,
,
are pairwise distinct real numbers satisfying 
Prove that
Let 
Prove that
and 
be positive integers.
and 
.
.
and 
be the two foci of ellipse 
.
is a point on 
.
,
of natural numbers with 
of 
if the product of the elements of 
,
Y by baopbc, don2001, Juraev, mathmaths, rkm0959, baladin, champion999, Davi-8191, Ankoganit, Problem_Penetrator, kk108, tenplusten, Tawan, Delray, megarnie, Adventure10, Rounak_iitr
Find all integers 
for which each cell of 
table can be filled with one of the letters 
and 
in such a way that:
table are each labelled 
to in a natural order. Thus each cell corresponds to a pair of positive integer 
with 
. For 
, the table has 
diagonals of two types. A diagonal of first type consists all cells for which 
is a constant, and the diagonal of this second type consists all cells for which 
is constant.
for which each cell of 
table can be filled with one of the letters 
and 
in such a way that:
- in each row and each column, one third of the entries are
, one third are 
and one third are ; and - in any diagonal, if the number of entries on the diagonal is a multiple of three, then one third of the entries are , one third are and one third are .
table are each labelled 
to in a natural order. Thus each cell corresponds to a pair of positive integer 
with 
. For 
, the table has 
diagonals of two types. A diagonal of first type consists all cells for which 
is a constant, and the diagonal of this second type consists all cells for which 
is constant.This post has been edited 2 times. Last edited by shinichiman, Jul 11, 2016, 6:40 AM
.
,
.
equal to 
if it is not.
is in an IMO-state if the following holds true.
is equal for all 
.
gives 
is in an IMO-state.
is in an IMO-state.
is in an IMO-state.
is in an IMO-state.
,
is an IMO-state.
,
.
,
.
of these tables, each row and column are guaranteed to have same number of 
.
![\[
\begin{array}{|ccc|ccc|ccc|} \hline
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\\hline
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\\hline
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\\hline
\end{array}
\]](http://latex.artofproblemsolving.com/a/0/0/a004e6d8a94929389ec03a2208f08aa47e828045.png)
,
,
sub-boxes (of size 
each). Consider
,
diagonals mentioned in the problem.
.
lucky cells contain an 
letters 
letters 
modulo 
by hypothesis. On the other hand, it is easy to check that the cells in these rows and columns are exactly the lucky and unlucky cells, and that the total number of 
.
and 
is a multiple of 3. The construction is straightforward. The diagram shows the lucky 
)
)![\[
\begin{array}{|c|c|c|c|c|c|} \hline
& \circ & & & \circ & \\\hline
\circ & \times & \circ & \circ & \times & \circ\\\hline
& \circ & & & \circ & \\\hline
& \circ & & & \circ & \\\hline
\circ & \times & \circ & \circ & \times & \circ\\\hline
& \circ & & & \circ & \\\hline
\end{array}
\]](http://latex.artofproblemsolving.com/9/d/c/9dcf3aef7c24efc4882b4527bac99e97f9850536.png)
with 
and then put 
in cell 
the proportion of 
modulo 
and 
.
.
and rows and columns 
.
or 
hence 
is divisible by 
)
and can be achieved as described in above posts.
,
.
be the number of 
.
so 
so 
as desired.
work.
.
times and the rest 
.
,
.
as desired. 
,
to be the square in column 
.
be the set of all squares 
be the set of all squares. Then let 
.
of its entries as 
,
,
,
be defined as follows
,
,
(AKA the set of all squares on a diagonal with a number of squares that is a multiple of 
(AKA the set of all squares on a diagonal with a number of squares that is a multiple of ![\[C+D+E+F=4A+B,\]](http://latex.artofproblemsolving.com/4/c/1/4c1625e67c9bf8d15bee00fc2bbbe9db1ed2d7a6.png)
and by problem conditions, since we summed together diagonals that had numbers of squares that were multiples of 
must have equal numbers of squares that have 
must satisfy the same condition.
is the set of the entire grid, by problem conditions, this implies that the set 
must have equal numbers of 
.![\[3\mid|A|=\frac{(3k)^2}{9}=k^2 \iff 3\mid k^2 \iff 3\mid k,\]](http://latex.artofproblemsolving.com/5/1/f/51f98d5d13bb0c8f3375e5a48996a21e74d7ea36.png)
meaning that 
,
grid with 
times with the following 
tile to get our construction.![[asy]
size(300,300);
draw((0,0)--(9,0)--(9,9)--(0,9)--(0,0));
draw((1,0)--(1,9)); draw((2,0)--(2,9)); draw((3,0)--(3,9)); draw((4,0)--(4,9)); draw((5,0)--(5,9)); draw((6,0)--(6,9)); draw((7,0)--(7,9)); draw((8,0)--(8,9));
draw((0,1)--(9,1)); draw((0,2)--(9,2)); draw((0,3)--(9,3)); draw((0,4)--(9,4)); draw((0,5)--(9,5)); draw((0,6)--(9,6)); draw((0,7)--(9,7)); draw((0,8)--(9,8));
label("$O$", (0.5,0.5)); label("$O$", (0.5,1.5)); label("$O$", (0.5,2.5)); label("$M$", (0.5,3.5)); label("$M$", (0.5,4.5)); label("$M$", (0.5,5.5)); label("$I$", (0.5,6.5)); label("$I$", (0.5,7.5)); label("$I$", (0.5,8.5));
label("$O$", (3.5,0.5)); label("$O$", (3.5,1.5)); label("$O$", (3.5,2.5)); label("$M$", (3.5,3.5)); label("$M$", (3.5,4.5)); label("$M$", (3.5,5.5)); label("$I$", (3.5,6.5)); label("$I$", (3.5,7.5)); label("$I$", (3.5,8.5));
label("$O$", (6.5,0.5)); label("$O$", (6.5,1.5)); label("$O$", (6.5,2.5)); label("$M$", (6.5,3.5)); label("$M$", (6.5,4.5)); label("$M$", (6.5,5.5)); label("$I$", (6.5,6.5)); label("$I$", (6.5,7.5)); label("$I$", (6.5,8.5));
label("$I$", (1.5,0.5)); label("$I$", (1.5,1.5)); label("$I$", (1.5,2.5)); label("$O$", (1.5,3.5)); label("$O$", (1.5,4.5)); label("$O$", (1.5,5.5)); label("$M$", (1.5,6.5)); label("$M$", (1.5,7.5)); label("$M$", (1.5,8.5));
label("$I$", (4.5,0.5)); label("$I$", (4.5,1.5)); label("$I$", (4.5,2.5)); label("$O$", (4.5,3.5)); label("$O$", (4.5,4.5)); label("$O$", (4.5,5.5)); label("$M$", (4.5,6.5)); label("$M$", (4.5,7.5)); label("$M$", (4.5,8.5));
label("$I$", (7.5,0.5)); label("$I$", (7.5,1.5)); label("$I$", (7.5,2.5)); label("$O$", (7.5,3.5)); label("$O$", (7.5,4.5)); label("$O$", (7.5,5.5)); label("$M$", (7.5,6.5)); label("$M$", (7.5,7.5)); label("$M$", (7.5,8.5));
label("$M$", (2.5,0.5)); label("$M$", (2.5,1.5)); label("$M$", (2.5,2.5)); label("$I$", (2.5,3.5)); label("$I$", (2.5,4.5)); label("$I$", (2.5,5.5)); label("$O$", (2.5,6.5)); label("$O$", (2.5,7.5)); label("$O$", (2.5,8.5));
label("$M$", (5.5,0.5)); label("$M$", (5.5,1.5)); label("$M$", (5.5,2.5)); label("$I$", (5.5,3.5)); label("$I$", (5.5,4.5)); label("$I$", (5.5,5.5)); label("$O$", (5.5,6.5)); label("$O$", (5.5,7.5)); label("$O$", (5.5,8.5));
label("$M$", (8.5,0.5)); label("$M$", (8.5,1.5)); label("$M$", (8.5,2.5)); label("$I$", (8.5,3.5)); label("$I$", (8.5,4.5)); label("$I$", (8.5,5.5)); label("$O$", (8.5,6.5)); label("$O$", (8.5,7.5)); label("$O$", (8.5,8.5));
[/asy]](http://latex.artofproblemsolving.com/0/3/a/03a64fe6ef9e6400c4b772a937997085f4986bf8.png)
.
,
:
.
,
,
squares.
to be an integer, so 
and consider the following ![\[\begin{matrix}
I & M & O & I & M & O & I & M & O\\
I & M & O & I & M & O & I & M & O\\
I & M & O & I & M & O & I & M & O\\
M & O & I & M & O & I & M & O & I\\
M & O & I & M & O & I & M & O & I\\
M & O & I & M & O & I & M & O & I\\
O & I & M & O & I & M & O & I & M\\
O & I & M & O & I & M & O & I & M\\
O & I & M & O & I & M & O & I & M\\
\end{matrix}\]](http://latex.artofproblemsolving.com/c/d/8/cd8fb020cc2102dc7602a707744638a59eee5caa.png)
Then consider the above as a singular unit and duplicate it as a 
array which yields a valid construction for 
.
.
cold cells contain an 
cells with an 
.
,
cells with 
,
,
implying 
construction is tiling an 
grid with the 
doesn't work first observe 
doesn't work because you can't have an equal amount of 
with 
,
,
,
denote the number of 
,
,
,
,
,
similarly. Summing the number of 
.
.
.
or 
.
a contradiction. 
works. for the construction, consider a 
block with the columns 
,
,
in that order. Let 
,
,
be the 
being 
,
,
respectively. Notice that both diagonals of each 
have an equal number of 
,
,
grid, add up the number of squares that are 
t
,
,
when 
,
,
is subtracted, where 
,
are the number of 
.
row and column (zero-indexed) and subtract the entire grid. We get that there are the same number of 
squares that are 
.
with 
.
![\[\begin{matrix}
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M
\end{matrix}\]](http://latex.artofproblemsolving.com/3/5/1/3519fd47b97af0b4d92020d030d2a347d5afbafb.png)
and loop it.
so that 
count the number of ![\[\begin{matrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{matrix}\]](http://latex.artofproblemsolving.com/7/0/f/70fae2c9d3a46168d05714fd07c1e714be904fa5.png)
Impose coordinates by letting the bottom-left corner be 
,
,
and interpolating. By summing over the diagonals of first type with number of cells divisible by three we get that ![\[I(1) + I(5) + I(9) = M(1) + M(5) + M(9) = O(1) + O(5) + O(9).\]](http://latex.artofproblemsolving.com/3/a/d/3ad5419db292f0e29f52a2fe99b48ae1b3d3ba6f.png)
Similarly, by summing over the diagonals of second type with number of cells divisible by three we get that ![\[I(3) + I(5) + I(7) = M(3) + M(5) + M(7) = O(3) + O(5) + O(7).\]](http://latex.artofproblemsolving.com/2/3/1/2311f2bca01659f4727780b09f32dc6b50a96275.png)
Additionally, we have that ![\[I(2) + I(5) + I(8) = M(2) + M(5) + M(8) = O(2) + O(5) + O(8)\]](http://latex.artofproblemsolving.com/9/7/0/970240a244c8a88bdc5a58a1dca0576c290d815a.png)
and ![\[I(4) + I(5) + I(6) = M(4) + M(5) + M(6) = O(4) + O(5) + O(6)\]](http://latex.artofproblemsolving.com/7/1/2/71250de33bc95261ef9233884b5eccf02b40d6aa.png)
by summing over specific columns and rows. Hence, ![\[\sum_{i = 1}^9 I(i) + 3I(5) = \sum_{i = 1}^9 M(i) + 3M(5) = \sum_{i = 1}^9 O(i) + 3O(5).\]](http://latex.artofproblemsolving.com/5/7/2/57250dabc1163a8b0b9a8dd8d4eb7247c44ad4d9.png)
But ![\[\sum_{i = 1}^9 I(i) = \sum_{i = 1}^9 M(i) = \sum_{i = 1}^9 O(i)\]](http://latex.artofproblemsolving.com/4/9/f/49f24b2a4785bcfd17b93aae8709150d4dfda5ee.png)
by summing over all columns, so it follows that ![\[3I(5) = 3M(5) = 3O(5)\]](http://latex.artofproblemsolving.com/3/f/7/3f79daca7f7f4a302c498e04d040f0edbe1b7aef.png)
by subtracting. This is only possible if 
and hence if 
.
(where 
,
(the sum of leads). If 
denote the no. of 
and 
.
which implies 
or 
:
so if 
grid of 
so 
denote 
for the first three rows, 
for the next three, and 
for the final three.
,
