与えられた10個の方程式について、整数解 $(x, y)$ を全て求める問題です。

## 解答

1. 問題の内容

与えられた10個の方程式について、整数解 $(x, y)$ を全て求める問題です。

2. 解き方の手順

**(1) $xy = 7$**

7の約数は1, 7のみであるので、

$x = 1, y = 7$

$x = 7, y = 1$

$x = -1, y = -7$

$x = -7, y = -1$

**(2) $xy = -5$**

-5の約数は-1, 1, -5, 5のみであるので、

$x = 1, y = -5$

$x = -1, y = 5$

$x = 5, y = -1$

$x = -5, y = 1$

**(3) $xy = 8$**

8の約数は1, 2, 4, 8のみであるので、

$x = 1, y = 8$

$x = 2, y = 4$

$x = 4, y = 2$

$x = 8, y = 1$

$x = -1, y = -8$

$x = -2, y = -4$

$x = -4, y = -2$

$x = -8, y = -1$

**(4) $(x+1)(y-4) = 3$**

3の約数は1, 3のみであるので、

$x+1 = 1, y-4 = 3 \implies x = 0, y = 7$

$x+1 = 3, y-4 = 1 \implies x = 2, y = 5$

$x+1 = -1, y-4 = -3 \implies x = -2, y = 1$

$x+1 = -3, y-4 = -1 \implies x = -4, y = 3$

**(5) $(x-1)(y+2) = -12$**

-12の約数の組み合わせを考える。

$x-1 = 1, y+2 = -12 \implies x = 2, y = -14$

$x-1 = -1, y+2 = 12 \implies x = 0, y = 10$

$x-1 = 2, y+2 = -6 \implies x = 3, y = -8$

$x-1 = -2, y+2 = 6 \implies x = -1, y = 4$

$x-1 = 3, y+2 = -4 \implies x = 4, y = -6$

$x-1 = -3, y+2 = 4 \implies x = -2, y = 2$

$x-1 = 4, y+2 = -3 \implies x = 5, y = -5$

$x-1 = -4, y+2 = 3 \implies x = -3, y = 1$

$x-1 = 6, y+2 = -2 \implies x = 7, y = -4$

$x-1 = -6, y+2 = 2 \implies x = -5, y = 0$

$x-1 = 12, y+2 = -1 \implies x = 13, y = -3$

$x-1 = -12, y+2 = 1 \implies x = -11, y = -1$

**(6) $xy - 2x + 5y = 1$**

$xy - 2x + 5y - 10 = 1 - 10$

$x(y-2) + 5(y-2) = -9$

$(x+5)(y-2) = -9$

-9の約数の組み合わせを考える。

$x+5 = 1, y-2 = -9 \implies x = -4, y = -7$

$x+5 = -1, y-2 = 9 \implies x = -6, y = 11$

$x+5 = 3, y-2 = -3 \implies x = -2, y = -1$

$x+5 = -3, y-2 = 3 \implies x = -8, y = 5$

$x+5 = 9, y-2 = -1 \implies x = 4, y = 1$

$x+5 = -9, y-2 = 1 \implies x = -14, y = 3$

**(7) $2xy + 4x + 3y = 0$**

$4xy + 8x + 6y = 0$

$2x(2y+4) + 3y = 0$

$2x(2y+3) + 4x + 3y = 0$

$4xy+8x+6y+12=12$

$2x(2y+4)+3(2y+4)=12$

$(2x+3)(2y+4)=12$

$(2x+3)(y+2)=6$

$2x+3 = 1, y+2=6 \implies x = -1, y=4$

$2x+3 = 3, y+2=2 \implies x=0, y=0$

$2x+3 = 6, y+2=1 \implies x=\frac{3}{2}, y=-1$ (xが整数でないので不可)

$2x+3 = -1, y+2=-6 \implies x = -2, y=-8$

$2x+3 = -3, y+2=-2 \implies x=-3, y=-4$

$2x+3 = -6, y+2=-1 \implies x = -\frac{9}{2}, y=-3$ (xが整数でないので不可)

**(8) $\frac{1}{x} + \frac{1}{y} = \frac{1}{5}$**

$\frac{x+y}{xy} = \frac{1}{5}$

$5(x+y) = xy$

$xy - 5x - 5y = 0$

$xy - 5x - 5y + 25 = 25$

$(x-5)(y-5) = 25$

25の約数の組み合わせを考える。

$x-5 = 1, y-5 = 25 \implies x = 6, y = 30$

$x-5 = 5, y-5 = 5 \implies x = 10, y = 10$

$x-5 = 25, y-5 = 1 \implies x = 30, y = 6$

$x-5 = -1, y-5 = -25 \implies x = 4, y = -20$

$x-5 = -5, y-5 = -5 \implies x = 0, y = 0$ (x,yが分母に来るので不可)

$x-5 = -25, y-5 = -1 \implies x = -20, y = 4$

**(9) $\frac{2}{x} + \frac{5}{y} = 1$**

$2y + 5x = xy$

$xy - 5x - 2y = 0$

$xy - 5x - 2y + 10 = 10$

$(x-2)(y-5) = 10$

10の約数の組み合わせを考える。

$x-2 = 1, y-5 = 10 \implies x = 3, y = 15$

$x-2 = 2, y-5 = 5 \implies x = 4, y = 10$

$x-2 = 5, y-5 = 2 \implies x = 7, y = 7$

$x-2 = 10, y-5 = 1 \implies x = 12, y = 6$

$x-2 = -1, y-5 = -10 \implies x = 1, y = -5$

$x-2 = -2, y-5 = -5 \implies x = 0, y = 0$ (x,yが分母に来るので不可)

$x-2 = -5, y-5 = -2 \implies x = -3, y = 3$

$x-2 = -10, y-5 = -1 \implies x = -8, y = 4$

**(10) $y = \frac{5}{x-1} + 2$**

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

$(x-1)(y-2) = 5$

5の約数の組み合わせを考える。

$x-1 = 1, y-2 = 5 \implies x = 2, y = 7$

$x-1 = 5, y-2 = 1 \implies x = 6, y = 3$

$x-1 = -1, y-2 = -5 \implies x = 0, y = -3$

$x-1 = -5, y-2 = -1 \implies x = -4, y = 1$

3. 最終的な答え

(1) $(1, 7), (7, 1), (-1, -7), (-7, -1)$

(2) $(1, -5), (-1, 5), (5, -1), (-5, 1)$

(3) $(1, 8), (2, 4), (4, 2), (8, 1), (-1, -8), (-2, -4), (-4, -2), (-8, -1)$

(4) $(0, 7), (2, 5), (-2, 1), (-4, 3)$

(5) $(2, -14), (0, 10), (3, -8), (-1, 4), (4, -6), (-2, 2), (5, -5), (-3, 1), (7, -4), (-5, 0), (13, -3), (-11, -1)$

(6) $(-4, -7), (-6, 11), (-2, -1), (-8, 5), (4, 1), (-14, 3)$

(7) $(-1, 4), (0, 0), (-2, -8), (-3, -4)$

(8) $(6, 30), (10, 10), (30, 6), (4, -20), (-20, 4)$

(9) $(3, 15), (4, 10), (7, 7), (12, 6), (1, -5), (-3, 3), (-8, 4)$

(10) $(2, 7), (6, 3), (0, -3), (-4, 1)$