Edited question Consider a vector field $\vec{A}(\vec{x})$ such that in one case $\nabla\cdot\vec{A}=0$. It looks like that this condition gives rise to a differential equation constraint $$\partial_xA_x+\partial_yA_y+\partial_zA_z=0.$$
Consider the next case where for a known vector $\vec{k}=k_x\hat{x}+k_y\hat{y}+k_z\hat{z}$, one finds that $\vec{k}\cdot\vec{A}=0$. Therefore, $$k_xA_x+k_yA_y+k_zA_z=0.$$
The latter case definitely reduces the number of independent components of $\vec{A}$ from 3 to 2. What about the first case? Does that also reduce the number of independent components of $\vec{A}$? If not, how to see that simply?
That the rotation is zero $∇\times \vec A=0$ means that locally (on small balls) you can find a potential function $\phi$ so that the vector field is its gradient, $\vec A = ∇\phi$. It may not be possible to extend the potential function to the full domain, especially in case it is not simply connected.
The second case $\vec k×\vec A=0$ can be stated even more restrictively, $\vec A$ has to be a multiple of $\vec k$, there is only one independent quantity.
More deeply, $∇×\vec A=0$ means that the one-form $\alpha=A_xdx+A_ydy+A_zdz$ is exact, $dα=0$, so that it is also locally closed, $α=dϕ$.
For the divergence something similar holds, $∇⋅\vec A=0$ is equivalent to the differential 2-form $α=A_xdy\wedge dz+A_ydz∧dx+A_zdx∧dy$ being exact, $dα=0$, so again locally closed, which means that $α=d\omega$ for some one-form $ω$ which is only determined up to gage transformations, as also $α=d(ω+dϕ)$. Or $\vec A=∇×(\vec V+∇ϕ)$. $ϕ$ can be chosen to fix one of the components of $\vec V$.