$x = \frac{\sqrt{5}+1}{2}$、$y = \frac{\sqrt{5}-1}{2}$ のとき、以下の値を求めます。 (1) $x^2 + y^2$ (2) $\frac{y}{x^2} + \frac{x}{y^2}$ (3) $x^5 + y^5$

1. 問題の内容

$x = \frac{\sqrt{5}+1}{2}$、$y = \frac{\sqrt{5}-1}{2}$ のとき、以下の値を求めます。

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

(2) $\frac{y}{x^2} + \frac{x}{y^2}$

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

2. 解き方の手順

(1) $x^2 + y^2$ を求めます。

$x^2 = \left(\frac{\sqrt{5}+1}{2}\right)^2 = \frac{5 + 2\sqrt{5} + 1}{4} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2}$

$y^2 = \left(\frac{\sqrt{5}-1}{2}\right)^2 = \frac{5 - 2\sqrt{5} + 1}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}$

$x^2 + y^2 = \frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} = \frac{3 + \sqrt{5} + 3 - \sqrt{5}}{2} = \frac{6}{2} = 3$

(2) $\frac{y}{x^2} + \frac{x}{y^2}$ を求めます。

$\frac{y}{x^2} + \frac{x}{y^2} = \frac{y^3 + x^3}{x^2 y^2}$

$x+y = \frac{\sqrt{5}+1}{2} + \frac{\sqrt{5}-1}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5}$

$xy = \frac{\sqrt{5}+1}{2} \times \frac{\sqrt{5}-1}{2} = \frac{5-1}{4} = \frac{4}{4} = 1$

$x^3 + y^3 = (x+y)^3 - 3xy(x+y) = (\sqrt{5})^3 - 3(1)(\sqrt{5}) = 5\sqrt{5} - 3\sqrt{5} = 2\sqrt{5}$

$x^2 y^2 = (xy)^2 = 1^2 = 1$

$\frac{y}{x^2} + \frac{x}{y^2} = \frac{x^3 + y^3}{x^2 y^2} = \frac{2\sqrt{5}}{1} = 2\sqrt{5}$

(3) $x^5 + y^5$ を求めます。

$x^2 + y^2 = 3$

$x^3 + y^3 = 2\sqrt{5}$

$(x^2+y^2)(x^3+y^3) = x^5 + x^2 y^3 + y^2 x^3 + y^5 = x^5 + y^5 + x^2 y^2 (y + x)$

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

$x^5 + y^5 = (3)(2\sqrt{5}) - (1)^2(\sqrt{5}) = 6\sqrt{5} - \sqrt{5} = 5\sqrt{5}$

3. 最終的な答え

(1) 3

(2) $2\sqrt{5}$

(3) $5\sqrt{5}$