Let $(M_n)_{n \in \mathbb{N}_0}$ be a stochastic process with $M_0 = 10$ such that both $(M_n)_{n \in \mathbb{N}_0}$ and $(M_n^2 )_{n \in \mathbb{N}_0}$ are martingales. Then $(M_n^5)_{n \in \mathbb{N}_0}$ is also a martingale.
Is the above statement true or false?
I was suggested by someone to disprove this statement by finding a counterexample. I was trying so hard but still cannot come out with a martingale which is itself and the square of itself martingale. It seems easy once we can come out with a suitable martingale.
Suppose that $(M_n)_{n \geq 0}$ is a martingale such that $M_0=10$ and $(M_n^2)_{n \in \mathbb{N}}$ is also a martingale. Then $$\mathbb{E}(M_n) = \mathbb{E}(M_0) = 10$$ and $$\mathbb{E}(M_n^2) = \mathbb{E}(M_0^2) = 100$$ for all $n \in \mathbb{N}$. Consequently, $$\text{var}(M_n) =\mathbb{E} \big[ (M_n-\mathbb{E}(M_n))^2 \big] = \mathbb{E}(M_n^2)-(\mathbb{E}(M_n))^2=0$$ for all $n \in \mathbb{N}$. Thus $M_n=\mathbb{E}(M_n)=10$ almost surely for all $n \in \mathbb{N}$. In particular, $(M_n^k)_{n \in \mathbb{N}}$ is a martingale for any $k \geq 1$.
In conclusion: yes, the statement is true.