$E(X^2)=E(X)=1$. Find $E(X^{100}).$

Question

$X$ is a random variable such that $E(X^2)=E(X)=1$. Find $E(X^{100}).$

My attempt: Assuming $X$ is discrete, we have $\sum x_i\mathbb P(X=x_i) = \sum x_i^2\mathbb P(X=x_i) = 1.$ We have something like $x_1p_1+\cdots+x_np_n=x_1^2p_1+\cdots+x_n^2p_n=1$. How do I find out $\sum x_i^{100}\mathbb P(X=x_i)$? I am completely blank. And, I am feeling uneasy as $X$ could well be continuous. How do I proceed then?

Answer 1

$$ 1 = \operatorname{E}(X^2) = (\operatorname{E}(X))^2 + \operatorname{var}(X) = 1 + \operatorname{var}(X). $$ Therefore $$ \operatorname{var}(X)=0. $$ So $\Pr(X=1)=1$.

Can you do the rest?