已知 $n\in\mathbb{N}$，用代数中的反证法证明：

反证时同时假设表达式是偶数、$n$ 也是偶数。写 $n=2k$，用代数说明表达式其实是奇数，得到矛盾。

We prove the statement by contradiction.

Assume that

is even, but $n$ is not odd.

Since $n\in\mathbb N$, this means $n$ is even.

So let

where $k$ is an integer.

Then

Now $4(k^2-2k)$ is even, and $5$ is odd.

Therefore

is odd.

So $n^2-4n+5$ is odd, contradicting the assumption that $n^2-4n+5$ is even.

Hence the assumption is false.

Therefore, if $n^2-4n+5$ is even, then $n$ is odd.

保持相同反证假设，先配方为 $(n-2)^2+1$，利用偶数的平方仍为偶数得到矛盾。

Assume, for a contradiction, that $n^2-4n+5$ is even and that $n$ is even.

Complete the square:

Since $n$ is even, write $n=2m$ with $m\in\mathbb Z$. Then

The first term is even, so this expression is odd. This contradicts the assumption that $n^2-4n+5$ is even.

Therefore, if $n^2-4n+5$ is even, $n$ must be odd.