A problem of colours of vertices in equilateral triangle I have a problem with this exercise. I spent a lot of time, but I didn't figure out anything at the moment. Maybe anyone could help me? 
An equilateral triangle, with side lenght $n$, is divided into $n^2$ equilateral triangles. Each vertex of these triangles, which form a triangular lattice, is coloured black or white. We do a sequence of moves. In one move we change a colour of all vertices, which lie on a line containing a side of one of these $n^2$ triangles. Compute $n \geq 2$, when it's possible to undergo from state "all vertices are white" to state "accurately one is black".
I find a procedure for $n=2$ and $n=3$, so it's possible for this numbers. For $n=4$ - I spent a lot of time and I don't figure anything, so I think it's impossible for $n \geq4$ and I was trying to prove that - i tried to find invariants, semivariants, but I don't have any idea now... 
I will be very, very pleased, if anyone could help me, because I'm depressed, I spent a few days thinking about it and I don't know, what I should do now...
 A: The $n=4$ case is indeed impossible. Here is a proof.
First, I want to label the different kinds of vertices with letters, but sinc I don't know how to draw a triangle using Mathjax, let me just describe it ... Since you have been drawing these triangles for a few days, you should have no trouble following what I mean.
Ok, so as you well know, the lines that you can draw either hit 2,3,4, or 5 vertices. The kind of line that hits 5 vertices is of course one of the outside lines of the triangle. Let's say it goes through vertices A,B,C,B,A in that order. Now let's label the other vertices on the outside likewise. And finally, label the 3 inside vertices D. So yes, by symmetry, there are 4 kinds of vertices: A,B,C, and D. Also, there are 4 kinds of lines: the ABCBA line, the BDDB line, the CDC line, and the BB line. Let's call these the 5-,4-,3-, and 2-lines respectively.
Now, here are two important insights: First, the order in which you use the lines makes no difference. Second,  if there is any solution at all, then there is a solution where you use each line in the triangle either once or not at all, since using the same line twice, no matter what you do in between, is of course exactly the same as not making those two moves at all. 
So, in our reasoning, once we have established that we have to use some line in order to accomplish something, then it follows that we cannot use that line again. Put a different way, we can always test whether there is a solution for any $n$ by considering any of the $3n$ lines that cn be used or not, and thus exhausting all $2^{3n}$ possibilities.
For $n=4$ that would give 4096 possibilities ... Fortunately we can prove that none of them will work a little more efficiently.
OK, so now I will prove that there is no solution that leaves exactly one vertex black of any of the 4 types.A: the only way to change the number of black A's is by using 5-lines, but any 5-line will change exactly the color of exactly two A's, so given that you start with 0 black A's you can only ever end up with an even number of black A's and so never with exactly one black A. (this reasonig works of course for any triangle of any size $n\geq 2$)
B: the only lines that change B's are the 5-, 4-, and 2-lines, but notice that each of those lines change exactly two B's. So, given that we start with 0 black B's, we can only get an even number of black B's, so never exactly one.( this reasoning also works for any triangle of size $n\geq 3$)
C: only 3- and 5-lines work. But 3-lines change exactly two C's so to end up with exactly one black C you would need to use a 5-line. But, once you use one of the 5-lines, you need to use the other two 5-lines as well, or else you end up with two black A's. OK, but after using all of the 5-lines, all B's and C's are black. To now get one black C, you need to use either one or two of the 3-lines. And either way (you have to draw a picture here see that this is true), in order to make all B's white again, you either have to use all of the 2-lines and none of the 4-lines, or all of the 4-lines and none of the 2-lines. But in all cases, you end up with at least one black D. So, you can't end up with one black C and the rest all white.
D: only 3- and 4- lines affect D's, and since 4-lines affect two D's you will have to use at least one 3-line. However, once you use a 3-line, you need to change the affected C's back to white. Using a 5-line doesn't help, since as we saw before, in order for all the A's to remain white you would then have to use all 5-lines, meaning that you would still end up with at least one black C if you don't use all 3-lines. So, the only way to get an odd number of black D's and have all C's be white is to use all 3-lines. And note that if you use all 3-lines, you can't use any of the 5-lines, or else you have a black C again. OK, so the only way to get one Black D andthe rest all white is to use all 3-lines, and no 5-lines. Now, if we want one black D, we either have to use one or two 4-lines (again, a picture at this point is needed to follow the rest of the reasoning). But either way, note that we end up with two pairs of two neighboring B's that have opposite colors, and since all we have left to use at this point are 2-lines, we can never get all the B's get to white. So, ending up with one black D and the rest all white is impossible.
So, we cannot up end with one black vertex of any type, so the goal is impossible for $n=4$.
I took a quick look at $n>4$ and I am guessing these are all impossible as well, but I foun no good proof yet (I was hoping to find that the non-solvability of some $n$ would impy the non-solvability of larger $n$'s but no luck yet...
