Case when random variable X and its square $X^2$ are independent I have a random variable $X$, and I need to give an example when $X$ and $X^2$ are independent.
I can choose whatever distribution I please, I can say $X$ is a Bernoulli, uniform, or any other distribution.
To prove this, I need to show that $P(X \cap X^2) = P(X)P(X^2)$, but I'm failing to do so.
I tired keeping it simple, taking:
$$X \sim Bernoulli(p)$$
In this case:
$$X = X^2$$ and $p$ is probability for succeess and $1-p$ for failure: $$P(X^2) = P(X) = p$$
I also said that $$P(X \cap X^2) = P(X)P(X^2) = P(X)P(X) = p^2$$
But what is $P(X \cap X^2)$? I need to show that it is also $p^2$ in order for this to work. Is my solution enough to prove the rule of independence $P(A \cap B)=P(A)P(B)$ ?
Thank you.

I want to add that a sufficient and necessary condition for $X$ and $X^2$ are independent is that $X^2$ is degenerate.
The sufficiency is trivial.  Conversely, suppose $X$ and $X^2$ are independent. It then follows by $X^2$ is a function of $X$ that $X^2$ and $X^2$ are independent, whence
\begin{align}
E[X^4] = E[X^2 \cdot X^2] = E[X^2]E[X^2],
\end{align}
i.e., $\operatorname{Var}(X^2) = 0$, which shows that $X^2$ is degenerate.
