Find the least positive integer , such that there is a polynomial with real coefficients that satisfies both of the following properties:
For it is .
There is a real number with .
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Clearly, for n = 2 0 1 4 we can just take a 2 k = 2 0 1 4 , a 2 k + 1 = 2 0 1 5 such that P ( − 1 ) = 0 .
Assume we have such a polynomial with degree 2 n < 2 ⋅ 2 0 1 4 which has a real root. Clearly P ( x ) ≥ 2 0 1 4 > 0 for x ≥ 0 and hence the root must be negative.
So Q ( x ) = P ( − x ) must have a positive real root.
But for any positive x we have Q ( x ) ≥ 2 0 1 4 x 2 n − 2 0 1 5 x 2 n − 1 + 2 0 1 4 x 2 n − 2 − … − 2 0 1 5 x + 2 0 1 4 = R ( x ) .
It is now sufficient to prove that R ( x ) > 0 for x > 0 .
We have R ( 1 ) = 2 0 1 4 ( n + 1 ) − 2 0 1 5 n = 2 0 1 4 − n > 0 . So now assume x = 1 .
Then R ( x ) > 0 is equivalent to 2 0 1 4 ( x 2 n + x 2 n − 2 + … + 1 ) ≥ 2 0 1 5 ( x 2 n − 1 + … + x ) which is likewise equivalent to 2 0 1 4 x 2 − 1 x 2 n + 2 − 1 ≥ 2 0 1 5 x x 2 − 1 x 2 n − 1 equivalent to x ( x 2 n − 1 ) x 2 n + 2 − 1 ≥ 1 + 2 0 1 4 1 .
Since n < 2 0 1 4 it is sufficient to prove x ( x 2 n − 1 ) x 2 n + 2 − 1 ≥ 1 + n 1 = n n + 1 which is equivalent (after the same transformations as above) to n x 2 n + n x 2 n − 2 + … + n x 2 + n ≥ ( n + 1 ) x 2 n − 1 + ( n + 1 ) x 2 n − 3 + … + ( n + 1 ) x for all positive x .
But for any k we have n x 2 k + 1 ≤ 2 n x 2 k + 2 n x 2 k + 2 by AM-GM inequality.
Additionally we have x k + x 2 n − k ≤ x 2 n + 1 (which is equivalent to ( x i − 1 ) ( x 2 n − i − 1 ) ≥ 0 which is clearly true).
Adding the first type of inequality for all k from 0 to n − 1 and the second type for all odd k from 1 to 2 n − 1 we get the desired result.
Hence for n < 2 0 1 4 such a polynomial can't exist. And thus, the answer is 2 0 1 4 .