Can we derive $N=aa^{\dagger}$ from these conditions? Let $[a,b]:=ab-ba$ for all $a,b\in X$ a non commutative ring. Suppose that $a,a^{\dagger},N$ are operators satisfying
$$\begin{align}
[a,a^{\dagger}]&=1 \\
[N,a]&=-a \\
[N,a^{\dagger}] &= a^{\dagger}
\end{align}$$
where $1$ is simply the identity operator and $a^{\dagger}$ is the physicist's adjoint of $a$. 

Does it follow that 
  $$
N=aa^{\dagger} 
$$
  just from these $3$ conditions?

Motivation: My friend ask me a related question, which can be easily solve if the above is true. The $3$ conditions are simply given at the very beginning of the exercise. If we interpret $N$ as the number operator, $a$ as the annihilation operator and $a^{\dagger}$ as the creation operator then it is generally known that $N=aa^{\dagger}$. So I wonder if his professor simply forgot to write that in the exercise sheet or if the fact can be derived. 

If we assume that  $N$ is positive, the second condition is just a restatement of the second one. If you multiply the first condition by $-a$ on the left you get the second condition for $N=a^{\vphantom\dagger} a^\dagger$. So $N=a^{\vphantom\dagger} a^\dagger$ is certainly a solution. The question is whether it is unique. 
This boils down to the question of whether $Xa-aX=0$ implies $X=0$. Such condition certainly doesn't follow from the given conditions. Actually, $X=\lambda I$ will always commute with $a$, so $N=a^{\vphantom\dagger} a^\dagger+\lambda I$ will satisfy the equations for all $\lambda\in\mathbb R$. 
