【数学A】例題4.2.6：方程式の整数解4（One More）★★★

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% 例題A4.2.6：方程式の整数解4（One More）★★★
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不定方程式$x+2y+3z=10$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A4.2.6の解答（One More）★★★
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与えられた不定方程式$x+2y+3z=10$を$z$について整理すると，

\[
3z=10-x-2y
\]

$x,y$は自然数であるから，$x\geqq 1,y\geqq 1$より，

\[
3z=10-x-2y\leqq 10-1-2\times 1=7
\]

したがって，$z\leqq\frac{7}{3}$より，$z=1,2$となる．

（ア）$z=1$のとき

$x+2y+3\times 1=10$より，$x+2y=7\cdots(\mathrm{i})$

$x\geqq 1$より，$2y=7-x\leqq 7-1=6$

したがって，$y\leqq 3$より，$y=1,2,3$

(i)より，$(x,y)=(5,1),(3,2),(1,3)$

（イ）$z=2$のとき

$x+2y+3\times 2=10$より，$x+2y=4\cdots(\mathrm{ii})$

$x\geqq 1$より，$2y=4-x\leqq 4-1=3$

したがって，$y\leqq\frac{3}{2}$より，$y=1$

(ii)より，$(x,y)=(2,1)$

よって，（ア），（イ）より，求める自然数の組は，

\[
(x,y,z)=(5,1,1),(3,2,1),(1,3,1),(2,1,2)
\]

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% 問題A4.2.6：方程式の整数解4（One More）★★★
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不定方程式$x+3y+4z=15$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A4.2.6の解答（One More）★★★
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与えられた不定方程式$x+3y+4z=15$を$z$について整理すると，$4z=15-x-3y$

$x,y$は自然数であるから，$x\geqq 1,y\geqq 1$より，

\[
4z=15-x-3y\leqq 15-1-3\times 1=11
\]

したがって，$z\leqq\frac{11}{4}$より，$z=1,2$となる．

（ア）$z=1$のとき

$x+3y+4\times 1=15$より，$x+3y=11\cdots(\mathrm{i})$

$x\geqq 1$より，$3y=11-x\leqq 11-1=10$

したがって，$y\leqq\frac{10}{3}$より，$y=1,2,3$

(i)より，$(x,y)=(8,1),(5,2),(2,3)$

（イ）$z=2$のとき

$x+3y+4\times 2=15$より，$x+3y=7\cdots(\mathrm{ii})$

$x\geqq 1$より，$3y=7-x\leqq 7-1=6$

したがって，$y\leqq 2$より，$y=1,2$

(ii)より，$(x,y)=(4,1),(1,2)$

よって，（ア），（イ）より，求める自然数の組は，

\[
(x,y,z)=(8,1,1),(5,2,1),(2,3,1),(4,1,2),(1,2,2)
\]

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% 例題A4.2.6の側注：方程式の整数解4（One More）★★★
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% 側注1
係数の大きい$3z$について注目して整理する．

% 側注2
$1\leqq z\leqq\frac{7}{3}$を満たす自然数$z$の値は，$z=1,2$である．

% 側注3
$3z$のときと同様に，不等式を利用して，係数の大きい$2y$について注目して整理する．

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% 問題A4.2.6の側注：方程式の整数解4（One More）★★★
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% 側注1
係数の大きい$4z$について注目して整理する．

% 側注2
$1\leqq z\leqq\frac{11}{4}$を満たす自然数$z$の値は，$z=1,2$である．

% 側注3
$4z$のときと同様に，不等式を利用して，係数の大きい$3y$について注目して整理する．

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% 例題A4.2.6の考え方・補足など：方程式の整数解4（One More）★★★
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% 考え方
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$x,y,z$は自然数であるから，$x\geqq 1,y\geqq 1,z\geqq 1$である．文字の値の範囲から，$x,y,z$の値を定めるとよい．例えば，係数の大きい$3z$に注目して，$x\geqq 1,y\geqq 1$であることを利用すると，$3z=10-x-2y\leqq 10-1-2\times 1=7$となる．つまり，$z\leqq\frac{7}{3}$より，$z$は自然数であるから，$z=1,2$となる．このように，不等式を上手く利用して，文字の値を定めていく．

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% 余談
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係数の大きい$3z$に注目せずに，他の文字について整理しても求めることができるが，場合分けに手間が掛かる．例えば，$y\geqq 1,z\geqq 1$であることを利用すると，$x=10-2y-3z\leqq 10-2-3=5$となる．つまり，$x\leqq 5$より，$x$は自然数であるから，$x=1,2,3,4,5$となり，5つの場合分けをしなければならなくなる．

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係数の大きい文字に注目して，不等式を利用することで文字の値を定める．