Assume that $H$ is a separable Hilbert space.
Is there a polynomial $p(z)\in \mathbb{C}[x]$ with $deg(p)>1$ with the following property?:


Every densely defined operator $A:D(A)\to D(A),\;D(A)\subset H$ with $p(A)=0$ is necessarily a bounded operator on $H$.


That is the polynomial-operator equation $p(A)=0$ has only bounded solution.



A question on unbounded operators
I do not think so. 


**Observation:** Without loss of generality, $p(x)$ can be taken to be monic (constant multiples won't affect either $p(A) = 0$ or boundedness). 


**Case 1:** $p$ is degree $2$.


By the above reduction, $p(x) = (x - \lambda)(x - \mu)$ for some $\lambda$ and $\mu$ in $\mathbb{C}$ (since $A$ clearly commutes with itself and with $I$, and since $A$ maps $D(A)$ to itself, this decomposition is reasonable). Yet if $p(A) = 0$, take the operator $\displaystyle B = A - \frac{\lambda + \mu}{2} I$ (with the same domain), and we see that for $\displaystyle \nu = \frac{\lambda - \mu}{2}$, $B$ satisfies $(B + \nu)(B - \nu) = 0$, or $B^2 - \nu^2 = 0$. We will show that an unbounded choice of $B$ exists, satisfying $B: D(B) \to D(B)$, hence an unbounded choice of $A$ exists, with $A: D(A) \to D(A)$. 


*Subcase 1:* $\nu = 0$. Then take $H = \ell^2(\mathbb{N})$, let $H\_0 = D(B)$ be the sequences with only finitely many nonzero elements, and let $B$ be the operator represented by the infinite matrix 


$$ \begin{pmatrix} 0 & 1 & & & & & \cdots \\
 0 & 0 & & & & & \cdots \\
 & & 0 & 2 & & & \cdots \\
 & & 0 & 0 & & & \cdots \\
 & & & & 0 & 3 & \cdots \\
 & & & & 0 & 0 & \ddots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots \end{pmatrix},$$
which is clearly well-defined on $H\_0$ and clearly maps $H\_0$ to itself. Then $B^2 = 0$, but letting $e\_j$ be the $j$th basis vector, $B e\_{2j} = j e\_{2j - 1}$, so clearly $B$ is unbounded.


*Subcase 2:* $\nu \neq 0$. Then again take $H = \ell^2(\mathbb{N})$, and $H\_0$ the almost-everywhere-$0$ sequences. We now define $B$ by the matrix 


$$ \begin{pmatrix} 0 & \nu & & & & & \cdots \\
 -\nu & 0 & & & & & \cdots \\
 & & 0 & 2\nu & & & \cdots \\
 & & -\frac{1}{2}\nu & 0 & & & \cdots \\
 & & & & 0 & 3\nu & \cdots \\
 & & & & -\frac{1}{3}\nu & 0 & \ddots \\
 \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots \end{pmatrix}.$$


Again, $B$ maps $H\_0$ to itself, and $B^2 e\_j = \nu^2 e\_j$ for all $j$, so $B^2 = \nu^2$ on $H\_0$. Yet $Be\_{2j} = j e\_{2j - 1}$, so $B$ is unbounded.


**Case 2:** $\deg p > 2$.


Well, then $p(x) = q\_1(x) q\_2(x)$, with $\deg q\_1 = 2$, $\deg q\_2 \geq 1$. Again take $H$ and $H\_0$ as above, and take $A$ to be an unbounded solution to $q\_1(A) = 0$, satisfying $A: D(A) \to D(A)$. Then $p(A) = q\_1(A) q\_2(A) = 0 q\_2(A) = 0$, and $A$ is unbounded. [Again, my naive factoring really depends on $D(A) \subseteq D(A^n)$, hence I am strongly using the $A: D(A) \to D(A)$ fact here.] QED.


Of course, Case 2 is sort of a cheat. I think there should be a "natural" example in general, since you can construct unbounded examples to $A^n = 0$ by just increasing the order of the nilpotency, and $A^n = I$ by taking positive real numbers $a\_1, \dotsc, a\_n$ with $a\_1 a\_2 \dotsc a\_n = 1$ and letting the building-block matrix be


$$ \begin{pmatrix} 0 & a\_1 & & & \cdots & & \\
 & 0 & a\_2 & & \cdots & & \\
 & & 0 & a\_3 & \cdots & & \\
 \vdots & \vdots & \vdots & \ddots & \ddots & &\\
\vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \\
 & & & & & 0 & a\_{n-1} \\
a\_n & 0 & 0 & & \dotsc & 0 & 0 \end{pmatrix} $$
Then again make a block-diagonal infinite matrix such that as we repeat the blocks, $a\_3, \dotsc, a\_n$ are constant, and $a\_1 \to \infty$ and $a\_2 \to 0$ (or somesuch). 


