AoPS Academy
are real numbers such that 
prove inequality 
chessboard is 
. (Two bishops are said to be attacking if they are on a common diagonal).
satisfying 
, where 
denotes the set of all real numbers.
be the set of sequences of length 2018 whose terms are in the set 
and sum to 3860. Prove that the cardinality of is at most![\[2^{3860} \cdot \left(\frac{2018}{2048}\right)^{2018}.\]](http://latex.artofproblemsolving.com/d/e/8/de857941803349e275e9c9aa968a201f2988cb8a.png)
where 
denotes the Fibonacci sequence.**
is a derivable function with the property 
, for every 
, then f is a polynomial function of degree smaller or equal than 1
be a function that admits a primitive 
such that 
if 
such that 
However I avoid details !
) that there is only one way of successfully attacking the problem, and that's what makes it beautiful -all other paths fail!
(and thus the solution)
![\[T_1 z = \frac{z + 2}{z + 3}, \qquad T_2 z = \frac z{z + 1},\]](http://latex.artofproblemsolving.com/b/7/2/b72b5e302cecdf2649c43a078aea3aa9a88ef2cc.png)
find 
and 
and 
in terms of 
and 
of 
.
pairwise disjoint, non-empty subsets 
such that 
and 
or 
in fact.
,
three element subsets of 
have equal sum by pigeonhole. Now, take these 
,
,
disjoint with 
and we are done. If not, by Turan's theorem it has at most 
edges, with the 
being an equality case. So therefore the average degree is at most 
,
.
,
,
,
,
.
and note that 
,
(WLOG 
to be the two element set consisting of the elements of 
,
,
and it's clear to see 
,
,
satisfy the given property so we are done in this case also.