SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Question

Let P(x) be a polynomial of degree n 2 with rational coefficients such that P(x) has n pairwise different real roots forming an arithmetic progression. Prove that among the roots of P(x) there are two that are also the roots of some polynomial of degree 2 with rational coefficients.

Accepted Answer

Denote x_1 < x_2 < < x_n as the roots of P(x). Let d = x_n - x_n-1 = = x_2 - x_1 > 0. Since P(x) has rational coefficients then by applying Vieta's theorem, we have _i=1^n x_i Q and _1 i < j n x_i x_j Q . Note that _i=1^n x_i = n(x_1 + x_n)2 so x_1 + x_n Q. On the other hand, _i=1^n x_i^2 = (_i=1^n x_i)^2 - 2 _1 i < j n x_i x_j Q and aligned _i=1^n x_i^2 & = _i=1^n (x_1 + (i-1)d)^2 = n x_1^2 + n(n-1) x_1 d + n(n-1)(2n-1)6 d^2 & = n(x_1 + (n-1)d2)^2 + n^2(n^2-1)12 d^2 & = n(x_1 + x_n)^24 + n(n^2-1)12 d^2 aligned From these, we can conclude that n(n^2-1)12 d^2 Q so d^2 Q. Thus x_1 x_n = (x_1 + x_n)^2 - (x_n - x_1)^24 = (x_1 + x_n)^2 - (n-1)^2 d^24 Q . Therefore, x_1 + x_n, x_1 x_n Q which implies that two numbers x_1, x_n satisfy.