[/quote]
,
be the lengths of the sides of a triangle. Prove that![\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]](http://latex.artofproblemsolving.com/7/4/b/74b7aa5621bc52977d9acd448d74293dc27a8e00.png)
and 
be arbitrary points on the sides 
and 
of triangle 
,
meets for the second time the circumcircle of 
at point 
.
meets line 
at point 
,
meets line 
at point 
.
and 
pass through the same point.
.
and 
intersect the altitude from 
to 
respectively. 
is the circumcenter of triangle 
and 
is the reflection of 
is the second intersection of circumcircles of triangles 
and 
.
are collinear.
and 
Find the value of 
and 
such that 
and 
is a prime number
such that 
for all 
be an even integer and let 
be real numbers satisfying 
.
and a prime number 
.
is nonconstant, 
,
for all integers 
.
.
in order. Ana wins if she can determine the value of 
Let 
.
of positive integers is called central if for every positive integer 
terms of the sequence is equal to 
of positive integers such that for every central sequence 
there are infinitely many positive integers 
.
Prove that
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 
