be an integer. Ivan writes the numbers 
each on different cards. He then shuffles these 
cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.
Find the minimum value of 
be real numbers such that
Prove
be the set of integers. Determine all functions 
such that, for all integers 
and 
,
Proposed by Liam Baker, South Africa
Y by sami1618
The numbers from 2 to 99 are written on a board. At each step, one of the following operations is performed:
Choose a natural number 
such that 
. If both numbers and 
are on the board, erase both.
Choose a natural number such that 
. If both numbers and 
are on the board, erase both.
By performing these operations, what is the maximum number of numbers that can be erased from the board?
Choose a natural number 
such that 
. If both numbers and 
are on the board, erase both.
Choose a natural number such that 
. If both numbers and 
are on the board, erase both.By performing these operations, what is the maximum number of numbers that can be erased from the board?
![[asy]
import graph;
size(120);
real f(real t) {return 2cos(7*t) - 11sin(2*t) + 3cos(t);}
path g=polargraph(f,-10pi,10pi,10000, operator --);
draw(rotate(87)*g,magenta);
draw(rotate(86)*g,heavygreen);
path p = (5,1)--(6,1)--(6,2)--(5,2)--cycle;
real r=sqrt(2);
draw(scale(r^0)*rotate( 0)*p,red+linewidth(0.9));
draw(scale(r^1)*rotate(30)*p,orange+linewidth(0.9));
draw(scale(r^2)*rotate(60)*p,heavycyan+linewidth(0.8));
draw(scale(r^3)*rotate(90)*p,blue+linewidth(0.7));[/asy]](http://latex.artofproblemsolving.com/texer/n/nxikambb.png)
shouldn't be divisible by 
.
numbers can be erased from the board.
numbers in a 
grid as shown:![[asy]
size(8cm, 8cm);
real cellSize = 0.8;
pair corner = (0,0);
for (int i = 0; i <= 10; ++i) {
draw(corner + (i*cellSize, 0) -- corner + (i*cellSize, -10*cellSize));
draw(corner + (0, -i*cellSize) -- corner + (10*cellSize, -i*cellSize));
}
for (int i = 0; i < 10; ++i) {
for (int j = 0; j < 10; ++j) {
pair bottomLeft = (j*cellSize, -(i+1)*cellSize);
pair topRight = ((j+1)*cellSize, -i*cellSize);
path cell = box(bottomLeft, topRight);
if ((i + j) % 2 == 0) {
fill(cell, lightgray);
} else {
fill(cell, white);
}
}
}
for (int i = 0; i < 10; ++i) {
for (int j = 1; j < 9; ++j) {
int n = 10*i + j + 1;
pair center = corner + ((j + 0.5)*cellSize, -(i + 0.5)*cellSize);
label(string(n), center);
}
}
for (int i = 1; i < 9; ++i) {
for (int j = 0; j < 10; ++j) {
int n = 10*i + j + 1;
pair center = corner + ((j + 0.5)*cellSize, -(i + 0.5)*cellSize);
label(string(n), center);
}
}
for (int i = 0; i < 1; ++i) {
for (int j = 9; j < 10; ++j) {
int n = 10*i + j + 1;
pair center = corner + ((j + 0.5)*cellSize, -(i + 0.5)*cellSize);
label(string(n), center);
}
}
for (int i = 9; i < 10; ++i) {
for (int j = 0; j < 1; ++j) {
int n = 10*i + j + 1;
pair center = corner + ((j + 0.5)*cellSize, -(i + 0.5)*cellSize);
label(string(n), center);
}
}
draw((1*cellSize,0)--(10*cellSize,0)--(10*cellSize,-9*cellSize)--(9*cellSize,-9*cellSize)--(9*cellSize,-10*cellSize)--(0,-10*cellSize)--(0,-1*cellSize)--(1*cellSize,-1*cellSize)--cycle, orange+linewidth(2));
[/asy]](http://latex.artofproblemsolving.com/6/9/b/69b725ccb091e18877942b7840994ffc491a44af.png)
erasing vertical dominoes and operations of type 
erasing horizontal dominoes). Placing a checkerboard pattern over the grid, it is clear that it can not be covered entirely by dominoes, and it is trivial to place 
dominoes, finishing the problem.
