Solving an equation with least residues and roots modulo $n$ [duplicate]

Question

I am facing the problem of proving $z^2 \equiv 1\pmod 8$ for all odd $z\in\mathbb{N}$.

What I know is that for each $z \in \mathbb{N}$, $z = m^2-1$, for $m\in\mathbb{N}$.
What I also know is that in a least residue system there are the following solutions for all $z\in\mathbb{N}$.

The solutions are: $$n \equiv 1,3,5,7$$

Which lead to solutions of the initial problem: $$n = 8 \cdot k + 1\\ n = 8 \cdot k + 3 \\ n = 8 \cdot k + 5 \\ n = 8 \cdot k + 7$$ where $k\in\mathbb{N}$.

My question now is how I can transfer my knowledge about the solutions for all $n\in\mathbb{N}$ to the key problem where $n$ must be an odd number and the proofs needs to be done for all odd $n$.

I also know that $8 = 2^3$ which leads to $$(2\;\cdot\;m\;-\;1)^2\;\equiv\;1\;\pmod 8$$ $$(2\;\cdot\;m\;-\;1)^2\;\equiv\;1\;\pmod {2^3}$$ but I do not know what this tells me right now.

Answer 1

$\bmod 8$ there are only eight numbers to check. All the odd $z\in \Bbb N$ are in one and only one of the residue classes $\bar1,\bar3,\bar5$ or $\bar7$. For each of these we can verify the condition is satisfied. Thus it is true for all odd naturals.

Answer 2

If $z$ is odd, then $z-1$ and $z+1$ are two consecutive even numbers;

both of them are divisible by $2$, and one of them is divisible by $4$.

Therefore, $(z+1)(z-1)=z^2-1$ is divisible by $8$; i.e., $z^2\equiv1\pmod8$.