# Let $Q$ be a polynomial of degree 23 such that $Q(x)=-Q(-x)$

I came across the above problem. I do not know how to approach the problem.Can someone point me in the right direction? Thanks in advance for your time.

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Hint: If two polynomials of degree 23 have the same value at at least 24 points, then $\ldots$ – Jyrki Lahtonen Jan 16 '13 at 12:54

Hint: Split integral into two parts $$\int_{-1}^1(Q(x)+c)dx=\int_{-1}^0(Q(x)+c)dx + \int_0^1(Q(x)+c)dx$$ and apply variable change $t=-x$ in one of them.
Additionally, try to explain why $Q(x) = -Q(-x)$ must be true not only for $|x| \geq 10$. – Yoni Rozenshein Jan 16 '13 at 12:27
@Adam Yes,i did it and see that the value of c is $2$.Am i right? – learner Jan 16 '13 at 12:45
@learner: yes. But of course first you need to show that $Q(x)=-Q(-x)$ for $|x|\leq1$ as Yoni said in his comment. – Adam Jan 16 '13 at 12:48
$Q$ is an odd function, so it's integral over a interval symmetric about zero is zero. You can take it from there.