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.
be an acute triangle, 
its orthocenter and 
the center of its nine point circle. Let 
be a point on the parallel through 
to 
such that 
and and 
are on different sides of 
and 
a point on the parallel through 
to 
such that 
and and are on different sides of . If 
and 
are the reflections of over and respectively, 
and 
are the intersections of 
and 
respectively with the circumcircle of , prove that the intersection of lines 
and 
lies on 
.
undergoes the following procedure in order to obtain the number 
: to the first position to obtain the numb er 
; to obtain the number 
; to the end to obtain the number 
.
.) For example, for 
, we get 
, 
, and 
.) for which 
.
show that there exists a prime 
such that one can choose distinct integers 
such that p divides 
for all 
.
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.
of positive integers is called central if for every positive integer 
, the arithmetic mean of the first 
terms of the sequence is equal to .
of positive integers such that for every central sequence 
there are infinitely many positive integers with 
.
be real numbers such that 
. Find the value of 
be real numbers such that 
and 
. Find the value of 
be a positive integer larger than 
. Build an infinite set 
of subsets of 
having the following properties: distinct sets of have exactly one common element;
distinct sets of have void intersection.
where 
are nine distinct integers. Prove that there exists an integer 
such that for all integers 
the number 
is divisible by a prime number greater than 20.
, no two of which are 
. Prove that:
is a positive integer. Call all positive divisors of which are different from and beautiful divisors.We call a special number when it has at least 
beautiful divisors and difference of any beautiful divisors divides as well. Find all special numbers.
is a positive integer. Call all positive divisors of which are different from and beautiful divisors.We call a special number when it has at least beautiful divisors and difference of any beautiful divisors divides as well. Find all special numbers.
is composite with at least two prime divisors, we may write 
for coprime 
. Then 
divides 
but also their gcd is (a common divisor dividing and also divides ), so 
and 
. If 
, then 
which works. If 
, then (as is even as a product of two consecutive integers) 
must divide 
, so 
, corresponding to 
. However, 
does not work with and 
, 
does not work with and 
, 
does not work with and 
, while 
works since all diiferences are 
.
for some prime , where 
from the problem statement. Then 
must divide 
, so 
divides 
, which implies 
. Now 
works (as 
), while 
does not work with 
. are 
.
is composite with at least two prime divisors, we may write for coprime . Then divides but also their gcd is (a common divisor dividing and also divides ), so and . If , then which works. If , then (as is even as a product of two consecutive integers) must divide , so , corresponding to . However, does not work with and , does not work with and , does not work with and , while works since all diiferences are . for some prime , where from the problem statement. Then must divide , so divides , which implies . Now works (as ), while does not work with . are .
cannot be odd, since the difference of any two (proper) divisors is even and hence cannot divide . If 
for a positive integer 
(note that and 
do not have two proper divisors), then from the proper divisors and we get 
, so 
. Hence 
, corresponding to 
, which indeed work.
Prove that
![\[
\text{Find all functions } f : \mathbb{R} \to \mathbb{R} \text{ such that:} \\
f(a^4 + a^2b^2 + b^4) = f\left((a^2 - f(ab) + b^2)(a^2 + f(ab) + b^2)\right)
\]](http://latex.artofproblemsolving.com/2/a/7/2a73a5e58eab4a994fbd83160fb79a0b00152951.png)
