AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
and 
be sequences of real numbers for which 
and
for all positive integers 
. Prove that is eventually increasing (that is, there exists a positive integer 
for which 
for all 
).
and 
be two concentric circles. Let 
and 
be any two equilateral triangles inscribed in and respectively. If 
and 
are any two points on and respectively, show that ![\[ P'A^2 + P'B^2 + P'C^2 = A'P^2 + B'P^2 + C'P^2. \]](http://latex.artofproblemsolving.com/2/a/7/2a73de753f0332329b743d494272d3fb85f60ef9.png)
(meaning 
takes pairs of positive real numbers to positive real numbers) such that for any real numbers 
,
holds; and
then 
.
, 
, 
, 
and 
. Extend 
to meet the circumcircle of triangle 
at 
. Prove that 
. people at a party. Prove that there are two people such that, of the remaining 
people, there are at least 
of them, each of whom either knows both or else knows neither of the two. Assume that knowing is a symmetric relation, and that 
denotes the greatest integer less than or equal to 
.
such that![\[2^{n!}+1\mid 2^{m!}+19\]](http://latex.artofproblemsolving.com/8/7/e/87ede5616cffcb33aa187f5d8a13827470f741b4.png)
and a positive integer 
is given.
and for each 
![\[f_n(k)=\frac{a_n+\dots+a_{k+1}}{n-k}-\frac{a_k+\dots+a_1}{k}\]](http://latex.artofproblemsolving.com/0/4/1/041f80cf63f17c0f62c166021af474a5d305eb87.png)
Then for every integer 
the condition![\[a_{n+1}=\max_{0\leq k<n}(f_n(k))\]](http://latex.artofproblemsolving.com/b/1/f/b1f5aff787de8fc9c113cf8441a6b2900a19f24f.png)
is satisfied. Prove that this sequence must be eventually constant.
be a positive integer. For every positive integer 
the sequence 
is defined, where 
are integers. Among these sequences, for at most how many of them does all the elements of the sequence give different remainders when divided by ?
with 
. Points 
and are on 
and 
respectively such that 
. Lines 
and 
meet at 
. 
is the incenter of 
and 
is the orthocenter of 
. 
and 
are circles with diameter 
and 
, respectively, intersecting each other at points 
and 
. Prove that and lie on 
.
with 
with all the unit squares are initially white coloured. Every move, we can turn the color (from white to black or otherwise) from the 5 unit squares that form this T-pentomino which can be rotated or reflexed (see the image below). Determine all natural numbers such that all unit squares on the board can be made into all black after a finite number of moves.
such that![\[f(f(f(\frac{x+y}{2}))+x+y)=f(x)+f(y)+f(\frac{x+y}{2})\]](http://latex.artofproblemsolving.com/6/2/4/624afa78b68f810aaaecf0e1e2251047beeddc36.png)
for all rational numbers 
.
, but your problem clearly stated 
.
Prove that
,
in term of a function 
such that 
Prove that
)
![\[ f(f(x)^2 + f(y)) = x f(x) + y \]](http://latex.artofproblemsolving.com/d/0/2/d022cecec0757daff00434c6a686225ccf384eef.png)
.
such that
.
to get 
for 
.
for 
and 
.
to yield
We then have
If 
and 
for nonzero 
,
This makes 
or 
,
to find 
.
is not constant, we have 
and 
.
denote the assertion, and observe that there are no constant solutions. By 
,
,
,
.
,
.
,
,
,
,
,
,
for some real 
is the only solution.
![\[ f(0) = 1 \]](http://latex.artofproblemsolving.com/8/7/8/8784a89d399b9480a41862bb77e7645094282249.png)
![\[ f^k(x) = f(x\sqrt{k}) \quad \forall k \in \mathbb{N}, \forall x \geq 0. \]](http://latex.artofproblemsolving.com/7/3/1/731d31076d43a297e6a9b249cfffd88bf5871f18.png)
,
and 
if 
gives injective and varying 
.
is a constant value, and since 
and therefore 
.
.
gives that 
,
implies that 
,
.
for some constant 
,
.
from 
to reals where 
and $f^k(x)=f(x\sqrt k).
and 
and 
,
Since 
and 
,
This makes 
.
.
with 
in the given functional equation, we get![\[
g(x)^k = g(kx).
\]](http://latex.artofproblemsolving.com/e/c/2/ec229f3427eadce49a2b42957cd6cf6ee4210516.png)
also we can prove that it's positive always by contradiction, assume 
a 
,
> 0 take a sequence 
as n tends to 
from this we can contradict continuity therefore 
![\[
h(x) = \log g(x).
\]](http://latex.artofproblemsolving.com/7/e/f/7ef39ee05c2bf39ad71ee6a16459195daa8cf2b6.png)
and
,
,
that forms an Arithmetic Progression
Now, set 
Subtracting the two equations, we get that:-
for some 
value and 
be the set of real numbers. Determine all functions ![\[ f(x^2 - y^2) = x f(x) - y f(y) \]](http://latex.artofproblemsolving.com/6/8/f/68f78e0df627f6ac04052433c01627bf16f0af8e.png)
for all pairs of real numbers 
such that 
and![\[
\Bigl(f(ab)\Bigr)^2 = \Bigl(af(b)+f(a)\Bigr)\Bigl(bf(a)+f(b)\Bigr) \quad \forall a,b \in (0,1).
\]](http://latex.artofproblemsolving.com/0/f/2/0f2d0a333bdd73ff468072cfd716f94865385710.png)
of pretty well known FEs 
we see 
.
and since 
for arbitrary 
we see 
.
and since 
is the identity.
works for all real 
.
twice to get 
which is equivalent to 
gives 
for nonnegative real 
gives 
(Cauchy FE) for nonnegative real 
so it still works for all nonnegative real 
for all nonnegative rational 
so 
be a positive rational
so 
which is equivalent to 
so 
.