prove that $\frac{n^2-1}{8}$ is an even integer if and only if $n \equiv 1\pmod 8$ or $n \equiv -1\pmod 8$ prove that $\frac{n^2-1}{8}$ is an even integer if and only if $n \equiv 1\pmod 8$ or $n \equiv -1\pmod 8$
so I know from what i've already solved that $8|n^2-1$
where can I go from here? any help would be greatly appreciated

Given that $8|n^2-1$ and to prove that if $8|n^2-1$ holds, then $16|n^2-1$ iff $n\equiv \pm 1 \pmod8$
PROOF:
Consider that $8|n^2-1 \Rightarrow 8|(n-1)(n+1)$ 
So $n-1$ and $n+1$ must be consecutive even numbers.
Then either $n-1$ or $n+1$ must be a multiple of $4$.
Now if $16|n^2-1  \Rightarrow 16|(n-1)(n+1)$, then both $n-1$ and $n+1$ cannot be multiples of $4$ since consecutive even numbers cannot be both multiples of $4$.
At most one of them will be a multiple of 8.
So either $$n-1 \equiv 0 \pmod 8$$ or $$n+1 \equiv 0 \pmod 8$$
Combining the two, we get the required condition: $$n\equiv \pm 1 \pmod8$$
