How do I denote the equivalence classes for $x^2 \equiv y^2 \pmod{7}$ for all integers. I know the possible results are $0, 1, 2$ and $4$.

$$x\equiv0 \pmod{7}, \space x^2\equiv0^2 \equiv 0 \pmod{7}$$ $$x\equiv1 \pmod{7}, \space x^2\equiv1^2 \equiv 1 \pmod{7}$$ $$x\equiv2 \pmod{7}, \space x^2\equiv2^2 \equiv 4 \pmod{7}$$ $$x\equiv3 \pmod{7}, \space x^2\equiv3^2 \equiv 2 \pmod{7}$$

But I do not understand how to denote it properly.


The equivalence class of an element $x$ is denoted by $[x]$.


So you should write like this: $$[0]=\{y\in\mathbb{Z}\colon 7\mid y^2\}=\{y\in\mathbb{Z}\colon 7\mid y\}=\{7k\colon k\in\mathbb{Z}\}$$ $$[1]=\{y\in\mathbb{Z}\colon 7\mid y^2-1\}=\{7k\pm 1\colon k\in\mathbb{Z}\}$$ etc. (of course it suffices to write it for 0,1,2,3,4,5,6, because $[x+7n]=[x]$ for any $x,n\in\mathbb{Z}$)

You can use shorter notation for these sets: $7\mathbb{Z}$, $7\mathbb{Z}\pm 1$ etc. where $AB=\{ab\colon a\in A, b\in B\}$.

Here is a similar topic: Congruence Class $[n]_5$ (Equivalence class of n wrt congruence mod 5) when n = $-3$, 2, 3, 6

  • $\begingroup$ So the equivalence classes would then be? [0]={y∈Z:7∣y^2}={y∈Z:7∣y}={7k:k∈Z} [1]={y∈Z:7∣y^2-1}={7k±1:k∈Z} [2]={y∈Z:7∣y^2-3}={7k±3:k∈Z} [4]={y∈Z:7∣y^2-2}={7k±2:k∈Z} Sorry for the terrible formatting. $\endgroup$ – user443044 May 3 '17 at 17:33

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