Equivalence class of $a^2-b^2$ is divisible by $5$

Question

I know that $a^2-b^2$ divisible by $5$ can be expressed as $a^2\equiv b^2 (\mod 5)$. I know the equivalence classes of this are $[0]$, $[1]$, $[2]$, $[3]$, and $[4]$. But, I am having trouble listing the elements for each equivalent class. Any help is appreciated.

Answer 1

We have that $x^2 \equiv (x+5)^2$ mod $5$, so we only need to consider the squares of $0,1, 2, 3, 4$, the $5$ equivalence classes mod $5$.

$0^2 \equiv 0$,

$1^2 \equiv 1$,

$2^2 \equiv 4$,

$3^3 \equiv 9 \equiv 4$,

$4^2 \equiv 16 \equiv 1$.

So with your equivalence relation, $a \sim b$ iff $a^2 \equiv b^2$ mod $5$, we have that $1 \sim 4$, $2 \sim 3$, and $0$ is in its own class.

Answer 2

Note that some books use $aRb$ to denote that $a$ is equivalent to $b$, while other books use $a\sim b$ to denote that $aRb$.

Assume that $a \sim b$ if and only if $5|(a^2-b^2)$ if and only if $a^2\equiv b^2 \mod 5$.

Let $[n] := \{ x\in \mathbb{Z}: x^2 \equiv n^2 \mod 5\}$, which is a subset of the set of integers. Here, $n$ is called a representative of the equivalence class $[n]$.

Then there are three distinct equivalence classes: $$ \begin{align*} [0] &= \{ \ldots,-15,-10,-5,0,5,10,15,\ldots \}, \\ [1] &= \{ \ldots,-4,-1,1,4,6,9,11, 14,\ldots \}=[4], \mbox{ and } \\ [2] &= \{ \ldots, ,-3,-2,2,3,7,8,12,13,\ldots\}=[3]. \end{align*} $$

Note that the equivalence class $[0]=\{x\in \mathbb{Z}:x^2 \equiv 0^2\mod 5 \}$. In other words, $[0]$ contains all those integers $x$ such that when you square it, it is congruent to $0\mod 5$. So let's take a look:

$[0]=\{x\in \mathbb{Z}: x^2 \equiv 0\mod 5 \} = \{\ldots, -5,0,5,10,15,\ldots \}$. It should be clear that these numbers are in the set $[0]$.

Next consider the equivalence class $[1]=\{ x\in \mathbb{Z}:x^2 \equiv 1^2\mod 5\}$. Systematically going through the positive integers first, we see that $1$ is in $[1]$. However, $2$ is not in $[1]$ because $2^2\not\equiv 1\mod 5$. Next consider the integer $3$. Then since $3^2\not\equiv 1\mod 5$, we see that $3\not\in[1]$. Next try $4$: since $4^2\equiv 1\mod 5$, we have $4\in [1]$.

Note that we can pick another representative for the equivalence class $[1]$ and write $[4]$ since the sets $[1]$ and $[4]$ are equal.

$\bf{Remark}$: notice that $1+5m, 4+5k\in [1]$ where $m,k\in \mathbb{Z}$.

Note that the set of equivalence classes partitions the set of integers.

And we say $\{ S_{\alpha}\}_{\alpha\in I}$ partitions the set $S$ if

1. $\bigcup_{\alpha\in I}S_{\alpha}=S$,
2. $S_{\alpha}\not=\varnothing$ for all $\alpha\in I$, and
3. $S_{\alpha}\cap S_{\beta}=\varnothing$ for all $\alpha\not=\beta$.