(i) 用一个反例说明下列命题为假：

“对所有 $n\in\mathbb N$，$n^2+3n+1$ 都是质数。”

（2分）

(ii) 用代数方法，通过穷尽所有情形证明：对所有 $n\in\mathbb N$，

“$n^2-2$ 不是 $4$ 的倍数。”

（4分）

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 $55$。

Take

Then

But

so $55$ is not prime.

Therefore, the statement is false.

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。

Every natural number is either even or odd. In the following cases, $k$ is an integer.

If $n$ is even, let

Then

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

If $n$ is odd, let

Then

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 $4$.

按除以 4 的余数，把所有自然数穷尽分为四类，逐类展开 $n^2-2$ 并写成 4 的倍数加 2 或 3。

For an integer $k$, every natural number has one of the forms $4k$, $4k+1$, $4k+2$, $4k+3$.

Case 1: $n=4k$.

Case 2: $n=4k+1$.

Case 3: $n=4k+2$.

Case 4: $n=4k+3$.

In every case the remainder on division by 4 is 2 or 3, never 0. Thus $n^2-2$ is not a multiple of 4 for any $n\in\mathbb N$. $\blacksquare$