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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
A two-variable & non-homogenous inequality that seems hard to me
MyLifeMyChoice   3
N 2 minutes ago by Radin_
Source: Developing from a larger, three-variable one
For $a,b>0$, prove/disprove the following claim: :maybe:

$a^3b^3+\frac{1}{a^3}+\frac{1}{b^3}+3\stackrel{?}{\ge}a^2b+b^2a+\frac{1}{a^2b}+\frac{1}{b^2a}+\frac{a}{b}+\frac{b}{a}$
3 replies
1 viewing
MyLifeMyChoice
Mar 13, 2025
Radin_
2 minutes ago
exponential diophantine with factorials
skellyrah   4
N 13 minutes ago by InftyByond
find all non negative integers (x,y) such that $$ x! + y! = 2025^x + xy$$
4 replies
skellyrah
Feb 24, 2025
InftyByond
13 minutes ago
Point satisfies triple property
62861   35
N 27 minutes ago by Sanjana42
Source: USA Winter Team Selection Test #2 for IMO 2018, Problem 2
Let $ABCD$ be a convex cyclic quadrilateral which is not a kite, but whose diagonals are perpendicular and meet at $H$. Denote by $M$ and $N$ the midpoints of $\overline{BC}$ and $\overline{CD}$. Rays $MH$ and $NH$ meet $\overline{AD}$ and $\overline{AB}$ at $S$ and $T$, respectively. Prove that there exists a point $E$, lying outside quadrilateral $ABCD$, such that
[list]
[*] ray $EH$ bisects both angles $\angle BES$, $\angle TED$, and
[*] $\angle BEN = \angle MED$.
[/list]

Proposed by Evan Chen
35 replies
1 viewing
62861
Jan 22, 2018
Sanjana42
27 minutes ago
Prove concyclic and tangency
syk0526   40
N an hour ago by Ilikeminecraft
Source: Japan Olympiad Finals 2014, #4
Let $ \Gamma $ be the circumcircle of triangle $ABC$, and let $l$ be the tangent line of $\Gamma $ passing $A$. Let $ D, E $ be the points each on side $AB, AC$ such that $ BD : DA= AE : EC $. Line $ DE $ meets $\Gamma $ at points $ F, G $. The line parallel to $AC$ passing $ D $ meets $l$ at $H$, the line parallel to $AB$ passing $E$ meets $l$ at $I$. Prove that there exists a circle passing four points $ F, G, H, I $ and tangent to line $ BC$.
40 replies
syk0526
May 17, 2014
Ilikeminecraft
an hour ago
No more topics!
Bulgarian National Olympiad 2018 Problem 5
Lamp909   6
N Mar 9, 2025 by Assassino9931
Given a polynomial $P(x)=a_{d}x^{d}+ \ldots +a_{2}x^{2}+a_{0}$ with positive integers for coefficients and degree $d\geq 2$. Consider the sequence defined by $$b_{1}=a_{0} ,b_{n+1}=P(b_{n}) $$for $n \geq 1$ . Prove that for all $n \geq 2$ there exists a prime $p$ such that $p$ divides $b_{n}$ but does not divide $b_{1}b_{2} \ldots b_{n-1}$.
6 replies
Lamp909
Apr 15, 2018
Assassino9931
Mar 9, 2025
Bulgarian National Olympiad 2018 Problem 5
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Lamp909
98 posts
#1 • 2 Y
Y by Adventure10, Mango247
Given a polynomial $P(x)=a_{d}x^{d}+ \ldots +a_{2}x^{2}+a_{0}$ with positive integers for coefficients and degree $d\geq 2$. Consider the sequence defined by $$b_{1}=a_{0} ,b_{n+1}=P(b_{n}) $$for $n \geq 1$ . Prove that for all $n \geq 2$ there exists a prime $p$ such that $p$ divides $b_{n}$ but does not divide $b_{1}b_{2} \ldots b_{n-1}$.
This post has been edited 2 times. Last edited by Lamp909, Apr 15, 2018, 3:32 PM
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ThE-dArK-lOrD
4071 posts
#3 • 2 Y
Y by Adventure10, Mango247
It's not hard to prove that: for every prime number $p$ that $p\mid b_n$ for some $n$, there exists index $i_0$ that
  • $p\mid b_{i_0}$ and $p\nmid b_j$ for all $1\leq j<i_0$, and
  • For all $n$ that $p\mid b_n$, we have $i_0\mid n$ and $\nu_p (b_n)=\nu_p (b_{i_0}).$
Note that this also implies $b_m\mid b_n$ for all $m\mid n$.
Now, back to the problem, suppose for every prime $p\mid b_n$, there exists $1\leq i<n$ that $p\mid b_i$.

If there exists prime number $a$ that $\nu_a (n)\geq 2$, we get that $d\mid \frac{n}{a}$ for all proper divisor $d$ of $n$.
This gives $\nu_p (b_{\frac{n}{a}}) =\nu_p (b_n)$ for all $p\mid b_n$.
So, $b_{\frac{n}{a}} \geq b_n$. But it's clear that the sequence is strictly increasing, hence contradiction, done.

If there's no such prime divisor $a$, i.e. $n$ can be written as product of distinct prime numbers.
If $n$ is prime, we get that $\nu_p (b_1)=\nu_p (b_n)$ for all $p\mid b_n$. So, $b_1\geq b_n$, contradiction.
Now, we're left with the case that $n$ has two distinct prime factors, say they're $u$ and $v$.
We get that $d\mid \frac{n}{u}$ or $d\mid \frac{n}{v}$ for all proper divisor $d$ of $n$.
Hence, for all $p\mid b_n$, either $\nu_p (b_{\frac{n}{u}}) =\nu_p (b_n)$ or $\nu_p (b_{\frac{n}{v}}) =\nu_p (b_n)$.
This gives
$$b_n\leq \mathrm{lcm} (b_{\frac{n}{u}} ,b_{\frac{n}{v}} )=\frac{ b_{\frac{n}{u}}b_{\frac{n}{v}}}{\gcd (b_{\frac{n}{u}},b_{\frac{n}{v}})}\leq b_{\frac{n}{u}}b_{\frac{n}{v}}.$$But it's clear that $b_{i+1}> b_i^2$ for all $i$, so $b_n>b_{\frac{n}{u}}b_{\frac{n}{v}}$, this contradict the above inequality, done.
This post has been edited 2 times. Last edited by ThE-dArK-lOrD, Apr 15, 2018, 7:54 PM
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MarkBcc168
1593 posts
#4 • 5 Y
Y by ThE-dArK-lOrD, Aryan-23, sabkx, Adventure10, Mango247
Just copy solution of ISL 2014 N7.
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TacH
64 posts
#5 • 2 Y
Y by Adventure10, Mango247
btw, just to give an alternate way to ThE-dArK-lOrD's sol
from
1) $p\mid b_{i_0}$ and $p\nmid b_j$ for all $1\leq j<i_0$, and
2) For all $n$ that $p\mid b_n$, we have $i_0\mid n$ and $\nu_p (b_n)=\nu_p (b_{i_0}).$

supposes we look at $b_n$. for every prime $p\mid b_n$, there exists $1\leq i<n$ that $p\mid b_i$ ,indeed $i\mid n$
we pair up each prime that divides $b_n$ with the index given by 1) condition ; note that the index divides $n$ hence, it is at most $\frac{n}{2}$
since $\nu_p (b_n)=\nu_p (b_{i_0}).$ the product of $b_i$ that include in the pairing must be not less than $b_n$
Hence, we must have $b_1b_2b_3...b_{\frac{n}{2}} \geq b_n$
but $b_{i} < b_{\frac{n}{2}}$ for $i < \frac{n}{2}$ . Thus $b_1b_2b_3...b_{\frac{n}{2}}  <  (b_{\frac{n}{2}})^{\frac{n}{2}}$
but $b_{i+1} \geq b_{i}^2$ . Hence $b_n > (b_{\frac{n}{2}})^{2^{\frac{n}{2}}}$
Hence $\frac{n}{2} > 2^{\frac{n}{2}}$ Which leads to a contradiction.
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liekkas
370 posts
#6 • 2 Y
Y by Adventure10, Mango247
Very similar to ISL 2014, N7.
It is also a generalization of 2016 China TST https://artofproblemsolving.com/community/c6h1212540p6016827
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mathaddiction
308 posts
#7 • 1 Y
Y by leozitz
Notice that $b_n=P^n(0)$, for convenience assume $b_0=0$.
Suppose on the contrary that there exists $n$ such that for every prime divisor of $b_n$ is also a prime divisor of $b_{n-1}b_{n-2}...b_1$.
CLAIM 1. For each integer $n$ we have
$$b_n>b_{n-1}b_{n-2}...b_1$$Proof.
We proceed by induction, the base case is obvious, now we have
$$b_n=P(b_{n-1})>b_{n-1}^2>b_{n-1}b_{n-2}...b_1$$by inductive hypothesis. $\blacksquare$
Therefore, there exists some prime $p$ such that $$v_p(b_n)>v_p(b_{n-1}b_{n-2}...b_1)\qquad(1)$$Let $k$ be the smallest positive integer such that $p|b_k$. Let $Q(x)=P^k(x)$. From an order argument we have $k|n$. Now notice that the coefficient of $x$ in $Q(x)$ is $0$, so $v_p(b_{2k})=Q(b_k)=v_p(b_k)$, and inductively we have $v_p(b_n)=v_p(b_k)$, whcih contradicts $(1)$, so we are done.
This post has been edited 1 time. Last edited by mathaddiction, Oct 23, 2020, 9:50 AM
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Assassino9931
1192 posts
#8
Y by
Very confusing formulation for me, $b_1 = a_0$ looks weird and it is hard to accept that there is no $x$ term. Anyway, a really great problem by Navid Safaei! Here is a better formulation in my opinion, and a solution:
Alternative formulation wrote:
Let $P(x)$ be a polynomial of degree at least $2$, where the coefficient at $x$ is zero and all other coefficients are positive integers. Consider the sequence $a_1 = P(0)$, $a_{n+1} = P(a_n)$. Prove that for any $n\geq 2$ there exists a prime $p$ which divides $a_n$, but does not divide any of $a_1,a_2,\ldots,a_{n-1}$.

Let $n$ be an arbitrary index, $p$ be a prime divisor of $a_n$ and let $\nu_p(a_n) = r \geq 1$. Then $a_{n+1} = Ka_n^2 + P(0) = Ka_n^2 + a_1$ for some integer $K$ and in particular $a_{n+1} \equiv a_1 \pmod {p^{r+1}}$. Since $P$ is a polynomial, it follows that $P(a_{n+1}) \equiv P(a_1) \pmod {p^{r+1}}$, i.e. $a_{n+2} \equiv a_2 \pmod {p^{r+1}}$. Continuing inductively, we obtain $a_{n+k} \equiv a_k \pmod {p^{r+1}}$ for all $k$. In particular, it follows that $a_n \equiv a_{sn} \pmod {p^{r+1}}$ for any positive integer $s$. Now since $\nu_p(a_n) = r$, it necessarily follows that $\nu_p(a_{sn}) = r$ for any $s$.

Now suppose $n$ does not satisfy the problem condition, i.e. for any prime divisor $p$ of $a_n$ there exists an index $i \leq n-1$ such that $p$ divides $a_i$. By the above paragraph it follows that $\nu_p(a_n) = \nu_p(a_{ni}) = \nu_p(a_i)$, hence $a_n$ divides $a_1a_2\cdots a_{n-1}$.

To reach a contradiction, we prove by induction that $a_n > a_1a_2\cdots a_{n-1}$ for any $n\geq 2$. For the base case, note that $a_2 = P(a_1) \geq a_1^2 + P(0) > a_1$. For the induction step, it suffices to prove $a_{n+1} > a_n^2$ to obtain the main inequality for $n+1$ from the main inequality for $n$. For the latter, note similarly that $a_{n+1} = P(a_n) \geq a_n^2 + P(0) > a_n^2$.
This post has been edited 1 time. Last edited by Assassino9931, Mar 10, 2025, 1:36 PM
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