magic square in the complex plane with equal integrals along every horizontal, vertical and diagonal What can you say about a function defined on a square region of the complex plane, if the integral of the function along any horizontal, vertical or diagonal of the square is equal ? - an analytic magic square.
 A: CORRECTION: many thanks to Joel David Hamkins and Willie Wong; yes, we should take $I_1 = I_2 = 0$ (now edited out and removed). For some reason "constant" and "zero" keep getting mixed up together in my head!!
You can get lots of examples satisfying the vertical and horizontal integral conditions by choosing $f(x,y) = g(x)h(y)$ where $\int_0^1 g(x) dx = \int_0^1 h(y) dy = 0$. The diagonal integrals being equal to $0$ gives you two equations relating $g,h$, but these still leave very many possibilities for $g,h$.
(Note: I wrote $I_1$, $I_2$ before, but I've removed these to make it clearer).
Finally, taking arbitrary finite linear combinations of such $f$ (and also infinite linear combinations, if you are careful with convergence) gives you yet more examples.
If you want to restrict $f$ to, say, analytic functions, then this won't work - but you didn't say this in your question! (Although I suppose maybe you meant this, since you do say "analytic" magic square!)
EDIT: I'm definitely not claiming that every continuous example $f$ can be obtained in this way!! I just want to show that there are many, many possibilities for $f$.
A: Since this is tagged complex analysis:
(1) Such a function does not exist if you assume it is complex analytic on its domain and the various integrals evaluate to non-zero values.  
Proof: apply Cauchy's integral theorem to the following contour: starting from the origin, go horizontally to $(1,0)$, go up to $(1,1)$, and travel back down diagonally to $(0,0)$. 

(2) An analytic function with your properties must also be periodic. That is: $f(0,y) = f(1,y)$ and $f(x,0) = f(x,1)$. 
Proof: consider the contour from the origin to $(1,0)$ to $(1,y)$ to $(0,y)$ and back to the origin. This shows that $\int_0^y f(0,t) dt = \int_0^t f(1,t)dt$ for every $t$. Apply the fundamental theorem of calculus. 

From (2), apply Rademacher's characterisation of analytic functions (see Theorem 10 in this), you have that a periodic extension of $f$ is an analytic function on the complex plane, which immediately implies that $f\equiv 0$. 
So, to summarize: if $f$ is a function with your properties, which in addition is analytic in $(0,1)\times(0,1)\subset\mathbb{C}$ and is continuous up to the boundary, then $f\equiv 0$. 
A: Absolutely nothing beyond the definition. The complex plane is just R x R with some additional operations on it, which are irrelevant by your definition. The function doesn't have to be differentiable anywhere, continuous anywhere, or integrate to zero anywhere other than on the edges or on the main diagonals. Sorry. If you limited the question to say, meromorphic functions you might be able to get a better answer. In addition, you haven's specified directions, but i've taken it as implicit that you have.
If you have any open subset of C that crosses all 6 relevant integrals, and define an integrable function everywhere except that open region, you can always define the function on that subset such that the complete function satisfies the magic square condition.
