@Michael Fan Zhang: could you please explain the relation between the two sets of equations. I can't see the equivalence.

Thank you. Yes, I was aware of it, but considered the contrary case to be trivial. I have now included it in my answer.

@ seno44: (28.10.15 14:04) unfortunately, I have just discovered a flaw in the "proof" of the recent comment. I try to mend it and return.

@ seno44: If the generalization I proposed in my last comment is correct, the Taylor method is easily generalized. As far as I can see it leads to an infinite set of equations of the form $(vx^k, vv) = 0 (k=0,1,2,...)$ where $vv = (v_{10}, ..., v_{N0})$ , $x^k = (x_{10}^k, ..., x_{N0}^k)$, and $(a,b)$ is the scalar product of the vectors $a$ and $b$. As the equations must hold for all k, the only solution for $vv$ is the trivial one. QED.

@ seno44: Ok. Please be more specific.Do you wish to generalize 0 = y_{10} + y_{20} to 0 = y_{10} + y_{20} + ... + y_{n0} ?

@ seno44: I agree, if you consider the Taylor Expansion "very dificult" which, honestly, I wouldn't. But, you are right, there should be some simple symmetry argument to give the proof (even simpler than in the answer of Fritz Veeman) because the contrary holds for a symmetric potential, i.e. you can have $d=0$ iff $x_{10}+x_{20}=0$ and $y_{10}+y_{20}=0$

@seno 44: Thanks for the hint to my trivial error. I have given a complete solution of the OP in the meantime.