be natural numbers such that: 
and 
.
?
,
and 
are taken on the legs 
and 
,
and 
.
and 
.
be the real polonomial such that all roots are real and distinct. Prove that there is a rational number 
that all roots of 
are real numbers
Y by GeoKing
Let 
be a positive integer. Initially, a 
grid has 
black cells and the rest white cells. The following two operations are allowed :
(1) If a
square has exactly three black cells, the fourth is changed to a black cell;
(2) If there are exactly two black cells in a
square, the black cells are changed to white and white to black.
Find the smallest positive integer such that for any configuration of the grid with black cells, all cells can be black after a finite number of operations.
be a positive integer. Initially, a 
grid has 
black cells and the rest white cells. The following two operations are allowed :(1) If a
square has exactly three black cells, the fourth is changed to a black cell;(2) If there are exactly two black cells in a
square, the black cells are changed to white and white to black.Find the smallest positive integer
such that for any configuration of the grid with black cells, all cells can be black after a finite number of operations.This post has been edited 3 times. Last edited by David-Vieta, Apr 1, 2023, 11:58 PM
vertical strip with 
and 
black cells where no black domino can be created.
.
can fail: I divide the board into 
2x2 squares. For squares above the main diagonal of 2x2 squares, paint the upper right square black and for squares below the main diagonal only paint the lower left square black.
works: if I can somehow get the number of black cells to increase, I clearly win. The objective is to get two black cells to share an edge.
squares on an 
square are painted black. Then I can get two squares to be adjacent.
are black and 
is one square below and to the right of 
,
are on the leftmost column or the bottom row. We can setup the coordinates for the grids such that if 
,
.
to 
(note down is positive), and 
to 
for 
.
be the number of black cells that are not on the leftmost row or bottom column. If 
,
cells on the leftmost row or bottom column, so the total number of black cells is at most 
.
.
So black squares must have assigned squares on the leftmost column or bottom row. Suppose the assigned squares divide the leftmost column and bottom row into "blocks" 
.
.
.
black squares, this is greater than 
.
,
"separators" from assigned squares. Observe that 
and if 
is the number of separators, 
.
be positive integers. If I can set 
.
is easy. Henceforth assume 
denote the maximum number of initial black squares so that the number of black squares never increases. It is easy to show that 
.
.
by symmetry.
we can apply inductive hypothesis on 
on rows 2 to 
because row 1 has at most 
black square. If row 1 has two black squares then I can move the "domino" right, create a "domino" on rows 1 and 2 and make the number of black squares increase. (*)
we note that row 1 has at most 
squares since there cannot be two adjacent black squares on row 1, or I can move them down and also move the two vertically adjacent black squares right to make the number of black squares increase. This implies that row 1 and row 2 in total can have at most 
squares, because if row 2 has a square, we can do (*) to see that it has at most one square and row 1 also has at most one square.
,
.
yields that there exists at most 
squares on rows 
to 
.
.
and we cannot move the domino up to rows 
,
such that row 
we are back to the 
,
on the first 
rows, and on 
on rows 
(note 
)
and row 3 has no black cells, because otherwise I can use (*) to go back to the 
cells. Applying i.h. on rows 
to 
![[asy]
size(150);
void tile(int a, int b){
fill(shift(a,b)*unitsquare,rgb(0.2,0.2,0.2));
D((a,b-1)--(a-1,b)--(a+1,b+2)--(a+2,b+1)--(a,b-1),linewidth(1)+blue);
}
int i, j;
for(i=0; i<8; ++i) for(j=0; j<8; ++j) if(i%2==1) if(j%2==0) if(i-j>0) tile(i, j);
for(i=0; i<8; ++i) for(j=0; j<8; ++j) if(i%2==0) if(j%2==1) if(j-i>0) tile(i, j);
D((0,0)--(8,0)--(8,8)--(0,8)--(0,0),linewidth(1.5));
[/asy]](http://latex.artofproblemsolving.com/a/2/d/a2d42e4b8611c8bae03323fc07ced1786059ba85.png)
works: we will assume we can't get 2 black squares adjacent and show 
(
![[asy]
size(120);
void tile(int a, int b, int c){
fill(shift(a,b)*unitsquare,rgb(0.2,0.2,0.2));
if(c==0) D((a+1,b-1)--(a-1,b+1)--(a,b+2)--(a+2,b)--(a+1,b-1),linewidth(1)+blue);
if(c==1) D((a,b-1)--(a-1,b)--(a+1,b+2)--(a+2,b+1)--(a,b-1),linewidth(1)+blue);
}
tile(0,1,1); tile(1,0,1); tile(2,2,0); tile(3,0,1); tile(5,0,1); tile(0,5,1); tile(1,4,1); tile(3,5,1); tile(4,3,0); tile(5,4,0);
D((-0.5,-0.5)--(4.5,-0.5)--(4.5,2.5)--(3.5,2.5)--(3.5,3.5)--(0.5,3.5)--(0.5,2.5)--(-0.5,2.5)--(-0.5,-0.5),linewidth(1)+red);
D((4.5,-0.5)--(6.5,-0.5)--(6.5,1.5)--(4.5,1.5)--(4.5,-0.5),linewidth(1)+red);
D((-0.5,3.5)--(-0.5,6.5)--(2.5,6.5)--(2.5,3.5)--(-0.5,3.5),linewidth(1)+red);
D((3.5,2.5)--(3.5,3.5)--(2.5,3.5)--(2.5,6.5)--(5.5,6.5)--(5.5,5.5)--(6.5,5.5)--(6.5,2.5)--(3.5,2.5),linewidth(1)+red);
D((0,0)--(6,0)--(6,6)--(0,6)--(0,0),linewidth(1.5));
[/asy]](http://latex.artofproblemsolving.com/2/6/4/26493f055f12014acc91411528340091a75c1fe7.png)
