Restriction of coefficients

Algebra Level 5

Find the least positive integer n n , such that there is a polynomial P ( x ) = a 2 n x 2 n + a 2 n 1 x 2 n 1 + + a 1 x + a 0 P(x) = a_{2n}x^{2n}+a_{2n-1}x^{2n-1}+\dots+a_1x+a_0 with real coefficients that satisfies both of the following properties:

  • For i = 0 , 1 , , 2 n i=0,1,\ldots,2n it is 2014 a i 2015 2014 \leq a_i \leq 2015 .

  • There is a real number ξ \xi with P ( ξ ) = 0 P(\xi)=0 .


The answer is 2014.

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1 solution

Clearly, for n = 2014 n=2014 we can just take a 2 k = 2014 , a 2 k + 1 = 2015 a_{2k}=2014, a_{2k+1}=2015 such that P ( 1 ) = 0 P(-1)=0 .

Assume we have such a polynomial with degree 2 n < 2 2014 2n<2\cdot2014 which has a real root. Clearly P ( x ) 2014 > 0 P(x) \ge 2014>0 for x 0 x \ge 0 and hence the root must be negative.

So Q ( x ) = P ( x ) Q(x)=P(-x) must have a positive real root.

But for any positive x x we have Q ( x ) 2014 x 2 n 2015 x 2 n 1 + 2014 x 2 n 2 2015 x + 2014 = R ( x ) Q(x) \ge 2014x^{2n}-2015x^{2n-1}+2014x^{2n-2}-\ldots-2015x+2014=R(x) .

It is now sufficient to prove that R ( x ) > 0 R(x)>0 for x > 0 x>0 .

We have R ( 1 ) = 2014 ( n + 1 ) 2015 n = 2014 n > 0 R(1)=2014(n+1)-2015n=2014-n>0 . So now assume x 1 x \ne 1 .

Then R ( x ) > 0 R(x)>0 is equivalent to 2014 ( x 2 n + x 2 n 2 + + 1 ) 2015 ( x 2 n 1 + + x ) 2014(x^{2n}+x^{2n-2}+\ldots+1) \ge 2015(x^{2n-1}+\ldots+x) which is likewise equivalent to 2014 x 2 n + 2 1 x 2 1 2015 x x 2 n 1 x 2 1 2014\dfrac{x^{2n+2}-1}{x^2-1} \ge 2015x\dfrac{x^{2n}-1}{x^2-1} equivalent to x 2 n + 2 1 x ( x 2 n 1 ) 1 + 1 2014 \dfrac{x^{2n+2}-1}{x(x^{2n}-1)} \ge 1+\dfrac{1}{2014} .

Since n < 2014 n<2014 it is sufficient to prove x 2 n + 2 1 x ( x 2 n 1 ) 1 + 1 n = n + 1 n \dfrac{x^{2n+2}-1}{x(x^{2n}-1)} \ge 1+\dfrac{1}{n}=\dfrac{n+1}{n} 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 nx^{2n}+nx^{2n-2}+\ldots+nx^2+n \ge (n+1)x^{2n-1}+(n+1)x^{2n-3}+\ldots+(n+1)x for all positive x x .

But for any k k we have n x 2 k + 1 n 2 x 2 k + n 2 x 2 k + 2 nx^{2k+1} \le \dfrac{n}{2}x^{2k}+\dfrac{n}{2}x^{2k+2} by AM-GM inequality.

Additionally we have x k + x 2 n k x 2 n + 1 x^k+x^{2n-k} \le x^{2n}+1 (which is equivalent to ( x i 1 ) ( x 2 n i 1 ) 0 (x^i-1)(x^{2n-i}-1) \ge 0 which is clearly true).

Adding the first type of inequality for all k k from 0 0 to n 1 n-1 and the second type for all odd k k from 1 1 to 2 n 1 2n-1 we get the desired result.

Hence for n < 2014 n<2014 such a polynomial can't exist. And thus, the answer is 2014 \boxed{2014} .

Great problem!

Julian Poon - 5 years, 8 months ago

Can you elaborate on your 4th statement (in which you introduce R(x))?

Abhishek Sharma - 5 years, 8 months ago

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