You want to prove that
\(n^2\equiv k\mod8,k\in\{0,1,4\}\)
for (presumably) all integers \(n\ge1\).
Let's consider some sub-cases.
Suppose \(n=2\ell-1\) is odd. Then
\(n^2=(2\ell-1)^2=4\ell^2-4\ell+1\)
If \(\ell\) is even, then so is \(\ell^2-\ell\), which means you can write \(4\ell^2-4\ell=8m\) for some integer \(m\), and this reduces to \(n^2\equiv1\mod8\).
If \(\ell\) is odd, the same thing happens; you get that \(\ell^2-\ell\) is still even, so \(4\ell^2-4\ell\equiv0\mod8\) and you're left again with \(n^2\equiv1\mod8\).
Now assume \(n=2\ell\) is even. Then \(n^2=4\ell^2\). If \(\ell=2j\) is even, then you will always be able to write
\(n^2=(2\ell)^2=4\ell^2=4(2j)^2=8\times2j^2\equiv0\mod8\)
Meanwhile, if \(\ell=2j-1\) is odd, then
\(n^2=4\ell^2=4(2j-1)^2=16j^2-16j+4\equiv4\mod8\)
So you conclude that
\(n^2\equiv\begin{cases}0\mod8&\text{for }n\in\{4,8,12,16,\ldots\}\\1\mod8&\text{for }n\in\{1,3,5,7,\ldots\}\\4\mod8&\text{for }n\in\{2,6,10,14,\ldots\}\end{cases}\)
