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

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