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be three non-negative real numbers satisfying 
such that ![\[f(x+f(x)+2f(y))+f(2f(x)-y)=4x+f(y)\]](http://latex.artofproblemsolving.com/4/4/2/4423651bb829b92debd65a23f6cee3d4d4b8530f.png)
holds for all reals 
and 
.
, such that: 
For all 
.
is 
, maybe even with rather elementary means? As usual, 
denotes the 
-th prime number. The problem of that limit came up in my partial solution of this problem: https://artofproblemsolving.com/community/c7h3495516.
. Define 
as the average of all 
for 
. Determine the value of ![\[ a = \lim_{k \to \infty} \sup_{n \geq k} \frac{a_n}{n^4} .\]](http://latex.artofproblemsolving.com/5/2/d/52d93af7aafa6657689246d9a74c8eb05d96efa3.png)
be a Chebyshev polynomial of the second kind. If n>2 and x > 2 is a integer, Could 
be a prime number?
be an open interval and 
a twice differentiable function such that 
for any 
Prove that 
for any
which verify relation![\[
\left( f'(x) \right)^2 + f''(x) \leq 0, \forall x \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/3/a/2/3a235e8be4c61c32f376999e61d64973db21dc75.png)
has property 
if for any real number 
there exists a 
such that 
for all 
that is not monotone on 
is nondecreasing.
is field. Define addition of elements of 
coordinate wise and for 
and 
define 
. is a vector space over field 
points?
points?
, with their second derivative being continuous, such that the following holds for all : ![\[(f(x)-g(y))(f'(x)-g'(y))(f''(x)-g''(y))=0\]](http://latex.artofproblemsolving.com/0/2/9/029a935e28ca0a9b36e65702689cca91eeb8f614.png)
board, show that there is a 
block on the board in which at least 3 of the squares are colored red.
squares of a board be colored red. This colouring will be called a k - saturation if and only if coloring any one of the remaining squares red will result in a block of 
squares at least 
of which are red. Find the least such that a k - saturation exists.
red in some 
board . Suppose otherwise. Note that we cannot have more than red squares in every 
adjacent columns. Since we have columns, we can have at most 
red squares. Contradiction.
, we cannot have 
or more red squares unless it is of the following configuration:
or more red squares. In a sub-board with rows and columns, suppose we can find two adjacent red squares in a row, then consider these two columns with either the immediate right column or left column. We can have at most red squares in these columns. Hence we have at least 
left for 
columns. Grouping them into blocks of , we have a contradiction. Hence we cannot find two adjacent red squares in a row, and hence it must be
red, we have 
left for 
rows. Contradiction. If not, remove the top rows. We will be removing at most red. Repeat this process to get a contradiction.
sub-rectangle of the 
square. It is made up of overlapping 
block. The maximum number of red squares in the 
rectangle is because if we choose any red squares, it is easy to show that of them are part of 
square. Hence, since there are 
non-overlapping rectangles of this configuration, with a total of 
red squares, it leaves row of uncolored squares at the bottom. All of these are not part of any considered 
square, and hence can be colored for a maximum of 
red squares.
,
,
,
.
.
red squares in 
is the best.