题目

Problem

In this question you must show detailed reasoning.

Use algebra to prove by exhaustion that,

for all positive integers $m$ that are not multiples of $3$ , the value of

m^2+3m+2

is always a multiple of $3$

(4)

解答

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$m$ 不是 $3$ 的倍数，所以除以 $3$ 的余数只能是 $1$ 或 $2$ 。因此只需检查 $m=3p+1$ 和 $m=3p+2$ 两类。

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Since $m$ is a positive integer that is not a multiple of $3$ , there are two cases:

m=3p+1 \quad\text{or}\quad m=3p+2,

where $p$ is a non-negative integer.

Case 1: $m=3p+1$ .

\begin{align*} m^2+3m+2 =&\,(3p+1)^2+3(3p+1)+2\\ =&\,9p^2+6p+1+9p+3+2\\ =&\,9p^2+15p+6\\ =&\,3(3p^2+5p+2). \end{align*}

So $m^2+3m+2$ is a multiple of $3$ in this case.

Case 2: $m=3p+2$ .

\begin{align*} m^2+3m+2 =&\,(3p+2)^2+3(3p+2)+2\\ =&\,9p^2+12p+4+9p+6+2\\ =&\,9p^2+21p+12\\ =&\,3(3p^2+7p+4). \end{align*}

So $m^2+3m+2$ is also a multiple of $3$ in this case.

Therefore, for all positive integers $m$ that are not multiples of $3$ , $m^2+3m+2$ is always a multiple of $3$ .

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先因式分解： $m^2+3m+2=(m+1)(m+2)$ 。如果 $m$ 不是 $3$ 的倍数，那么接下来的两个整数 $m+1,m+2$ 之中必有一个是 $3$ 的倍数。

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Factorise:

\begin{align*} m^2+3m+2=(m+1)(m+2). \end{align*}

Since $m$ is not a multiple of $3$ , $m$ has remainder $1$ or $2$ when divided by $3$ .

If $m=3p+1$ , then

\begin{align*} m+2=3p+3=3(p+1), \end{align*}

so $m+2$ is a multiple of $3$ .

If $m=3p+2$ , then

\begin{align*} m+1=3p+3=3(p+1), \end{align*}

so $m+1$ is a multiple of $3$ .

In both cases, one factor of

\begin{align*} (m+1)(m+2) \end{align*}

is a multiple of $3$ .

Therefore,

\begin{align*} m^2+3m+2 \end{align*}

is always a multiple of $3$ .