【数学A】例題4.2.7：方程式の整数解5（One More）★★★

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% 例題A4.2.7：方程式の整数解5（One More）★★★
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$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1,x\leqq y\leqq z$を満たす自然数の組$(x,y,z)$をすべて求めよ．

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% 例題A4.2.7の解答（One More）★★★
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$0<x\leqq y\leqq z$より，$\frac{1}{z}\leqq\frac{1}{y}\leqq\frac{1}{x}$である．

これより，$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leqq\frac{1}{x}+\frac{1}{x}+\frac{1}{x}$

したがって，$1\leqq\frac{3}{x}$より，$x\leqq 3$

(i) $x=1$のとき

$\frac{1}{1}+\frac{1}{y}+\frac{1}{z}=1$より，$\frac{1}{y}+\frac{1}{z}=0$

これは，$y,z$は正の整数であるので不適である．

(ii) $x=2$のとき

$\frac{1}{2}+\frac{1}{y}+\frac{1}{z}=1$より，$\frac{1}{y}+\frac{1}{z}=\frac{1}{2}$

ここで，$\frac{1}{y}+\frac{1}{z}\leqq\frac{1}{y}+\frac{1}{y}$より，$\frac{1}{2}\leqq\frac{2}{y}$

したがって，$y\leqq 4$

ゆえに，$x\leqq y$より，$y=2,3,4$

$y=2$のとき，$\frac{1}{2}+\frac{1}{z}=\frac{1}{2}$

これは，$z$は正の整数であるので不適である．

$y=3$のとき，$\frac{1}{3}+\frac{1}{z}=\frac{1}{2}$より，$z=6$

$y=4$のとき，$\frac{1}{4}+\frac{1}{z}=\frac{1}{2}$より，$z=4$

(iii) $x=3$のとき，$\frac{1}{3}+\frac{1}{y}+\frac{1}{z}=1$より，$\frac{1}{y}+\frac{1}{z}=\frac{2}{3}$

ここで，$\frac{1}{y}+\frac{1}{z}\leqq\frac{1}{y}+\frac{1}{y}$より，$\frac{2}{3}\leqq\frac{2}{y}$

したがって，$y\leqq 3$

ゆえに，$x\leqq y$より，$y=3$

このとき，$\frac{1}{3}+\frac{1}{z}=\frac{2}{3}$より，$z=3$

よって，(i)〜(iii)より，$(x,y,z)=(2,3,6),(2,4,4),(3,3,3)$

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% 問題A4.2.7：方程式の整数解5（One More）★★★
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$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2},x\leqq y\leqq z$を満たす自然数の組$(x,y,z)$をすべて求めよ．

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% 問題A4.2.7の解答（One More）★★★
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$0<x\leqq y\leqq z$より，$\frac{1}{z}\leqq\frac{1}{y}\leqq\frac{1}{x}$である．

これより，$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leqq\frac{1}{x}+\frac{1}{x}+\frac{1}{x}$

したがって，$\frac{3}{2}\leqq\frac{3}{x}$より，$x\leqq 2$

(i) $x=1$のとき

$\frac{1}{1}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2}$より，$\frac{1}{y}+\frac{1}{z}=\frac{1}{2}$

ここで，$\frac{1}{y}+\frac{1}{z}\leqq\frac{1}{y}+\frac{1}{y}$より，$\frac{1}{2}\leqq\frac{2}{y}$

したがって，$y\leqq 4$

ゆえに，$x\leqq y$より，$y=1,2,3,4$である．

$y=1$のとき，

\[
1+\frac{1}{z}=\frac{1}{2}
\]

となり不適である．

$y=2$のとき，$\frac{1}{2}+\frac{1}{z}=\frac{1}{2}$

これは，$z$は正の整数であるので不適である．

$y=3$のとき，$\frac{1}{3}+\frac{1}{z}=\frac{1}{2}$より，$z=6$

$y=4$のとき，$\frac{1}{4}+\frac{1}{z}=\frac{1}{2}$より，$z=4$

(ii) $x=2$のとき，$\frac{1}{2}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2}$より，$\frac{1}{y}+\frac{1}{z}=1$

ここで，$\frac{1}{y}+\frac{1}{z}\leqq\frac{1}{y}+\frac{1}{y}$より，$1\leqq\frac{2}{y}$

したがって，$y\leqq 2$

ゆえに，$x\leqq y$より，$y=2$

このとき，$\frac{1}{2}+\frac{1}{z}=1$より，$z=2$

よって，(i)，(ii)より，$(x,y,z)=(1,3,6),(1,4,4),(2,2,2)$

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% 例題A4.2.7の側注：方程式の整数解5（One More）★★★
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両辺に$x(>0)$を掛ける．

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$x=1,2,3$

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$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$に$x=1$を代入する．

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両辺に$y(>0)$を掛ける．

% 側注5
両辺に$y(>0)$を掛ける．

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% 問題A4.2.7の側注：方程式の整数解5（One More）★★★
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% 側注1
両辺に$x(>0)$を掛ける．

% 側注2
$x=1,2$

% 側注3
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2}$に$x=1$を代入する．

% 側注4
両辺に$y(>0)$を掛ける．

% 側注5
両辺に$y(>0)$を掛ける．

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% 例題A4.2.7の考え方・補足など：方程式の整数解5（One More）★★★
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% 考え方
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$0<x\leqq y\leqq z$より，$\frac{1}{z}\leqq\frac{1}{y}\leqq\frac{1}{x}$である．文字の値の範囲から，$x,y,z$の値を定めるとよい．$x\leqq y\leqq z$より，$x$か$z$に注目することを考える．

例えば，$z$に注目して，$1=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geqq\frac{1}{z}+\frac{1}{z}+\frac{1}{z}=\frac{3}{z}$より，$z\geqq 3$としても，上手く文字の値を定めることができない．ここでは，$x$に注目して，$1=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leqq\frac{1}{x}+\frac{1}{x}+\frac{1}{x}=\frac{3}{x}$より，$x\leqq 3$とすると，上手く文字の値を定めることができる．