$x = \frac{\sqrt{3}+1}{\sqrt{3}-1}$ , $y = \frac{\sqrt{3}-1}{\sqrt{3}+1}$ のとき、以下の式の値を求める。 (1) $x+y$ (2) $xy$ (3) $x^2+y^2$ (4) $x^3+y^3$ (5) $x^5+y^5$

1. 問題の内容

$x = \frac{\sqrt{3}+1}{\sqrt{3}-1}$ , $y = \frac{\sqrt{3}-1}{\sqrt{3}+1}$ のとき、以下の式の値を求める。

(1) $x+y$

(2) $xy$

(3) $x^2+y^2$

(4) $x^3+y^3$

(5) $x^5+y^5$

2. 解き方の手順

まず、$x$と$y$を簡単にします。

$x = \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$

$y = \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$

(1) $x + y = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$

(2) $xy = (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$

(3) $x^2 + y^2 = (x+y)^2 - 2xy = 4^2 - 2(1) = 16 - 2 = 14$

(4) $x^3 + y^3 = (x+y)(x^2 - xy + y^2) = (x+y)((x+y)^2 - 3xy) = 4(4^2 - 3(1)) = 4(16 - 3) = 4(13) = 52$

(5) $x^5 + y^5 = (x^2+y^2)(x^3+y^3) - x^2y^2(x+y) = 14(52) - 1^2(4) = 728 - 4 = 724$

または、別の解法：

$x^5+y^5 = (x+y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4) = (x+y)(x^4 + y^4 -xy(x^2+y^2) + (xy)^2)$

ここで、$x^4 + y^4 = (x^2+y^2)^2 - 2(xy)^2 = 14^2 - 2(1)^2 = 196 - 2 = 194$

よって、$x^5+y^5 = 4(194 - (1)(14) + (1)^2) = 4(194 - 14 + 1) = 4(181) = 724$

3. 最終的な答え

(1) $x+y = 4$

(2) $xy = 1$

(3) $x^2+y^2 = 14$

(4) $x^3+y^3 = 52$

(5) $x^5+y^5 = 724$