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C-Cube The problem

to 8 º mathematical challenge of the newspaper "El País". It is to be placing on each vertex of a cube (a die mathematically speaking, that we understand) a value can be 1 or -1. With this we have 8 values, one for each corner, we're going to add. The result we add a number for each face of the cube that correspond to the product of its vertices (which must be 1 or -1 depending on the number of -1 that is in the vertices). Adding these 14 values, girls who proposed the problem ask that this amount is 0 by changing the values \u200b\u200bof the vertices at will.
The 256 possible cases are less in reality because many cases are symmetries or rotations of other cases. Advised not to try all cases for one simple reason: there is no solution. Oops, I escaped. Well, as I have annoyed saying there is no solution, we must now show why there is not no test all possible cases. I leave
This entry scheduled for 23:30 with my proposed solution in case someone wants to copy and send. This is the link of "country" with the proposal of the problem in video. The poor kid will mess up the end, but as they had recorded almost the entire presentation, someone with evil intentions decided to leave it to people laugh a little of it. We are involved around 2000 people, which says those who are really interested in this sort of thing. I will not say that Spain is participating because many people outside our borders.

response is no solution.
Keep in mind that the 256 possible combinations are not really so since some of them are rotations or symmetries of others, but still calculate all time consuming.
Let's name to the upper edges of the A1, A2, A3 and A4, and the underside B1, B2, B3 and B4.
Let's split the load into 2 groups, first the sum of the values \u200b\u200bof the edges, which is adding Ai and Bi with i = 1,2,3,4, and second the sum of the products of the edges of each side, 6 total. A1A2A3A4, B1B2B3B4, A1A2B1B2, A3A4B3B4, A1A3B1B3 and A2A4B2B4.
Suppose we change a value, such as the A1 rotations and symmetries gives equal value to change. Change A1 in the first group involves moving from -1 to 1 or vice versa, thus the result of the sum would be -2 or +2. A1 Change in the second group would like to change the polarity of 3 of these products. Possible cases would be the following:
-1, -1, -1 -> 1,1,1 => +6
-1, -1.1 -> 1.1, -1 => +2
-1,1,1 -> 1, -1, -1 => -2
1,1,1 -> -1, -1, -1 => ; -6
The result of the variation of the sum-case scenario each time you change an edge would then be:
a) -2 +4 +6 =
b) +2 +8 +6 =
c) -2 +2 = 0
d) +2 +4 +2 =
e) -2 to 2 =- 4
f) +2-2 = 0
g) -2 to 6 =- 8
h) +2-6 =- 4
With this we can deduce that each variation is to increase or reduce the amount in a multiple of 4.
Take now one of the possible cases. Adding individual variations of its edges can reach every possible combination and we know that these variations are multiples of 4. Let us start with the one in which all edges are 1 and we find that the sum of the first group is 8, the total number of edges, and the sum of the second group, being the product in the 6 +1 possible cases is 6, bringing the total would be 14.
As we reach the case where the sum is 0, we deduce that it is not possible to reach that solution as there is no integer x such that 14 +4 X = 0.