Question

Show that every perfect square is congruent to 0, 1, or 4 modulo 8

Answer 1

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

$n^2\equiv\begin{cases}0\mod8&\text{for }n\in\{4,8,12,16,\ldots\}\\1\mod8&\text{for }n\in\{1,3,5,7,\ldots\}\\4\mod8&\text{for }n\in\{2,6,10,14,\ldots\}\end{cases}$