問題の解答
検索用コード（LaTeX）
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% 例題A4.1.7：互いに素に関する証明1（One More）★★
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$n$を自然数とする．$n+2$が5の倍数であり，$n+3$が7の倍数であるとき，$n+17$が35の倍数であることを証明せよ．
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% 例題A4.1.7の解答（One More）★★
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$n+2,n+3$は，それぞれ$n+2=5k,n+3=7l(k,l\text{は自然数})$とおける．
$n+17$を$k,l$を用いて表すと，
\[
n+17=(n+2)+15=5k+15=5(k+3)\cdots(\mathrm{i}),
\]
\[
n+17=(n+3)+14=7l+14=7(l+2)
\]
したがって，
\[
5(k+3)=7(l+2)
\]
5と7は互いに素であるから，$k+3$は7の倍数である．
ゆえに，$k+3=7m(m\text{は自然数})$と表せる．
(i)より，$n+17=5(k+3)=5\cdot 7m=35m$
よって，$n+17$は35の倍数である．$\blacksquare$
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% 問題A4.1.7：互いに素に関する証明1（One More）★★
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$n$を自然数とする．$n+3$が6の倍数であり，$n+1$が8の倍数であるとき，$n+9$が24の倍数であることを証明せよ．
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% 問題A4.1.7の解答（One More）★★
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$n+3,n+1$は，それぞれ$n+3=6k,n+1=8l(k,l\text{は自然数})$とおける．
$n+9$を$k,l$を用いて表すと，
\[
n+9=(n+3)+6=6k+6=6(k+1)\cdots(\mathrm{i}),
\]
\[
n+9=(n+1)+8=8l+8=8(l+1)
\]
したがって，$6(k+1)=8(l+1)$
すなわち，$3(k+1)=4(l+1)$であり，3と4は互いに素であるから，$k+1$は4の倍数である．
ゆえに，$k+1=4m(m\text{は自然数})$と表せる．
(i)より，$n+9=6(k+1)=6\cdot 4m=24m$
よって，$n+9$は24の倍数である．$\blacksquare$
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