【数学A】例題4.2.9：方程式の整数解7（One More）★★★

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% 例題A4.2.9：方程式の整数解7（One More）★★★
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次の方程式を満たす整数の組$(x,y)$をすべて求めよ．

(1) $xy-2x+3y=0$

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

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% 例題A4.2.9の解答（One More）★★★
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(1) $xy-2x+3y=0$より，$(x+3)(y-2)=-6$

$x,y$は整数であるから，$x+3,y-2$も整数である．

したがって，

\[
(x+3,y-2)=(1,-6),(6,-1),(2,-3),(3,-2),(-1,6),(-6,1),(-2,3),(-3,2)
\]

よって，

\[
(x,y)=(-2,-4),(3,1),(-1,-1),(0,0),(-4,8),(-9,3),(-5,5),(-6,4)
\]

(2) $\frac{2}{x}+\frac{1}{y}=1$より，$xy=2y+x$

したがって，$xy-x-2y=0$であるから，$(x-2)(y-1)=2$

$x,y$は整数であるから，$x-2,y-1$も整数である．

ここで，$x\neq 0,y\neq 0$より，$x-2\neq-2,y-1\neq-1$であるから，

\[
(x-2,y-1)=(2,1),(1,2),(-1,-2)
\]

よって，

\[
(x,y)=(4,2),(3,3),(1,-1)
\]

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% 問題A4.2.9：方程式の整数解7（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

次の方程式を満たす整数の組$(x,y)$をすべて求めよ．

(1) $xy+x+2y=0$

(2) $\frac{3}{x}+\frac{1}{y}=1$

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% 問題A4.2.9の解答（One More）★★★
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(1) $xy+x+2y=0$より，$(x+2)(y+1)=2$

$x,y$は整数であるから，$x+2,y+1$も整数である．

したがって，

\[
(x+2,y+1)=(1,2),(2,1),(-1,-2),(-2,-1)
\]

よって，

\[
(x,y)=(-1,1),(0,0),(-3,-3),(-4,-2)
\]

(2) $\frac{3}{x}+\frac{1}{y}=1$より，$xy=3y+x$

したがって，$xy-x-3y=0$であるから，$(x-3)(y-1)=3$

$x,y$は整数であるから，$x-3,y-1$も整数である．

ここで，$x\neq 0,y\neq 0$より，$x-3\neq-3,y-1\neq-1$であるから，

\[
(x-3,y-1)=(3,1),(1,3),(-1,-3)
\]

よって，

\[
(x,y)=(6,2),(4,4),(2,-2)
\]

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% 例題A4.2.9の側注：方程式の整数解7（One More）★★★
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% 側注1
与えられた式の左辺の係数から，$(x+3)(y-2)$を作る．

% 側注2
掛けて$-6$になる整数の組を求める．

% 側注3
両辺に$xy$を掛ける．

% 側注4
$\frac{2}{x}+\frac{1}{y}=1$の分母は0ではない．

% 側注5
$x-2\neq-2,y-1\neq-1$より，$(x-2,y-1)=(-2,-1)$は不適であることに注意すること．

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% 問題A4.2.9の側注：方程式の整数解7（One More）★★★
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% 側注1
与えられた式の左辺の係数から，$(x+2)(y+1)$を作る．

% 側注2
掛けて$2$になる整数の組を求める．

% 側注3
両辺に$xy$を掛ける．

% 側注4
$\frac{3}{x}+\frac{1}{y}=1$の分母は0ではない．

% 側注5
$x-3\neq-3,y-1\neq-1$より，$(x-3,y-1)=(-3,-1)$は不適であることに注意すること．

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% 例題A4.2.9の考え方・補足など：方程式の整数解7（One More）★★★
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% 考え方
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(1) $xy+ax+by=(x+b)(y+a)-ab$であることから，$xy+ax+by=c(a,b,c$は定数$)$の形の式は，$(x+b)(y+a)=c+ab$と式変形できることを利用する．そこで，$xy+ax+by$の係数$a,b$から，$(x+b)(y+a)$を作ることを考える．

(2) 両辺に$xy$を掛けて，(1)と同様に考えるとよい．このとき，$x\neq 0,y\neq 0$より，$x-2\neq-2,y-1\neq-1$であることに注意すること．

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$xy+ax+by=(x+b)(y+a)-ab$を利用する．