用反证法证明：当 $n$ 是整数时，

永远不可能被 $4$ 整除。

假设存在整数 $n$，使 $n^2-2$ 能被 $4$ 整除。这个假设会先推出 $n^2$ 为偶数，从而 $n$ 为偶数；但偶数的平方必为 $4$ 的倍数，这与假设所得等式矛盾。

Suppose, for a contradiction, that there is an integer $n$ such that $n^2-2$ is divisible by $4$.

Then, for some integer $k$,

Hence

so $n^2$ is even. Therefore $n$ is even, since the square of an odd integer is odd.

Thus $n=2m$ for some integer $m$, and so

Combining this with $n^2=4k+2$ gives

and hence

This is impossible because the left-hand side is even whereas the right-hand side is odd. This contradiction shows that the original assumption is false.

Therefore, $n^2-2$ is never divisible by $4$ for any integer $n$.

也可以从根式看同一个矛盾。由原假设可得

若右边的根式是整数 $r$，则

当 $r$ 为奇数时，右边不是整数；当 $r=2s$ 为偶数时，右边为偶数 $2s^2$，也不可能等于左边的奇数。因此这个根式不可能是整数，与 $|n|$ 为整数矛盾。

假设商 $\frac{n^2-2}{4}$ 是整数，再分别令 $n=2m$ 和 $n=2m+1$。两种情况下，商都是一个整数减去非整数的小数部分，从而与假设矛盾。

Suppose, for a contradiction, that $n^2-2$ is divisible by $4$ for some integer $n$. Then

must be an integer.

If $n$ is even, write $n=2m$, where $m$ is an integer. Then

Since $m^2$ is an integer, $q$ is not an integer, a contradiction.

If $n$ is odd, write $n=2m+1$, where $m$ is an integer. Then

Since $m^2+m$ is an integer, $q$ is again not an integer, a contradiction.

Every integer is either even or odd, so the assumption is false in both cases. Therefore, $n^2-2$ is never divisible by $4$ for any integer $n$.

上述分类也可以直接比较整除关系：若 $n$ 为偶数，则 $n^2-2=4m^2-2$。原假设与 $4m^2$ 能被 $4$ 整除，会迫使它们的差 $2$ 也被 $4$ 整除，不可能。若 $n$ 为奇数，则

同理，原假设会迫使 $1$ 被 $4$ 整除，也不可能。两个分支都得到矛盾，便否定了原假设。