Suppose that has the property that for anyevenwe have no real roots. We can easy show that has property that is has exactly one real root. Note that where is even, and so the derivative is non-vanishing for , thus it is an increasing odd polynomial, thus, it has exactly one root.

There is an elementary way to prove that for even, which will complete what you want to show. Using this we can prove this. Let then if is even. This mean .

I will try to come up with an approximating solution to this problem also, since I like putting bounds on functions.