For which ${n\in{\Bbb Z}}$ does there exist a matrix $P\in{\Bbb C}^{4\times 4}$ such that $P^n=M$? 
Consider the matrix
  $$
M=\left(\begin{matrix}
0&0&0&1\\
0&0&0&0\\
0&0&0&0\\
0&0&0&0\\
\end{matrix}\right).
$$
  For which ${n\in{\Bbb Z}}$ does there exist a matrix $P\in{\Bbb C}^{4\times 4}$ such that $P^n=M$?


Since $P$ must be not invertible, one must have $n\geq 1$. For $n=1$, it is trivial. 
When $n=3$, we have (thanks to answers to this question) $P^3=M$ where $P$ is the $4\times 4$ Jordan black with $\lambda=0$:
$$
P=\left(\begin{matrix}
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
0&0&0&0\\
\end{matrix}\right).
$$
How can I deal with the general cases?
 A: You saw that for $n \leq 3$ you can find a matrix such that $P^n=M$ (Just shift the matrix $P^3$ further). 
There is a simple argument that for $n \geq 4$ such a matrix can't exist:
Note that $P$ must be nilpotent as $P^{2n}=P^n P^n=M^2=0$. But if $P$ is nilpotent and $P \in \mathbb{C}^{4 \times 4}$ then $P^4=0$ as the minimal polynomial always has degree $\leq 4$ and is of the form $\mu(X)=X^k$ for some $k \leq 4$.
A: The case $n=1$ is trivial. When $n\ge4$, as pointed out in Listing's answer, no feasible solution exists. Now you can characterise all $P$ such that $P^n=M$ for $n=2,3$.
When $n=2$, the Jordan form of such $P$ must be $\pmatrix{0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0}$, which is similar to $J=\pmatrix{0&1&0&0\\ 0&0&0&1\\ 0&0&0&0\\ 0&0&0&0}$. Thus the solutions $P$ are given by $P=SJS^{-1}$, where $S$ is any invertible matrix such that $M=P^2=SJ^2S^{-1}=SMS^{-1}$. Solving $MS=SM$, we see that $S$ must be of the form
$$\pmatrix{x&\ast&\ast&\ast\\ 0&a&b&\ast\\ 0&c&d&\ast\\ 0&0&0&x},\tag{$\star$}$$
where $x\ne0$ and $\pmatrix{a&b\\ c&d}$ is invertible.
When $n=3$, the Jordan form of $P$ must be the $4\times4$ nilpotent Jordan block $J_4(0)$. So, the solutions $P$ are given by $P=SJ_4(0)S^{-1}$. Since we need $M=P^3=SJ_4(0)^3S^{-1}=SMS^{-1}$, $S$ again must be in the form of $(\star)$.
