I'm solving a problem in which it gives me some conditions:
$\vec{F}$ is a continuous vector field in a volume V and in the surface S which limits V;
I know that $\nabla\cdot\vec{F}=\rho(\vec{r})$ and that $\nabla\times\vec{F}=\mu(\vec{r})$ and both $\rho(\vec{r})$ and $\mu(\vec{r})$ are known, where $\vec{r}$ is the position vector;
Also, I know $\vec{F}\cdot\hat{n}$ in every point in space, where $\hat{n}$ is the normal vector to the surface S.
The problem asks me to prove whether there are infinite solutions to $\vec{F}$ or if it's unique given these conditions.
What I tried doing was to suppose there are two $\vec{F}$'s, $\vec{F_1}=\vec{F_1}$ and $\vec{F_2}=\vec{F_1}+\vec{A}$, where $\vec{A}$ is any vector field. My idea was to prove by contradiction that $\vec{F}$ is unique.
However, when I calcualted their respective divergents and curls, I got that
$\nabla\cdot\vec{A}=0 \implies \vec{A}=\nabla\times\vec{B}$, where $\vec{B}$ is a potential vector;
$\nabla\times\vec{A}=0 \implies \vec{A}=\nabla\phi$, where $\phi$ is a scalar function.
And I don't know what that means. I thought maybe the only possible outcome was that $\vec{A}=0$, because as the gradient of a function has the direction of the function's fastest increase, it's perpendicular to the function's level curves, so it shouldn't have any curl. But I'm not sure if this reasoning is correct.