【数学A】例題4.1.11：余りによる場合分け1（One More）★★

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% 例題A4.1.11：余りによる場合分け1（One More）★★
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$n$を整数とするとき，次の問いに答えよ．

(1) $n^2+3n+2$は偶数であることを証明せよ．

(2) $n^2-2$は3の倍数ではないことを証明せよ．

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% 例題A4.1.11の解答（One More）★★
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(1) すべての整数$n$は，$n=2k,n=2k+1(k$は整数)のいずれかの形で表される．

(i) $n=2k$のとき

\[
n^2+3n+2=(2k)^2+3\cdot 2k+2=4k^2+6k+2=2(2k^2+3k+1)
\]

(ii) $n=2k+1$のとき

\[
n^2+3n+2=(2k+1)^2+3(2k+1)+2=4k^2+10k+6=2(2k^2+5k+3)
\]

よって，(i)，(ii)より，$n^2+3n+2$は偶数である．$\blacksquare$

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% 例題A4.1.11の別解
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$n^2+3n+2=n(n+1)+2(n+1)$

連続する2つの整数の積$n(n+1)$は偶数であり，$2(n+1)$も偶数であるから，$n^2+3n+2$は偶数である．$\blacksquare$

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% 例題A4.1.11の解答（One More）★★
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(2) すべての整数$n$は，$n=3k,n=3k+1,n=3k+2$（$k$は整数）のいずれかの形で表される．

(i) $n=3k$のとき

$n^2-2=(3k)^2-2=9k^2-2=3\cdot 3k^2-2$

(ii) $n=3k+1$のとき

$n^2-2=(3k+1)^2-2=9k^2+6k-1=3(3k^2+2k)-1$

(iii) $n=3k+2$のとき

$n^2-2=(3k+2)^2-2=9k^2+12k+2=3(3k^2+4k)+2$

よって，(i)〜(iii)より，$n^2-2$は3の倍数ではない．$\blacksquare$

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% 問題A4.1.11：余りによる場合分け1（One More）★★
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$n$を整数とするとき，次の問いに答えよ．

(1) $n^2-5n+4$は偶数であることを証明せよ．

(2) $n^3+2n+1$を3で割った余りが1であることを証明せよ．

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% 問題A4.1.11の解答（One More）★★
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(1) すべての整数$n$は，$n=2k,n=2k+1(k$は整数)のいずれかの形で表される．

(i) $n=2k$のとき

\[
n^2-5n+4=(2k)^2-5\cdot 2k+4=4k^2-10k+4=2(2k^2-5k+2)
\]

(ii) $n=2k+1$のとき

\[
n^2-5n+4=(2k+1)^2-5(2k+1)+4=4k^2-6k=2(2k^2-3k)
\]

よって，(i)，(ii)より，$n^2-5n+4$は偶数である．$\blacksquare$

(2) すべての整数$n$は，$n=3k,n=3k+1,n=3k+2(k$は整数)のいずれかの形で表される．

(i) $n=3k$のとき

\begin{align*}
n^3+2n+1&=(3k)^3+2\cdot 3k+1\\
&=27k^3+6k+1\\
&=3(9k^3+2k)+1
\end{align*}

(ii) $n=3k+1$のとき

\begin{align*}
n^3+2n+1&=(3k+1)^3+2(3k+1)+1\\
&=27k^3+27k^2+15k+4\\
&=3(9k^3+9k^2+5k+1)+1
\end{align*}

(iii) $n=3k+2$のとき

\begin{align*}
n^3+2n+1&=(3k+2)^3+2(3k+2)+1\\
&=27k^3+54k^2+42k+13\\
&=3(9k^3+18k^2+14k+4)+1
\end{align*}

よって，(i)〜(iii)より，$n^3+2n+1$を3で割ると余りは1である．$\blacksquare$

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% 例題A4.1.11の側注：余りによる場合分け1（One More）★★
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% 側注1
偶数である（2で割り切れる）．

% 側注2
偶数である（2で割り切れる）．

% 側注3
連続する2つの整数の積$n(n+1)$を作るように，式変形する．

% 側注4
$3k\pm 1,3k$などと表してもよい（場合分けが楽になる）．なお，$3k\pm 1$と表した場合，

\begin{align*}
&n^2-2\\
&=(3k\pm 1)^2-2\\
&=3(3k^2\pm 2k)-1(\text{複号同順})
\end{align*}

% 側注5
3で割った余りが0ではない．

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% 問題A4.1.11の側注：余りによる場合分け1（One More）★★
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% 側注1
偶数である（2で割り切れる）．

% 側注2
偶数である（2で割り切れる）．

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% 例題A4.1.11の考え方・補足など：余りによる場合分け1（One More）★★
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% 考え方
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(1) $n^2+3n+2$は偶数であることを示すので，2で割ったときの余りで分類して考える．整数を$n=2k,2k+1(k$は整数)のときに分けて，それぞれ偶数であることを示す．

(2) $n^2-2$は3の倍数ではないことを示すので，3で割ったときの余りで分類して考える．整数を$3k,3k+1,3k+2(k$は整数)のときに分けて，それぞれ3の倍数ではないことを示す．

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すべての整数は，正の整数$m$を用いて，次のいずれかの形で表される．

\[
mk,mk+1,mk+2,\cdots,mk+(m-1)(k\text{は整数})
\]