题目

Problem

(i) Use a counter example to show that the following statement is false

\text{``}n^2+3n+1\text{ is prime for all }n\in\mathbb{N}\text{''}

(2)

(ii) Use algebra to prove by exhaustion that for all $n\in\mathbb{N}$

\text{``}n^2-2\text{ is not a multiple of }4\text{''}

(4)

解答

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反例只需要找到一个自然数使表达式不是质数。这里可以试 $n=6$ ，因为算出来是 $55$ 。

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Take

\begin{align*} n=6. \end{align*}

Then

\begin{align*} n^2+3n+1=&\,6^2+3(6)+1\\ =&\,36+18+1\\ =&\,55. \end{align*}

But

\begin{align*} 55=5\times11, \end{align*}

so $55$ is not prime.

Therefore, the statement is false.

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任意自然数不是奇数就是偶数，所以只需分两类讨论。这就是 exhaustion：把所有可能情况列完。

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Every natural number is either even or odd.

If $n$ is even, let

\begin{align*} n=2k. \end{align*}

Then

\begin{align*} n^2-2=&\,(2k)^2-2\\ =&\,4k^2-2. \end{align*}

This is $2$ less than a multiple of $4$ , so it is not a multiple of $4$ .

If $n$ is odd, let

\begin{align*} n=2k+1. \end{align*}

Then

\begin{align*} n^2-2=&\,(2k+1)^2-2\\ =&\,4k^2+4k+1-2\\ =&\,4k^2+4k-1\\ =&\,4(k^2+k)-1. \end{align*}

This is $1$ less than a multiple of $4$ , so it is not a multiple of $4$ .

Therefore, in all cases, $n^2-2$ is not a multiple of $4$ .