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.
+1 w
denote the set of positive integers
such that for all 
it holds that 
holds for all real numbers 
.
be positive real numbers such that 
Prove that:
and 
be two sequences defined by:![\[
a_{n+1} = \frac{1 + a_n + a_n b_n}{b_n} \quad \text{and} \quad b_{n+1} = \frac{1 + b_n + a_n b_n}{a_n}
\]](http://latex.artofproblemsolving.com/a/a/0/aa0a5c5acedc990eb345fb50a8395db05307a4e7.png)
for all 
, with initial values 
and 
.![\[
a_{2024} < 5.
\]](http://latex.artofproblemsolving.com/f/c/a/fca6e6630f33429b5a7ea867d639a2e29ff881f2.png)
and write the numbers 
down on a board. A move consists in erasing any two numbers 
and 
, then writing down the numbers 
and 
on the board, and then removing repetitions (e.g., if the board contained the numbers 
, then one could choose the numbers 
and 
, obtaining the board with numbers 
). For all integers , determine whether it is possible to be left with exactly two numbers on the board after a finite number of moves. and a polynomial 
with real coefficients of degree 
(
) such that has no real roots. Prove that the polynomial![\[
Q(x) \;=\; P(x) \;+\; a\,P'(x) \;+\; a^2\,P''(x) \;+\; \dots \;+\; a^n\,P^{(n)}(x)
\]](http://latex.artofproblemsolving.com/6/5/e/65e5e4eefcadd5d0019b4929b3472cdf203c13aa.png)
has no real roots.
and 
, with 
, a set 
the pair 
if for any elements 
we have 
. Let 
be the greatest possible number of elements of a set that escapes the pair . Find the maximum and minimum values of 
.
with incircle 
, let 
be the incenter and 
be the point where touches 
. Let 
be the point on 
with 
and 
be the orthocenter of 
, so that 
on 
also satisfies 
. Prove that 
, the reflection of over the midpoint of 
, lies on .
satisfying 
the following inequality holds:
of the triangle consider a mobile point 
. Let 
the orthogonal projection of 
on 
. If the mobile points 
and 
are such that 
is a paralellogram, find the locus of point 
when goes through 
.
such that the average of any nonempty subset is an integer.
, where 
.
. By the Chinese Remainder Theorem, since these moduli are pairwise coprime, there exists a unique solution modulo 
. Since there are 11 numbers in , the largest must be at least 
. However, all numbers in must be at most four digits long, leading to a contradiction! 
satisfying 
and 
be any two numbers in 
of 
that excludes both 
.
be the sum of the elements in 
and 
must be divisible by ![\[
X + a \equiv 0 \pmod{m} \quad \text{and} \quad X + b \equiv 0 \pmod{m} \implies a \equiv b \pmod{m}
\]](http://latex.artofproblemsolving.com/e/5/3/e534bb5b0a8f0d72b35819f1f64adf73fb614364.png)
Applying this argument to all pairs 
in