Expected value and standard deviation of $X^2$ in terms of moments of $X$

If $$X$$ is a discrete random variable with expected value $$\mu$$ and standard deviation $$\sigma$$, and $$Y = X^2$$, how can we describe the expected value and standard deviation of $$Y$$?

Would expected value of $$Y = \mu^2$$?

For the mean, the answer is simple: $$E[Y]=E[X^2]=\operatorname{var}(X)+E[X]^2=\sigma^2+\mu^2$$
But, you can't find the variance of $$Y$$, without the fourth moment of $$X$$ because: $$\operatorname{var}(Y)=E[Y^2]-E[Y]^2=E[X^4]-(\sigma^2+\mu^2)^2$$
The variance is given in this related question. The mean of $$X^2$$ depends on the first two moments of $$X$$ and the variance of $$X^2$$ depends on the first four moments of $$X$$. These are:
\begin{align} \mathbb{E}(X^2) &= \mu^2 + \sigma^2, \\[6pt] \mathbb{V}(X^2) &= 4 \mu^2 \sigma^2 + 4 \mu \gamma \sigma^3 + (\kappa-1) \sigma^4, \\[6pt] \end{align}
where $$\gamma$$ and $$\kappa$$ are the skewness and kurtosis of $$X$$. (To get the standard deviation you just take the square-root of the variance.)