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Let the relation $µ$ on $Z$ (set of integers) be defined by $xµy$ if and only if $x^2 ≡ y^2(mod 4).$

I have already proven that the relation $µ$ is an equivalence relation, but I am currently struggling with determining the equivalence classes

Do I just testing values for $x \in Z$?

For example:

If $x ≡ 0 (mod 4)$, then $x^2 ≡ 0^2 ≡ 0 (mod 4)$. So, the equivalence class [x] contains all integers that are congruent to $0$ modulo $4$.

If $x ≡ 1 (mod 4)$, then $x^2 ≡ 1^2 ≡ 1 (mod 4).$ The equivalence class [x] contains all integers that are congruent to $1$ modulo $4$.

Therefore, the equivalence classes of the relation $µ$ on $Z$ are:

$[0] = \{..., -8, -4, 0, 4, 8, ...\}$

$[1] = \{..., -7, -3, 1, 5, 9, ...\}$

Is this correct? But I believe that my equivalence classes need to complete $Z$? Not sure if I am doing the right thing. And where is $2$?

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2 Answers 2

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Actually, $$ x\mathrel\mu 0\iff x^2\equiv0\pmod4\iff x\text{ is even}, $$ and therefore $[0]$ is the set of all even integers.

And $$ x\mathrel\mu 1\iff x^2\equiv1\pmod4\iff x\text{ is odd,} $$ and therefore $[1]$ is the set of all odd integers.

Since every integer is even or odd, you are done: these are the only equivalence classes.


Here's another way of reaching the same conclusion. If $x,y\in\mathbb{Z}$, then \begin{align} x\mathrel\mu y&\iff x^2\equiv y^2\pmod4\\ &\iff4\mid(x-y)(x+y). \end{align} Now, when a product of integers is a multiple of $4$, then either both factors are even or one of them is odd whereas the other one is a multiple of $4$. But the current situation the second possibility cannot occur. Indeed, if, say $x+y$ is odd, the $x-y$ is odd too, since it is equal to $(x+y)-2y$. So, both $x+y$ and $x-y$ are even, and this means that $x$ and $y$ have the same parity. So, \begin{align} [x]&=\{y\in\mathbb{Z}\,|\,x\text{ and }y\text{ have the same parity}\}\\ &=\begin{cases} \{\text{even integers}\}&\text{ if $x$ is even}\\ \{\text{odd integers}\}&\text{ if $x$ is odd.} \end{cases} \end{align}

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  • $\begingroup$ Thank you for your answer, just one more question, which is kinda unrelated, is μ anti symmetry? From my proof, I think so, but I am not sure. $\endgroup$ Commented May 20, 2023 at 11:44
  • $\begingroup$ No, it is not: $0\mathrel\mu2$ and $2\mathrel\mu0$, but $0\ne2$. $\endgroup$ Commented May 20, 2023 at 12:05
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We denote $$[\ell]_{\mathbb{Z}_4}:=\{n\in\mathbb{Z}:n\equiv \ell\mbox{ mod } 4\},\qquad\ell=0,1,2,3.$$ Then for each $m\in[\ell]_{\mathbb{Z}_4}$, there is $k\in\mathbb{Z}$ such that $m=4k+\ell$. Therefore $$m^2=(4k+\ell)^2=\ell^2.$$ Denote $$[p]_{\mathbb{Z}_4,2}:=\{n\in\mathbb{Z}:n^2\equiv p^2 \mbox{ mod }4\}.$$ It follows from $0^2\equiv 2^2\equiv 0 \mbox{ mod 4}$ and $1^2\equiv 3^2\equiv 1 \mbox{ mod 4}$ that \begin{equation*} \begin{cases} [0]_{\mathbb{Z}_4,2}=[0]_{\mathbb{Z}_4}\bigcup [2]_{\mathbb{Z}_4}\\ [1]_{\mathbb{Z}_4,2}=[1]_{\mathbb{Z}_4}\bigcup [3]_{\mathbb{Z}_4} \end{cases} \end{equation*} and $$[0]_{\mathbb{Z}_4,2}\bigcup[1]_{\mathbb{Z}_4,2}=\left([0]_{\mathbb{Z}_4}\bigcup [2]_{\mathbb{Z}_4}\right)\bigcup\left([1]_{\mathbb{Z}_4}\bigcup [3]_{\mathbb{Z}_4}\right)=\mathbb{Z}.$$

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