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.

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 $\{-1, 1 \}$.
