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.
attaining all positive values and positive numbers 
are given. Numbers 
satisfy
for every 
. Prove that there exists a positive integer 
satisfying 
.
?
and colors are given. We will say that a set of 
points on a plane is 
, if it contains exactly 2 points of each color and if lines connecting every two points of the same color are pairwise distinct. Find, in terms of the least integer 
such that: in every set of 
points of a plane, no three of which are collinear, consisting of points of every color there exists a subset.
and a set 
consisting of 
disting positive integers smaller than 
are given. Prove that there exists a positive integer 
that can be written in the form 
, for 
in at least 
different ways.
, let 
be the set of all positive integers not exceeding that are relatively prime to . Consider the polynomial![\[
P_n(x) = \sum_{k \in S_n} x^{k - 1}
\]](http://latex.artofproblemsolving.com/3/a/c/3ac60cb70baf4dfcf297e9af87480a8ee34ba70c.png)
and a polynomial 
with integer coefficients such that![\[
P_n(x) = (x^{r_n} + 1) Q_n(x).
\]](http://latex.artofproblemsolving.com/7/0/e/70e19a9d3e0a7e97abc7aa08d3f23126c14ea7d0.png)
such that 
is irreducible in ![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
.
integers are such that 
is prime and 
. Prove that there exists a positive integer 
such that 
.
board, each square is colored either white or black, with all the squares on the border of the board being black. Additionally, no 
square within the board has all four squares of the same color. Prove that the board contains a square colored like a chessboard.
, where 
, then the number of integers in the range of 
is __
of 
in 
is the second intersection of 
with the circumcircle, the lines 
and 
are parallel if and only if the pencil 
is harmonic. The conclusion follows from the fact that 
, where 
is the foot of the 
-altitude in triangle 
.
and P and Q are homologues points, hence the conclusion follows.
the altitudes 
and 
intersect at 
is the circumcenter, and the midpoint of side 
. The line intersects side at 
, while the lines 
and 
meet at 
. Prove that the lines and are parallel.
. Let 
through 
intersects 
at 
But, 
Hence, 
be the foot of the 
-altitude, and let 
be the -antipode; it suffices to show 
. Notice that 
, so 
Also, 
It is evident now that 
, as needed.
and 
are similar parts in the inverse similarity 
, which explains a lot.
be reals such that 
and 
.
prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
then 
.
,
.
.
.
is a parrallelogram. We have :![$ \frac{BP}{PH}=\frac{[BMN]}{[HMN]}=\frac{BM.BN}{HM.HN}$](http://latex.artofproblemsolving.com/6/4/b/64b73edf9c3451ef0429789c2d955ac19c66f495.png)
![$ \frac{BQ}{BT}=\frac{[BAC]}{[TAC]}=\frac{BA.BC}{HA.HC}$](http://latex.artofproblemsolving.com/9/0/6/90681899f18414b7af02a06d801945331ce62d77.png)
which reflects a point in the bisector of 
and dilates with a factor of 
wrt to 
.
denote the 
.
since triangle 
.
goes to 
.
from where it follows that 
.
are collinear this finishes.