[/quote]
,
and, for each 
,
be the unique integer in the range 
for which 
is divisible by 
the obtained sequence is 
.
eventually becomes constant.
,
lie on sides 
,
.
meet at orthocenter 
.
lie on segment 
such that 
and 
.
and 
intersect at point 
.
.
.
is the midpoint of side 
And 
.
over 
intersects lines 
and 
and 
and 
.
lies on 
.
such that 
prove that the triangle must have a right angle.
such that for any real numbers 
and 
,![\[
f\left(x^2+2y f(x)\right) + (f(y))^2 \leq (f(x+y))^2
\]](http://latex.artofproblemsolving.com/8/4/3/8437449cd3f8f0da24d17a60b55dcb563e6b5655.png)
defined by 
and![\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/9/9/4/994aa754cc1288ce4f28a95a0276e64282fb5f66.png)
a) Prove that there exist infinitely many positive integers 
such that 
.![\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]](http://latex.artofproblemsolving.com/f/f/1/ff174e7431cbfbc17c650d109651241286756a1a.png)
is a cyclic quadrilateral. Let 
and 
.
and 
the orthocenters of triangles 
and 
,
,
be a triangle. A circle passing through 
and 
intersects the sides 
and 
again at 
and 
respectively. Prove that 
,
and 
are concurrent, where 
and 
are the orthocentres of triangles 
respectively.
.
and 
.
=
.
is the foot of the altitude from 
in triangle 
,
,
+ 
+ 
= 
,
,
+ 
+
= 0 holds
= 
= 
+ 
+ 
.
Y by
Given a polynomial 
and a positive integer 
, we denote the -fold composition of by 
. A polynomial with real coefficients is called perfect if for each integer 
there is a positive integer so that 
is an integer. Is it true that for each perfect polynomial , there exists a positive 
so that for each integer there is 
for which is an integer?
and a positive integer 
, we denote the -fold composition of by 
. A polynomial with real coefficients is called perfect if for each integer 
there is a positive integer so that 
is an integer. Is it true that for each perfect polynomial , there exists a positive 
so that for each integer there is 
for which is an integer?This post has been edited 1 time. Last edited by Phorphyrion, Apr 5, 2023, 1:48 PM
and 
is a fixed point.
,
such that the ![\[
P(x) = \frac{4x^3 + 6x^2 + 10x + 3}{2}
\]](http://latex.artofproblemsolving.com/e/9/5/e9576fb4152ac11bea39f043b1f050b0fd81c159.png)
evaluated at ![\[
P\left(\frac{n-1}{2}\right) = \frac{n(n^2 + 7)}{4} - \frac{1}{2}.
\]](http://latex.artofproblemsolving.com/6/f/a/6fa565afb529efed4a90a79c5ddc2977bce75e71.png)
.
must be a half-integer, i.e., 
for some odd integer 
.![\[
2P(n) + 1,\quad 2P(P(n)) + 1,\quad 2P(P(P(n))) + 1, \ldots
\]](http://latex.artofproblemsolving.com/9/b/3/9b3324db81c2cce4cf53ce05ea7fb76515fe42bb.png)
![\[
m \mapsto (m^3 + 7m)/2.
\]](http://latex.artofproblemsolving.com/d/e/e/dee7010b7d067c28abf03240e9b31d2825660684.png)
Hence, the sequence follows the recurrence:![\[
a_{i+1} = (a_i^3 + 7a_i)/2.
\]](http://latex.artofproblemsolving.com/3/f/7/3f76a3ede66c73f59e183a0e70aaa41cc5ba1d3f.png)
![\[
v_2((m^3 + 7m)/2) = v_2(m) - 1,
\]](http://latex.artofproblemsolving.com/3/3/6/3361ac1d7359aa70f7fd71fd34e5a4db106c543d.png)
where 
denotes the 2-adic valuation. That is, each iteration reduces the 2-adic valuation by 1. Therefore, after exactly 
steps, the resulting term in the sequence will be odd. Since the input to 
at that point is an integer, and 
for odd inputs, the composition 
.
.![\[
2P\left(\frac{n-1}{2}\right) + 1 = n(n^2 + 7),
\]](http://latex.artofproblemsolving.com/b/b/9/bb9167c42684c20e5c4941bddc513e1de22a69fc.png)
we note that 
for odd 
is arbitrarily large. It is known that 
is a quadratic residue modulo any power of 2, so there exist such 
