If the curl of some vector function = 0, Is it a must that this vector function is the gradient of some other scalar function? I know of course that If the curl of a vector function is equal to zero, then the vector function is the gradient of some other scalar function, but is this a must?

if so, please give mathematical proof.
First I would like to note the difference between the following two statements:

If a vector field is the gradient of a scalar function then the curl of that vector field is zero.

If the curl of some vector field is zero then that vector field is a the gradient of some scalar field.

I have seen some trying to prove the first where I think you are asking for the second

I apologize for not giving full details on math here because I'm doing this on my tablet.

Anyway, if the curl of a vector field $F$ is zero $(\nabla\times F=0)$ then the surface integral of the resulting vector field over any arbitrary surface $S$ is also zero.

Stokes theorem (read the Wikipedia article on Kelvin-Stokes theorem) the surface integral of the curl of any vector field is equal to the closed line integral over the boundary curve.

Then since $\nabla\times F=0$ which implies that the surface integral of that vector field is zero then (BY STOKES theorem) the closed line integral of the boundary curve of that (arbitrary) selected surface is also zero.

Since the selection of the surface is arbitrary the. We can say the closed line integral of $F$ over any arbitrary closed curve is zero.

This implies that the line integral of the vector field $F$ is path independent which means the line integral over any curve only depending the initial and final position (not necessarily a closed curve)

To prove this just divide your closed path into two paths from point $P_{1}$ two point $P_{2}$, call those paths $A$ and $B$, the line integral over a closed path $C$ is equal to the summation of the line integral over paths $A$ and $B$ so:

$$
\oint\limits_C F \mathrm{d}\ell = \int\limits_{A_{P_1 \to P_2}} F \mathrm{d}\ell + \int\limits_{B_{P_2 \to P_1}} F \mathrm{d}\ell =0
$$
Then
$$
\begin{split}
\int\limits_{A_{P_1 \to P_2}}\! F \mathrm{d}\ell &amp;=- \int\limits_{B_{P_2 \to P_1}} F \mathrm{d}\ell\\
&amp;\Updownarrow\\
\int\limits_{A_{P_1 \to P_2}}\! F \mathrm{d}\ell &amp;= \int\limits_{B_{P_1 \to P_2}} F \mathrm{d}\ell
\end{split}
$$
( please note that those are alkaline integral and that there should be a for product sign between the two f and dl)

This latter equality implies that it doesn't matter your choice of the path $A$ or $B$ or any path because the result will be the same and it will only depend on the vector field $F$ and the two end points.

And since it only depends on the two points $P_{1}$ and $P_{2}$, then we can DEFINE a scalar field  $\Phi(P)$ (note that the points $P_{1}$ and $P_2$ are position vectors) such that 
$$\Phi(P_{2}) - \Phi(P_{1}) = \int\limits_{P_{1}}^{P_{2}}F \mathrm{d}\ell$$

(Note that the integral doesn't depend on the path and that is the only reason we can write it this way).

Now from the gradient theorem ( look for the Wikipedia article on gradient theorem ) $F=\nabla \Phi$.

Also look for the Wikipedia article on conservative fields.
It is rather sufficient to prove that the curl of a vector function $\mathbf{F}$ which is the gradient of a scalar-function $\phi$ is 0.

Let $\phi(x,y,z)$ be a scalar-function. Then its gradient will be $$\nabla\phi(x,y,z) = \frac{\partial \phi(x,y,z)}{\partial x}\hat{\mathbf{x}} + \frac{\partial \phi(x,y,z)}{\partial y}\hat{\mathbf{y}} + \frac{\partial \phi(x,y,z)}{\partial z}\hat{\mathbf{z}} \quad.$$ By assumption, $$\mathbf{F} = \nabla \phi$$. We have to show that $$\nabla \times \nabla \phi(x,y,z) = \mathbf{0}$$. 

Since $\nabla \times \mathbf{h} = \text{vector}$, we can write the components of this vector by the rule of cross-product i.e. $$(\nabla \times \mathbf{h})_x = \nabla_y h_z - \nabla_z h_y$$. So, we can write $$\text{curl}\mathbf{F} = \left(\frac{\partial F_z}{\partial y} - \frac{ F_y}{\partial z}\right)\hat{\mathbf{x}} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right)\hat{\mathbf{y}} +\left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\hat{\mathbf{z}} $$. But we know $$F_x = \nabla\phi_x = \frac{\partial \phi(x,y,z)}{\partial x} \\\\\ F_y = \nabla\phi_y =\frac{\partial \phi(x,y,z)}{\partial y} \\\\\\\ F_z = \nabla \phi_z =  \frac{\partial \phi(x,y,z)}{\partial z}$$. Substituting these in the definition of $\text{curl} \mathbf{F}$, we get $$\text{curl} \mathbf{F} = \nabla \times \nabla \phi = \left(\frac{\partial^2 \phi}{\partial y \partial z} - \frac{\partial^2 \phi}{\partial z \partial y}\right)\hat{\mathbf{x}} + \left(\frac{\partial^2 \phi}{\partial z \partial x} - \frac{\partial^2 \phi}{\partial x \partial z}\right)\hat{\mathbf{y}} + \left(\frac{\partial^2 \phi}{\partial x \partial y} - \frac{\partial^2 \phi}{\partial y \partial x}\right)\hat{\mathbf{z}} = \mathbf{0} \qquad .$$