計算第 1 題
f(x) = Q_1(x)(x - 1)^2 + ax + b
= Q_2(x)(x + 1)^2 + bx + a
= Q_3(x)(x - 1)^2(x + 1)^2 + cx^3 + dx^2 + ex
ac ≠ 0
f(1) = a + b = c + d + e
f(-1) = a - b = - c + d - e
可解出 a = d，b = c + e
f '(1) = a = 3c + 2d + e
f '(-1) = b = 3c - 2d + e
d = 3c + 2d + e
c + e = 3c - 2d + e
可解出 d = c，e = -4c
R(x) = cx^3 + dx^2 + ex = cx^3 + cx^2 - 4cx = 0
x(x^2 + x - 4) = 0
x = 0 or (-1 ± √17)/2
計算第 2 題
(1) 當 n + 1 為完全平方數，且有 m 個正因數，易知 m 是正奇數
[√(n + 1)] = [√n] + 1
Σ[(n + 1) / k] (k = 1 ~ n + 1) = Σ[n / k] (k = 1 ~ n) + m
[√(n + 1)] + Σ[(n + 1) / k] (k = 1 ~ n + 1) = [√n] + Σ[n / k] (k = 1 ~ n) + (m + 1)
[√(n + 1)] + Σ[(n + 1) / k] (k = 1 ~ n + 1) 和 [√n] + Σ[n / k] (k = 1 ~ n) 同奇或同偶
(2) 當 n + 1 不為完全平方數，且有 m 個正因數，易知 m 是正偶數
[√(n + 1)] = [√n]
Σ[(n + 1) / k] (k = 1 ~ n + 1) = Σ[n / k] (k = 1 ~ n) + m
[√(n + 1)] + Σ[(n + 1) / k] (k = 1 ~ n + 1) = [√n] + Σ[n / k] (k = 1 ~ n) + m
[√(n + 1)] + Σ[(n + 1) / k] (k = 1 ~ n + 1) 和 [√n] + Σ[n / k] (k = 1 ~ n) 同奇或同偶
(3) 而 n = 1 時，[√1] + [1 / 1] = 2，是偶數
故對任意正整數 n， [√n] + Σ[n / k] (k = 1 ~ n) 必為偶數
