2004 AMC 12B 第 23 题

The integer zero is the sum of the other two, which must be conjugates $\dfrac{a}{2} \pm r$

The integer zero is half of $2004,$ so it is $1002;$ then count valid $r^2$

Let the integer zero be $a.$ The other two zeros are irrational conjugates $\dfrac{a}{2} \pm r,$ whose sum $a$ equals the integer zero. Vieta’s formula on the $x^2$ coefficient gives $a + a = 2004,$ so $a = 1002$ and the conjugate pair is $501 \pm r.$

The coefficients are integers exactly when $r^2$ is a positive integer, and the zeros are positive and distinct when $1 \le r^2 \le 501^2 - 1 = 251{,}000.$ Since $r$ cannot be an integer, we exclude the $500$ perfect-square values $r^2 = 1^2, \ldots, 500^2,$ leaving $251{,}000 - 500 = 250{,}500$ values of $n.$

Thus, the correct answer is C.