r_n)=a_n(1-r_1)(1-r_2)...(1-r_n)$

Since all the roots $r_k$ where ${1}\le{k}\le{n}$ are odd, $1-r_k$ must be even, and $p(1)$ must be even.

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