[/quote]
,
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 
.
such that if any 
elements are removed from the set 
,
,
be two sequences of positive real numbers. Suppose 
is a sequence of positive real numbers such that 
for all 
.
.![\[
\left( \frac{M+z_2+\dots+z_{2n}}{2n} \right)^2
\ge
\left( \frac{x_1+\dots+x_n}{n} \right)
\left( \frac{y_1+\dots+y_n}{n} \right). \]](http://latex.artofproblemsolving.com/8/c/4/8c4a72f57363eb1ef114825c8d36191bd5f42cb7.png)
be a triangle. On side 
take points 
and 
such that 
take points 
and 
and on side 
take points 
and 
such that 
Let 
and 
are concurrent.
that verify the following equation :
lie the following points: 
and 
on 
,
on 
,
on 
.![\[
P = AM\cap BN,\quad
R = KM\cap LN,\quad
S = KN\cap LM,
\]](http://latex.artofproblemsolving.com/7/e/a/7ea90a0f1261ed7445bbac58fb0cdd76810259f3.png)
and let the line 
meet 
.
,
are collinear.
be inscribed in the circle 
.
intersects 
and 
and 
,
of triangle 
lies on the circle 
and 
coincide.
be a triangle and 
and 
be two points on 
,
such that 
and let 
.
to the interior angle bisector of 
intersect 
in 
.
.
,
be its incenter. Incircle touches sides 
at 
,
to 
, respectively. Rays 
intersect 
at 
,
touches incircle at 
.
and circumcircle 
.
be the midpoints of sides 
,
,
and let 
be the intersections of line 
with line 
and line 
,
be the midpoint of arc 
of 
.
bisects 
.
.
and 
,
.
,
is composite, then all positive divisors of 
Y by
Let 
be a set of integers with the following properties:
contains all positive integers.
be a set of integers with the following properties:
.- If
and 
, then 
. - If for some
, 
is composite, then all positive divisors of are in .
contains all positive integers.This post has been edited 1 time. Last edited by pikapika007, Mar 21, 2025, 12:12 PM
Reason: wrong year
Reason: wrong year
,
composite spam rule #2![$\sqrt[3]{k} < p < \sqrt{k}$](http://latex.artofproblemsolving.com/c/e/6/ce6e509d7a7df8ccce799cf56b09518c4ae9939f.png)
such that 
be inverse of 
'
so we get 
also in 
works for fermat primes but i didnt realize that 
be the odd primes in 
.
is composite, so all of its prime divisors are in 
.
is composite and in 
.
be the smallest positive integer not in 
in 
for some residue 
.
is in 
gives that 
,
?
)
then 
,
is composite so assume that 
are all on 
and 
(because 
is smol :c), base cases are given and for the inductive step just note that 
and therefore 
and also then 
but also note that 
so it can't be prime therefore it is also in 
for all 
composite and that posses one odd factor but then considering the rest of divisors of this amount by making 
for all positive integers 
for all positive integers 
are on 
for 
and using euler theorem we get that all odd prime powers are on 
.
with the base cases being 
already given.
:
is composite we're already done from the third bullet.
is an (odd) prime number. We say a number is good if the prime powers in its prime factorization are all less than 
nor 
is a power of ![\[ s \coloneq p^2-1 = (p-1)(p+1) \]](http://latex.artofproblemsolving.com/6/b/7/6b76522a095ce72e0f2002ca8394a31d0bcab5b1.png)
is good. Indeed, one of the numbers has only a single factor of 
is certainly less than 
is good. As 
,
.
,
is odd. First we contend that the number ![\[ s' \coloneq 2^{q+1} - 1 = \left( 2^{(q+1)/2}-1 \right) \left( 2^{(q+1)/2}+1 \right) \]](http://latex.artofproblemsolving.com/1/6/a/16a703361bf5076dc25201a4ace6db41df5811b4.png)
is good. Indeed, this follows from the two factors being coprime and both less than 
is in ![\[ s \coloneq p^2-1 = (p-1)(p+1) \in S \]](http://latex.artofproblemsolving.com/8/c/d/8cdaf54033ce3efb72c46ec28d5c2af8a9f4f527.png)
as we did in the previous case, because the largest power of 
will be exactly 
which is known to be in 
,
is a Fermat prime. Then in particular, 
.![\[ s \coloneq 2p-1 \equiv 0 \pmod 3 \]](http://latex.artofproblemsolving.com/e/6/4/e64d203eb4f3d7fcc962507123ea7c8e1444eb49.png)
and moreover 
(it would imply 
,
by Zsigmondy/Mihailescu/etc.). So 
is good, and since 
,
is sufficient as a base case; however this requires a bit more work to check. Here is how:
,
.
we get 
.
we get 
.
we get 
.
,
are sufficient as base cases for the earlier cases to finish the rest of the problem. (Case 2 works once 
,
.
is not sufficient; for example 
satisfies all the problem conditions.
.
)
)
composite? WHAT! Help Me!![$[x] \cap S = [x]$](http://latex.artofproblemsolving.com/9/4/f/94f7de57826641ffcf953b59af1f41ca0bf0a26c.png)
.
!
is NOT PRIME NUGGET! By theorem: 
WHERE: 
,
.
IS PRIME NUGGET! Consider case:
!
.
and 
is 1! Notice: 
!
!
is NOT PRIME! Because divide by prime nugget 
.
!
.
!
.
!
!
.
NUGGETS DIVIDE 
!
!
so 
.
!
!
!![\[ 2^{\mathcal G + 1} \nu \mu \in S!!!!!!!!! \]](http://latex.artofproblemsolving.com/1/f/8/1f87397cdea89bde8d071fa1136ecf7931cd5ad8.png)
Gegagedi finish: Nugget point 3 give 
divisors in 
is in 
and had the main idea for the proof of that in my mersenne prime section (in 
)
How many pts?
.
and 
are in 
.
and 
is composite, all positive divisors of 
must be in 
in ascending order. Say the set of maximal primes powers dividing elements of 
.
,
lies in 
and all its divisors lie in 
.
for some 
,
dividing 
lies in 
,
.
clearly lies in 
.
lies in 
lies in 
