[/quote]
,
such that ![\[ f( xf(x) + f(y) ) = f^2(x) + y \]](http://latex.artofproblemsolving.com/9/7/5/975510041aa2af0cfbfefac1ad76f0a4cf12b0ee.png)
for all 
.
such that the sequence 
,
,
such that the sequence 
, for 
, is periodic modulo 
be an integer. Isaac wants to cover an 
board with some tiles, without overlaps and without going off the board. For this purpose, Isaac has many tiles of the form:
squares of the board. Isaac is allowed to choose any tile and, with a straight cut, divide the tile into two new tiles. Each new tile must be able to cover exactly an integer number of squares of the board. For which values of 
is it possible for Isaac to achieve his goal by making at most two cuts?
such that 
for any 
are real numbers. Prove that 
for all real numbers 
.
board, players 
and 
take turns in a game. On each turn, a player selects an edge of the board (not on the board border) and makes a cut along that edge. If after the move the board is split into more than one piece, then that player loses the game.
.
and ![\[
\left| \alpha - \frac{m}{n} \right| < \frac{1}{m}.
\]](http://latex.artofproblemsolving.com/d/c/3/dc3a3ebc402a7396e2675c4e640add2e7949e3cd.png)
find all 
satisfying 
:![\[g(f(x)-y)=f(g(y))+x.\]](http://latex.artofproblemsolving.com/3/d/7/3d7dd5cd43f2936d6e743474744dcd883633bcca.png)
be a positive real number. The scorpion tries to catch the flea on a 
chessboard. The length of the side of each small square of the chessboard is 
.
vertexes of the chessboard and, in each turn, can jump from the vertex where it is to one of the adjacent vertexes. The scorpion moves on the boundary line of the chessboard, and, in each turn, it can walk along any path of length less than 
,
an alphabet with 
satisfying the following conditions:
,
.
,
.
such that 
and 
.![$P, Q \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/c/3/0/c301578bade5b93056d77454921cba0a9812212c.png)
be relatively prime nonconstant polynomials. Show that there can be at most three real numbers 
such that 
is the square of a polynomial.
rectangle can be tiled using 
ways to file a fixed 
rectangle 
with 
pieces. (symmetric constructions are supposed to be different.)
be non-negative real numbers such that 
.![\[
6xyz \leq x(1 - x) + y(1 - y) + z(1 - z),
\]](http://latex.artofproblemsolving.com/f/5/b/f5b208ea6439994cb40f6b7f0806c216d9868e86.png)
and determine when equality holds.
Y by GammaBetaAlpha, Adventure10, jhu08, Mango247, GA34-261
An 
-term sequence 
in which each term is either 0 or 1 is called a binary sequence of length . Let 
be the number of binary sequences of length containing no three consecutive terms equal to 0, 1, 0 in that order. Let 
be the number of binary sequences of length that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that 
for all positive integers .
-term sequence 
in which each term is either 0 or 1 is called a binary sequence of length . Let 
be the number of binary sequences of length containing no three consecutive terms equal to 0, 1, 0 in that order. Let 
be the number of binary sequences of length that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that 
for all positive integers .
-
in it, we can define two 
as follows:
t
or 
in them: furthermore, when we vary the sequence 
be the binary sequences of length 
be the binary sequences of length 
or 
in that order.
for all positive integers 
.
are trivial. Since the sequences are too short for any restrictions to apply, 
because there are two choices 
possible binary sequences of length 
,
since only one of these is eliminated. There are 
possible binary sequences of length 
since only two of these are eliminated. Thus, 
.
for all positive integers through 
:
is the number of 
.
is the number of 
or 
.
is the number of 
is the number of 
is the number of 
is the number of 
or 
.
is the number of 
is the number of 
since we can add 
since we can add 
for any 
by the induction hypothesis.
.
since we can get an 
with the exception that we can’t add 
since we can get a 
.
.
.
since if 
ends in 
.
choices for 
.
since 
does not end in 
since 
.
and 
,
.
construct another sequence 
of length 
and 
.It is easy to see that such a constuction is bijective.Also the sequence of the first type has a subsequence 
the sequence of the second type has a subsquence 
or 
.
can be seen analogously by interchanging the 
denote the number of strings counted by 
,
denote the number of strings counted by 
or 
.
.
.
choices for the first 
bits and 
strings ending with ![\[a_n = 2a_{n-1} - c_{n-1}.\]](http://latex.artofproblemsolving.com/5/0/5/5050025a04fe9be3b7d0a340435842f61016bba5.png)
Count 
,![\[b_n = 2b_{n-1} - d_{n-1}.\]](http://latex.artofproblemsolving.com/f/1/4/f14277cafc10428fbacadc097634175b902ca27a.png)
To count 
choices for the first 
bits and must append 
-![\[c_n = a_{n-2} - c_{n-2}.\]](http://latex.artofproblemsolving.com/8/f/1/8f137efaa4269df399e17568b2281b97865b394b.png)
Count 
choices for the first 
t
and ![\[d_n = b_{n-2} - d_{n-2}.\]](http://latex.artofproblemsolving.com/7/1/4/71431493098fb156313d722a6fb3f88aefde5a4f.png)
Now apply induction using these recursions to finish.
there are 
there are 
.![\[ (y_1, \dots, y_{n+1}) \mapsto (y_1 + y_2, y_2 + y_3, \dots, y_n + y_{n+1}). \]](http://latex.artofproblemsolving.com/8/d/8/8d8cd5ae85593ff97ef72119ad02a352f49e733d.png)
It is 
.
"
. We proceed using induction. Base case clearly works.
and then we would only have to handle the condition where no 
.
denote the number of 
.
denote the number of 
.
.
and 
.
where the last equality follows due to induction. It just leaves us to prove 
.
We now have 
where the last equality again follows due to induction. This simply follows from that the digit after 
,
where 
.
that does not contain the string 
is the same as the number of 
as desired.
,
via 
.
where 
.
and 
and they are also the only ones which do. Thus, any sequence avoiding 
.
to 
if 
and 
if 
,
![\[(y_1, \dots, y_{n+1}) \to (y_1 + y_2, y_2 + y_3, \dots, y_n + y_{n+1}). \]](http://latex.artofproblemsolving.com/5/8/b/58b956de416fd29ff9a5920fdca0ba446262b6bb.png)
Obviously, each sequence is considered twice, so we are done.
