問題の解答
検索用コード（LaTeX）
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% 例題A4.2.4：方程式の整数解2（One More）★★
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不定方程式$9x-5y=1$の整数解をすべて求めよ．
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% 例題A4.2.4の解答（One More）★★
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$9\times(-1)-5\times(-2)=1$であるから，$x=-1,y=-2$は$9x-5y=1$を満たす整数解の1つである．
したがって，
\[
9x-5y=1\cdots(\mathrm{i}),9\times(-1)-5\times(-2)=1\cdots(\mathrm{ii})
\]
とすると，$(\mathrm{i})-(\mathrm{ii})$より，$9(x+1)-5(y+2)=0$
したがって，$9(x+1)=5(y+2)\cdots(\mathrm{iii})$
ここで，9と5は互いに素であるから，$x+1$は5の倍数となり，$k$を整数とすると，$x+1=5k$，すなわち，$x=5k-1$
これを(iii)に代入すると，$9\times 5k=5(y+2)$
$9k=y+2$より，$y=9k-2$
よって，一般解は，$x=5k-1,y=9k-2(k$は整数)
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% 問題A4.2.4：方程式の整数解2（One More）★★
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不定方程式$4x+7y=1$の整数解をすべて求めよ．
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% 問題A4.2.4の解答（One More）★★
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$4\times 2+7\times(-1)=1$であるから，$x=2,y=-1$は$4x+7y=1$を満たす整数解の1つである．
したがって，
\[
4x+7y=1\cdots(\mathrm{i}),4\times 2+7\times(-1)=1\cdots(\mathrm{ii})
\]
とすると，$(\mathrm{i})-(\mathrm{ii})$より，$4(x-2)+7(y+1)=0$
したがって，$4(x-2)=-7(y+1)\cdots(\mathrm{iii})$
ここで，4と7は互いに素であるから，$x-2$は7の倍数となり，$k$を整数とすると，$x-2=7k$，すなわち，$x=7k+2$
これを(iii)に代入すると，$4\times 7k=-7(y+1)$
$4k=-(y+1)$より，$y=-4k-1$
よって，一般解は，$x=7k+2,y=-4k-1(k$は整数)
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