Neutral points in a system of charges on the vertices of a square There are four positive charges of equal magnitude placed at the four vertices of a square. Is there any point where the electric field vanishes (neutral point) within the square and in its plane, other than its center?
 A: I believe there is such a point. 
Let us assume that unit charges are located in points $[\pm 1,\pm 1]$ of a plane with coordinates $x,y$. Let us consider the field on the abscissa ($y=0$). The $y$-component of the field is zero on the abscissa due to a symmetry. It is obvious that the $x$-component of the field at the "left infinity" ($x\rightarrow-\infty$) is directed to the left ($E_x<0$)) (provided the charges are positive and repel a positive charge). According to my calculation, the field at point,say, $[-\frac{1}{2},0]$ is directed to the right. Therefore, there is an intermediate point on the abscissa between $-\infty$ and $-\frac{1}{2}$ (therefore, not in the center) where the field vanishes. Symmetry will give three more such points.
By the way, it looks like the expression for the field given by @Haru Fujimura is not correct.

EDIT (9/18/2016): So let us calculate the $x$-component of the field at point $[-\frac{1}{2},0]$, assuming each charge equals +1. The distance from this point to the left bottom charge is $\sqrt{1+\frac{1}{4}}=\frac{\sqrt{5}}{2}$, the relevant cosine is $\frac{1}{\sqrt{5}}$, so the contribution of this charge to the $x$-component of the electric field is $\frac{1}{\sqrt{5}}\frac{4}{5}\approx 0.36$.
The distance from point $[-\frac{1}{2},0]$ to the right bottom charge is $\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}$, the relevant cosine is $\frac{3}{\sqrt{13}}$, so the contribution of this charge to the $x$-component of the electric field is -$\frac{3}{\sqrt{13}}\frac{4}{13}\approx -0.26$  (the minus sign arises because the contributions of the left and right charges have opposite signs). The calculation of the contributions of the top charges yields the same results. Thus, the field at this point is directed to the right.

EDIT [9/19/2016]: It is clear from the edited question that the point is required to be not just in the plane of the square, but also within the square. To adjust the proof, one can just consider point [-1,0] instead of [-infinity,0]: it is obvious that the field is directed to the left in this point. Therefore, there is an intermediate point between [-1,0] and [-1/2,0], i.e., within the square, where the field vanishes.
A: There are four extra points within the square where field vanishes. Try to solve for such a point, you may want to solve it numerically.
A: No.
Think of this Geometrically. 
if each charge generates a field: $ \vec{E}_i = {kq \over {|\vec{r}-\vec{r}_i|}^2} \hat{({\vec{r}-\vec{r}_i})} $ one can see that the center of the square is the only point where $\Sigma \vec{E}_i =0$. This is because of the fact that any point, to the left or to the right, up or down, from the center, has a nonzero field. Now, the direction of the field would depend on where that point is. 
Perhaps if you drew some field lines it would help you understand, finding a field plotting software would be most helpful here, but I know not of one which i could recommend. 
