Sum of cells on infinite board is even Let $a,b,c$ be pairwise relatively prime positive integers. In an infinite checker board (infinite in all directions), each cell contains an integer. The sum of the integers in any $a\times a$ square is even, in any $b\times b$ square is even, and in any $c\times c$ square also even. Does it always follow that all numbers on the board are even?
If $a=1$ then all the numbers must be even. For higher $a,b,c$, we probably have to take overlapping squares of sizes $a,b,c$ to deduce that some cells are even. But it is not clear which ones.

Suppose that we put a light under each cell in this infinite grid, all of the lights starting off. We can, at will, switch all the state of all the lights in any $a\times a$, $b\times b$, or $c\times c$ square - that is, any light in this square that was on is turned off, and any light that was off is turned on. So, by choosing appropriate squares in succession, we may try to obtain various complex patterns of cells. However, what is of note here, is that the sum of the values in the lighted squares is always even, since, working in $\mathbb{Z}/\mathbb{2Z}$ - that is, considering only the parity of values - adding a value is the same as subtracting it, so when we flip all the lights in a square, the change in parity is the same regardless of whether lights were flipped on or off.
Now, if we can use this process to produce a lighted $1\times 1$ square, we win, since this implies that the sum of the lighted squares - of which there is but one - is even, and hence that square is even. To do this, notice that by switching $b$ side-by-side $a\times a$ squares, we could create a $ab\times a$ rectangle of lighted points. Similarly, we can create a $ab\times b$ rectangle by switching $a$ side-by-side $b\times b$ squares. Suppose we overlayed these such that two of their corners matched up and one was within the other - now we have a $ab\times|b-a|$ rectangle, since we basically lighted the bigger rectangle, then switched off lights in the smaller rectangle, but they had the same width, and hence yielded a rectangle. If we stacked one rectangle atop the other, we would have an $ab\times a+b$ rectangle. By similar reasoning, we can construct an $ab\times |k_1a + k_2b|$ rectangle for any $k_1,k_2\in \mathbb{Z}$. Since $a$ and $b$ are coprime, there exists a pair of integers $k_1$ and $k_2$ such that $k_1a+k_2b=1$, thus, by layering a lot of $a\times a$ and $b\times b$ squares, we get a $ab\times 1$ rectangle. Similarly, we may construct $bc\times 1$ and $ca\times 1$ rectangles. Then, when we layer these, we can get any rectangle expressible as, for $k_1,k_2,k_3\in\mathbb{Z}$:
$$(k_1ab + k_2bc + k_3ca)\times 1$$
and since $\gcd(ab,bc,ca)=1$, there exist $k_1,k_2,k_3$ such that the above simplifies to
$$1\times 1$$
implying that we can, by switching lights this way, yield a $1\times 1$ lighted square which must contain an even number. By translating this solution around, we can prove this of every square.
