How can I finish off this argument? Today I'm here because I need your help. I've got this problem:

Consider a cross like in the figure but with size $2021$. Every square have a $+1$. Every minute we select a sub-cross of size $3$ and multiply their squares by $-1$. It's posible achieve that all the squares of the cross with size $2021$ have a $-1$?


Here you can see that the subcross of size 3 is on the image. I was trying this problem for a while and I think I have a very good idea. After many attempts, with different sized cross, I could see that this seems to be imposible. Let me share my idea.

If you can get a configuration with all the squares in a cross having a $-1$ then we'll call this configuration a "Good configuration". If a Square have a $-1$ it will be coulored  white and black if has a $+1$. So, a good configuration muts be composed of only white squares. Now, consider a 2021 cross, if you color this cross like a chessboard, you can see that the sub-cross-3 always covers 4 squares of one color and 1 of the other, so the 2021 cross cannot be perfectly cover by the sub-cross-3 (and also, the number of squares in the 2021 cross is $\equiv$ $1$ $\pmod{5}$ ) .Thus since for getting a good configuration all the squares must be hit, there should be overlapping.

Now here is where I need some help because I don't know how to finnish this idea.

We can see that after a finite number of transformations, we must reach a point in which the sub-crosses, should start overlapping by the above argument, so that you can never get a good configuration since there is overlapping and either 2 or 1  squares of an overlapped cross must change color. Now think about like, the best possible configuration, in which we could have at most white squares as possible and at least black squares as possible. Even if only one movement else were needed, since the overlapping, we would know that at least one square will remain black and never could we get a good configuration.

The only problem is that I think this last argument lacks rigour, and I believe I should give an example on how to reach this "Best possible configuration" thing that I don't know how to do. Any help or idea would be helpful. (:
 A: You're playing Lights Out on the cross, which means that the order of pressing (selecting) sub-crosses doesn't matter, as well as whether any sub-cross is pressed $\{0,2,4\dots\}$ or $\{1,3,5\}$ times – each sub-cross only needs to be pressed (at its centre) $0$ or $1$ times.
Suppose there are $n$ squares on the northeast edge of the large cross, $S_1,S_2,\dots,S_n$ in order. They can only be toggled by crosses centred on the diagonal of squares immediately below them – $T_1,T_2,\dots,T_{n-1}$ where $T_i$ borders and can toggle $S_i$ and $S_{i+1}$. We use the same variable names $T_i$ to denote whether the sub-cross centred there is pressed; since we want $T_1,\dots,T_n$ to be all flipped (pressed) as well, we get these linear equations $\bmod2$:
$$S_1=T_1=1$$
$$S_2=T_1+T_2=1$$
$$\cdots$$
$$S_{n-1}=T_{n-2}+T_{n-1}=1$$
$$S_n=T_{n-1}=1$$
If you now solve these equations you will find they are inconsistent if $n$ is odd. A cross of size $2021$, howeve, has $n=1011$, so the task is impossible.
