Prove for any odd integer $n$, $n^2-1\equiv 1\pmod{8}$

Question

Let, $\exists k \in \mathbb{Z}, n=2k+1\implies n^2-1= 4k(k+1) \pmod{8}$
Now, if $k$ is odd, $k+1$ is even, and vice-versa, so $4k(k+1) \pmod{8}=\forall k\, \exists k' \in \mathbb{Z},\,8k' \pmod{8}$

Unable to proceed further.

Answer 1

The source of confusion stated the following:

$2\mid n \,\Rightarrow\, 8\mid \color{#c00}{n^3},\,$ else $\,(n,2)=1 \,\Rightarrow\, 8\mid \color{#0a0}{n^2-1}\ $ by $\,{\rm odd}^2\equiv \{\pm1,\pm3\}^2\equiv 1 \pmod{\!8}$

That is they claim that if $n$ is odd, then $8|n^2-1$.

$8|n^2-1$ is equivalent to $$n^2-1 \equiv 0 \pmod{8}$$

Their argument is if $n$ is odd, that is $n \in \{ \pm1, \pm 3\}$, then we have $n^2\equiv 1 \pmod{8}$.

Remark:

My own understanding of $n^2-1$ is $n^2-1=(n-1)(n+1)$, which is a product of two consecutive even numbers, of which one of them must be a multiple of $4$. Hence the product must be a multiple of $8$.

Answer 2

You've shown that any odd integer satisfies $n^2 - 1 \equiv 0 \pmod 8$. The original statement to be proved is wrong of course, as can be easily checked taking, say, $n=1$ or $n=3$.

Answer 3

The statement should have been$$ n^2-1\equiv 0\pmod{8}$$

Let $n=2k+1$, then $n^2 -1 =(2k+1)^2-1 = 4k^2 + 4k = 4k(k+1)$

Note that $ k(k+1)$ is always even.

Thus $n^2-1 = 4k(k+1)$ is a multiple of 8.

That implies$$ n^2-1\equiv 0\pmod{8}$$