Difference between revisions of "2020 INMO Problems/Problem 2"
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Latest revision as of 13:02, 5 November 2020
PROBLEM
Suppose is a polynomial with real coefficients, satisfying the condition , for every real . Prove that can be expressed in the formfor some real numbers and non-negative integer .
SOLUTION(1)
Assume to the contrary. Suppose satisfies for all real , and is of minimal degree and not of the prescribed form.
For some , we have .
Note that . Set . Then as the latter vanishes at both . Now let for some .
Then holds for all , by plugging in the original equation, since we have the identities and .
(Subtlety for beginners: while the equation in only holds for away from roots of , since these form a discrete subset of , the equation extends to these as is continuous.)
In particular, plugging we get and so , hence . Thus, as desired.
Finally, we see that and so has the prescribed form. But then also has the prescribed form, and our result follows. ~anantmdgal09