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.
, the function 
is defined as, 
and for all 
, if the prime factorization of is 
then 
. We have an arithmetic sequence 
. If for a positive integer 
, the sequence 
is also an arithmetic sequence, show that the sequence 
has to be constant.
, 
are coprime positive integers. Find all 
such that:
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 
.
vertices, each edge has one of the colors 
, 
, or 
. For each 
, if the vertices can be divided into 
groups such that any two vertices connected by an edge of color 
are in different groups, find the minimum possible value of 
.
+1 w
be given pairwise coprime positive integers where 
. Let 
be positive integers. We call 
to be a grandson of if and only if, for all possible piles of stones whose total mass adds up to and consist of stones with masses , it's possible to take some of the stones out from this pile in a way that in the end, we can obtain a new pile of stones with total mass of . Find the greatest possible number that doesn't have any grandsons.
, for which 
for all functions 
, who satisfy the equation 
for all non-negative real numbers 
and 
.
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)
in term of a function 
such that 
Prove that
Find the value of 
Find the value of 
Find the value of 
Prove that 
,
,
be defined as![\[
x=\frac{a}{b-c},\quad y=\frac{b}{c-a},\quad z=\frac{c}{a-b}.
\]](http://latex.artofproblemsolving.com/c/6/b/c6b2629dfafd73409a6d911d1dd984ff4c70a94c.png)
Then, the system of equations![\[
a - xb + xc = 0,
\]](http://latex.artofproblemsolving.com/d/9/b/d9b088945fa707c2bfd4cad278308b35642dbc49.png)
![\[
ya + b - yc = 0,
\]](http://latex.artofproblemsolving.com/1/6/e/16eace37b2c4c09aa59968e0d299e1f982e0213d.png)
![\[
-za + zb + c = 0,
\]](http://latex.artofproblemsolving.com/8/6/f/86f4f7da9c109d2621815fef961b7f16c5d164fc.png)
admits a nonzero solution 
.![\[
\begin{vmatrix}
1 & -x & x \\
y & 1 & -y \\
-z & z & 1
\end{vmatrix}
= xy + yz + zx + 1 = 0.
\]](http://latex.artofproblemsolving.com/3/a/1/3a10ec43739f905365053fb18558037bda2cce8e.png)
It follows that![\[
(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=2-2=0.
\]](http://latex.artofproblemsolving.com/4/6/0/4605588d4b8c63e3de8fdbeec43959607c06c5c3.png)
