Is it possible that X and Y are uncorrelated but X can significantly predict Y? I was just wondering is it possible that X and Y are uncorrelated but X can significantly predict Y?
If so, how would you explain and interpret that?

Let's prove the fact that @whuber gave in a comment to the question:

Let $X$ be symmetrically distributed around $0$ and let $Y=X^2$. The
latter is perfectly predictable from $X$ but the correlation is zero.

We first note that $X$ obviously determines its square $X^2=Y$. The proof of the second part is more interesting:
By definition, a random variable $X$ has a symmetric distribution about $\mu$ if $X-\mu$ has the same distribution as $\mu-X$. For symmetry about $\mu=0$ we get that $X-0=X$ has the same distribution as $0-X=-X$.
Using that $X$ and $-X$ have the same distribution and thus the same moments, the fact that $(-1)^a=-1$ for any odd number $a$, and linearity of expectation, we get
$$\mathbb{E}(X^a)=\mathbb{E}[(-X)^a]=\mathbb{E}(-X^a)=-\mathbb{E}(X^a)$$ and hence $\mathbb{E}(X^a)=0$ for any odd number $a$.
For the covariance between $X$ and $Y$ we have
$$\mathrm{Cov}(X,Y)=\mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)=\mathbb{E}(X^3)-\mathbb{E}(X^1)\mathbb{E}(X^2)$$
and using the previous result $\mathbb{E}(X^a)=0$ for $a\in\{1,3\}$ it follows that $\mathrm{Cov}(X,Y)=0$, which is equivalent to $\mathrm{Corr}(X,Y)=0$.
This is not particularly surprising as the covariance and correlation are measures of linear association between two random variables.
