Equivalence class for the following relation A relation of $\mathbb{R}$ is defined as $a\sim b : a^4-b^2=b^4-a^2$
Show that $\sim$ is equivalence relation (I have done this part)
Determine the equivalence class $[-1]_\sim$
Prove or disprove: Every equivalence class in $\mathbb{R}/\sim$ contains exactly 2 real numbers.
I am facing difficulty in second and third part of the question.
For the second one I think it is $[-1]_\sim=\{-1,1\}$ as $-1\sim x: (-1)^4-x^2 =x^4-(-1)^2 $ so both $-1$ and $1$ are proving the equality.
I think the third statement is true but how can I prove it?
Please help.
 A: Rearrange: $a^4 - b^4 = b^2 - a^2$
But $a^4 - b^4 = (a^2 - b^2)(a^2 + b^2)$ (DOTS)
So $(a^2 - b^2)(a^2 + b^2) = b^2 - a^2$
(1) Assume $a^2 - b^2 \neq 0$
Then divide both sides by $a^2 - b^2$
$a^2 + b^2 = -1$
Since $a^2$ and $b^2$ are both positive in $\Bbb R$, this is a contradiction.
(2) Therefore $a^2 - b^2 = 0$, or $a^2 = b^2$.
Therefore $b = \pm a$ and each equivalence class of a comprises $\{a, -a \}$.
The equivalence class of $-1$ is $\{-1, 1 \}$.
A: Follow definitions.
$[-1] = \{b\in \mathbb R|(-1)^2 - b^2=b^4 -(-1)^2\}=$
$\{b\in \mathbb R| 1-b^2 = b^4 -1\}=$
$\{b\in \mathbb R| b^4+b^2-2 = 0\}=$
$\{b\in \mathbb R| (b^2+2)(b-1)(b+1)=0\}=$
$\{b\in \mathbb R| b^2 = -2\lor b=1\lor b=-1\}=$
$\{1,-1\}$
As for the second:
$[a] = \{b\in \mathbb R| a^4-b^2 = b^4 -a^2\}=$
$\{b \in \mathbb R|b^4+b^2 -(a^4+a^2) = 0\}$
Now potentially $b^4 + b^2-(a^4 + a^2)=0$ may have up to four solutions.
Indeed in the case of $[-1]$ we found $b \in [-1]$ if $(b^2 + 2)(b+1)(b-1)=0$ had two elements because $b^2 + 2=0$ is impossible.
We can use quadratic formula to solve $b^4 + b^2 -(a^4+a^2)$ for $b$ but it's probably easier to factor:
$b^4 + b^2-(a^4 + a^2)=0$
$(b^4 -a^4) + (b^2 - a^2) = 0$
$(b^2 +a^2)(b^2 - a^2)+(b^2 -a^2) =0$
$(b^2 + a^2 + 1)(b^2-a^2)=0$
$(b^2  +a^2 + 1)(b-a)(b+a)=0$.
Well, $b^2 + a^2 + 1 > 0$ so the only solutions are $b=\pm a$ and that will always be the case that $[a] = \{a,-a\}$ so.....
but what if $a = 0$.  Then $[0] = \{0,-0\} =\{0\}$ has only one element.
And we can show that directly $0^4 -b^2 = b^4 - 0^2$
$b^4 + b^2 = 0$ so $b^2(b^2 + 1)=0$ so $b^2 = 0$ or $b^2 =-1$ so $b = 0$.
So the third is false but only because $[0]$ has one element. For all others it is true.
Not  $\sim$ is exactly the same as the relationship $a = \pm b$.
Note:  $a^4 -b^2 = b^4 - a^2 \iff a =\pm b$
In hindsight had we proven that from the very beginning we wouldn't have need to do so much work.
Live and learn.
