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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Reflections of lines through reflections of excenters
cjquines0   39
N 10 minutes ago by awesomeming327.
Source: 2016 IMO Shortlist G7
Let $I$ be the incentre of a non-equilateral triangle $ABC$, $I_A$ be the $A$-excentre, $I'_A$ be the reflection of $I_A$ in $BC$, and $l_A$ be the reflection of line $AI'_A$ in $AI$. Define points $I_B$, $I'_B$ and line $l_B$ analogously. Let $P$ be the intersection point of $l_A$ and $l_B$.
[list=a]
[*] Prove that $P$ lies on line $OI$ where $O$ is the circumcentre of triangle $ABC$.
[*] Let one of the tangents from $P$ to the incircle of triangle $ABC$ meet the circumcircle at points $X$ and $Y$. Show that $\angle XIY = 120^{\circ}$.
[/list]
39 replies
+1 w
cjquines0
Jul 19, 2017
awesomeming327.
10 minutes ago
2xy is perfect square and x^2 + y^2 is prime
parmenides51   4
N 11 minutes ago by LeYohan
Source: Dutch NMO 2020 p4
Determine all pairs of integers $(x, y)$ such that $2xy$ is a perfect square and $x^2 + y^2$ is a prime number.
4 replies
parmenides51
Nov 23, 2020
LeYohan
11 minutes ago
Really classical inequatily from canada
shobber   79
N 14 minutes ago by sharknavy75
Source: Canada 2002
Prove that for all positive real numbers $a$, $b$, and $c$,
\[ \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c \]
and determine when equality occurs.
79 replies
shobber
Mar 5, 2006
sharknavy75
14 minutes ago
Functional equation
Pmshw   18
N 15 minutes ago by jasperE3
Source: Iran 2nd round 2022 P2
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for any real value of $x,y$ we have:
$$f(xf(y)+f(x)+y)=xy+f(x)+f(y)$$
18 replies
Pmshw
May 8, 2022
jasperE3
15 minutes ago
f(x)f(yf(x)) = f(x+y)
ISHO95   5
N 35 minutes ago by jasperE3
Find all functions $f:\mathbb R^+ \to \mathbb R^+$, for all $x,y \in \mathbb R^+$, \[ f(x)f(yf(x))=f(x+y). \]
5 replies
ISHO95
Jan 14, 2013
jasperE3
35 minutes ago
Two players want to obtain a number divisible by 2023
a_507_bc   3
N 39 minutes ago by fathalishah
Source: All-Russian MO 2023 Final stage 11.5
Initially, $10$ ones are written on a blackboard. Grisha and Gleb are playing game, by taking turns; Grisha goes first. On one move Grisha squares some $5$ numbers on the board. On his move, Gleb picks a few (perhaps none) numbers on the board and increases each of them by $1$. If in $10,000$ moves on the board a number divisible by $2023$ appears, Gleb wins, otherwise Grisha wins. Which of the players has a winning strategy?
3 replies
a_507_bc
Apr 23, 2023
fathalishah
39 minutes ago
Points on a lattice path lies on a line
navi_09220114   1
N 39 minutes ago by pbornsztein
Source: TASIMO 2025 Day 1 Problem 3
Let $S$ be a nonempty subset of the points in the Cartesian plane such that for each $x\in S$ exactly one of $x+(0,1)$ or $x+(1,0)$ also belongs to $S$. Prove that for each positive integer $k$ there is a line in the plane (possibly different lines for different $k$) which contains at least $k$ points of $S$.
1 reply
navi_09220114
Today at 11:43 AM
pbornsztein
39 minutes ago
Functional inequality
Jackson0423   2
N an hour ago by nitride
Show that there does not exist a function \( f : \mathbb{R}^+ \to \mathbb{R}^+ \) such that for all positive real numbers \( x, y \),
\[
f^2(x) \geq f(x+y)\left(f(x) + y\right).
\]
2 replies
Jackson0423
5 hours ago
nitride
an hour ago
Find all integers
velmurugan   3
N 2 hours ago by grupyorum
Source: Titu and Dorin Book Problem
Find all positive integers $(x,n)$ such that $x^n + 2^n + 1$ is a divisor of $x^{n+1} + 2^{n+1} +1 $ .
3 replies
velmurugan
Jul 30, 2015
grupyorum
2 hours ago
Graph Process Problem
Maximilian113   10
N 2 hours ago by Ru83n05
Source: CMO 2025 P1
The $n$ players of a hockey team gather to select their team captain. Initially, they stand in a circle, and each person votes for the person on their left.

The players will update their votes via a series of rounds. In one round, each player $a$ updates their vote, one at a time, according to the following procedure: At the time of the update, if $a$ is voting for $b,$ and $b$ is voting for $c,$ then $a$ updates their vote to $c.$ (Note that $a, b,$ and $c$ need not be distinct; if $b=c$ then $a$'s vote does not change for this update.) Every player updates their vote exactly once in each round, in an order determined by the players (possibly different across different rounds).

They repeat this updating procedure for $n$ rounds. Prove that at this time, all $n$ players will unanimously vote for the same person.
10 replies
Maximilian113
Mar 7, 2025
Ru83n05
2 hours ago
Congrats to former two perfect scorer in IMO
mszew   0
2 hours ago
Source: Where should it be posted?
Congrats to the new president of Romania...Mr. Nicuşor Dan

https://en.wikipedia.org/wiki/Nicu%C8%99or_Dan

https://www.imo-official.org/participant_r.aspx?id=1571
0 replies
mszew
2 hours ago
0 replies
Austrian Regional MO 2025 P4
BR1F1SZ   3
N 2 hours ago by LeYohan
Source: Austrian Regional MO
Let $z$ be a positive integer that is not divisible by $8$. Furthermore, let $n \geqslant 2$ be a positive integer. Prove that none of the numbers of the form $z^n + z + 1$ is a square number.

(Walther Janous)
3 replies
BR1F1SZ
Apr 18, 2025
LeYohan
2 hours ago
Nice concurrency
navi_09220114   3
N 2 hours ago by sami1618
Source: TASIMO 2025 Day 1 Problem 2
Four points $A$, $B$, $C$, $D$ lie on a semicircle $\omega$ in this order with diameter $AD$, and $AD$ is not parallel to $BC$. Points $X$ and $Y$ lie on segments $AC$ and $BD$ respectively such that $BX\parallel AD$ and $CY\perp AD$. A circle $\Gamma$ passes through $D$ and $Y$ is tangent to $AD$, and intersects $\omega$ again at $Z\neq D$. Prove that the lines $AZ$, $BC$ and $XY$ are concurrent.
3 replies
navi_09220114
Today at 11:42 AM
sami1618
2 hours ago
system in R+, four equations/variables
jasperE3   2
N 2 hours ago by Yiyj
Source: Bulgaria 1972 P2
Solve the system of equations:
$$\begin{cases}\sqrt{\frac{y(t-y)}{t-x}-\frac4x}+\sqrt{\frac{z(t-z)}{t-x}-\frac4x}=\sqrt x\\\sqrt{\frac{z(t-z)}{t-y}-\frac4y}+\sqrt{\frac{x(t-x)}{t-y}-\frac4y}=\sqrt y\\\sqrt{\frac{x(t-x)}{t-z}-\frac4z}+\sqrt{\frac{y(t-y)}{t-z}-\frac4z}=\sqrt z\\x+y+z=2t\end{cases}$$if the following conditions are satisfied: $0<x<t$, $0<y<t$, $0<z<t$.

H. Lesov
2 replies
jasperE3
Jun 21, 2021
Yiyj
2 hours ago
Fun.... polynomial divisors
DottedCaculator   12
N Apr 29, 2025 by bin_sherlo
Source: USEMO 2022/5
Let $\tau(n)$ denote the number of positive integer divisors of a positive integer $n$ (for example, $\tau(2022) = 8$). Given a polynomial $P(X)$ with integer coefficients, we define a sequence $a_1, a_2,\ldots$ of nonnegative integers by setting
\[a_n =\begin{cases}\gcd(P(n), \tau (P(n)))&\text{if }P(n) > 0\\0 &\text{if }P(n) \leq0\end{cases}\]for each positive integer $n$. We then say the sequence has limit infinity if every integer occurs in this sequence only finitely many times (possibly not at all).

Does there exist a choice of $P(X)$ for which the sequence $a_1$, $a_2$, . . . has limit infinity?

Jovan Vuković
12 replies
DottedCaculator
Oct 23, 2022
bin_sherlo
Apr 29, 2025
Fun.... polynomial divisors
G H J
Source: USEMO 2022/5
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DottedCaculator
7356 posts
#1 • 6 Y
Y by LLL2019, rama1728, Kingsbane2139, centslordm, Mango247, Mango247
Let $\tau(n)$ denote the number of positive integer divisors of a positive integer $n$ (for example, $\tau(2022) = 8$). Given a polynomial $P(X)$ with integer coefficients, we define a sequence $a_1, a_2,\ldots$ of nonnegative integers by setting
\[a_n =\begin{cases}\gcd(P(n), \tau (P(n)))&\text{if }P(n) > 0\\0 &\text{if }P(n) \leq0\end{cases}\]for each positive integer $n$. We then say the sequence has limit infinity if every integer occurs in this sequence only finitely many times (possibly not at all).

Does there exist a choice of $P(X)$ for which the sequence $a_1$, $a_2$, . . . has limit infinity?

Jovan Vuković
This post has been edited 8 times. Last edited by DottedCaculator, Oct 24, 2022, 12:29 AM
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CANBANKAN
1301 posts
#2 • 2 Y
Y by Sross314, LLL2019
Solution
This post has been edited 2 times. Last edited by CANBANKAN, Oct 24, 2022, 12:55 AM
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v_Enhance
6877 posts
#4 • 6 Y
Y by ApraTrip, bjump, HamstPan38825, PROA200, levimpcbranco, Kingsbane2139
We claim the answer is no, such $P$ does not exist.

Clearly we may assume $P$ is nonconstant with positive leading coefficient. Fix $P$ and fix constants $n_0, c > 0$ such that $c = P(n_0) > 0$. We are going to prove that infinitely many terms of the sequence are at most $c$.

We start with the following lemma.
Claim: For each integer $n \ge 2$, there exists an integer $r = r(n)$ such that
  • For any prime $p$ which is at most $n$, we have $\nu_p(P(r)) = \nu_p(c)$.
  • We have \[ c \cdot \prod_{\text{prime }p \le n} 			\le r \le 2c \cdot \prod_{\text{prime }p \le n} p. \]
Proof. This follows by the Chinese remainder theorem: for each $p \le n$ we require $r \equiv n_0 \pmod{p^{\nu_p(c)+1}}$, which guarantees $\nu_p(P(r)) = \nu_p(P(n_0)) = \nu_p(c)$. Then there exists such an $r$ modulo $\prod_{p \le n} p^{\nu_p(c)+1}$ as needed. $\blacksquare$
Assume for contradiction that all $a_i$ are eventually larger than $c$. Take $n$ large enough that $n > c$ and $r = r(n)$ has $a_r > c$. Then consider the term $a_r$:
  • Using the conditions in the lemma it follows there exists a prime $p_n > n$ which divides $a_r = \gcd(P(r), \tau(P(r)))$ (otherwise $a_r$, which divides $P(r)$, is at most $c$).
  • As $p_n$ divides $\tau(P(r))$, this forces $P(r)$ to be divisible by (at least) $q_n^{p_n-1}$ for some prime $q_n$.
  • For the small primes $p$ at most $n$, we have $\nu_p(P(r)) = \nu_p(c) < c < n \le p_n - 1$. It follows that $q_n > n$.
  • Ergo, \[ P(r) \ge q_r^{p_r-1} > n^n. \]
In other words, for large enough $n$, we have the asymptotic estimate \begin{align*} 	n^n &< P(r) = O(1) \cdot r^{\deg P} \\ 	&= O(1) \cdot c^{\deg P} \cdot \prod_{\text{prime } p \le n} p^{\deg p} \\ 	&< O(1) \cdot n^{\deg P \cdot \pi(n)} \end{align*}where $\pi(n)$ denotes the number of primes less than $n$. For large enough $n$ this is impossible since the primes have zero density: \[ \lim_{n \to \infty} \frac{\pi(n)}{n} = 0. \]
Remark: For completeness, we outline a short elementary proof that $\lim_{n \to \infty} \frac{\pi(n)}{n} = 0$. For integers $M > 0$ define \[ \delta(M) \coloneqq \prod_{p \le M} \left( 1 - \frac 1p  \right). \]Then $\pi(n) < \delta(M)n + O(M)$, so it suffices to check that $\lim_{M \to \infty} \delta(M) = 0$. But \[ 		\frac{1}{\delta(M)} 		= \prod_{p \le M} \left( 1 - \frac 1p  \right)^{-1} 		= \prod_{p \le M} \left( 1 + \frac1p + \frac{1}{p^2} + \dots \right) 		\ge 1 + \frac 12 + \dots + \frac 1M 	\]which diverges for large $M$.
This post has been edited 1 time. Last edited by v_Enhance, Oct 24, 2022, 12:31 AM
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Remon
20 posts
#5 • 1 Y
Y by Mango247
https://artofproblemsolving.com/community/c6h539544p3104308 6floor ,this method can help us find n such that almost all the prime factors of P(n) are large.
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SnowPanda
186 posts
#6
Y by
Solution
This post has been edited 1 time. Last edited by SnowPanda, Oct 24, 2022, 12:02 PM
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primesarespecial
364 posts
#8 • 1 Y
Y by PRMOisTheHardestExam
Solution attached below..(The second pic is useless basically, I don't know why I even wrote bezout,the existence for (M+1) is obvious anyways and for primes greater than p_n,the existence of non roots mod p,is obvious by lagrange.
Attachments:
This post has been edited 3 times. Last edited by primesarespecial, Nov 7, 2022, 12:23 PM
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CahitArf
80 posts
#9
Y by
Please someone can check my solution?
Attachments:
USEMO 2022 P5.pdf (87kb)
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blackbluecar
303 posts
#10 • 1 Y
Y by IAmTheHazard
We will prove the general result for any $P \colon \mathbb{Z} \to \mathbb{Z}$ but not necessarily $P \in \mathbb{Z}[X]$

Suppose $d$ divides $P(n)$ for all $n \in \mathbb{Z}$. Thus, we let $P(x) = d \cdot Q(x)$ where $Q \colon \mathbb{Z} \to \mathbb{Z}$. By CRT on the prime divisors of $d$ there exists a positive integer $a$ where $\gcd(d, Q(dn+a))=1$ for every $n \in \mathbb{Z}$. So, we will let $f(x)=Q(dx+a)$.

So, there is no number $>1$ dividing every $f(n)$. Let $p_1<p_2< \cdots$ be all the primes dividing at least one $f(n)$. If we let $N$ be a sufficiently large positive integer, then by CRT there is an $M<p_1p_2 \cdots p_N$ where $p_i$ does not divide $f(M)$ for any $i \leq N$. Thus, let $f(M)=q_1^{\alpha_1}q_2^{\alpha_2} \cdots q_k^{\alpha_k}$. We should see that \[ \gcd(q_1^{\alpha_1}q_2^{\alpha_2} \cdots q_k^{\alpha_k},(\alpha_1+1)(\alpha_2+1) \cdots (\alpha_k+1)) \]becomes unbounded. So, there must be some $\alpha_i$ and $q_j$ where $\alpha_1 > q_j$. This clearly implies that \[ f(M) > q_1^{q_1} > p_N^{p_N}\]but this gets too big by prime number theorem. We look back at $d \cdot f$ by realizing $\tau(d \cdot f(n)) = \tau(d) \cdot \tau(f(n))$ so \[ \gcd(n,\tau(d) \cdot f(n))<\tau(d) \cdot \gcd(n,f(n)) \]Which we proved cannot grow at a sufficient speed. Done.
This post has been edited 1 time. Last edited by blackbluecar, Apr 4, 2023, 11:46 PM
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IAmTheHazard
5003 posts
#11 • 1 Y
Y by centslordm
Here is a... suboptimal solution in terms of elegance


The answer is no. Suppose otherwise; clearly $P$ is nonconstant, so let $\deg P=d>0$.

Write
$$P=f_1^{e_1}\ldots f_k^{e_k}$$where for all $1 \leq i \leq k$, $f_i$ is an irreducible integer polynomial, $e_i\geq 1$, and all the $f_i$ are pairwise distinct (hence coprime). We will construct (by CRT) a positive integer $m$ satisfying a number of desirable properties. Consider massive positive integers $A<B<C$ with $B$ and $C$ to be selected later, such that
  • For all $n\geq A$, $P(A)>0$,
  • For all (finitely many) primes $p$ dividing every value of $P(n)$, $p<A$,
  • By Bezout's, for all $n \geq A$ and all $i$, we have $\gcd(f_i(n),f_i'(n))<A$, since $f_i$ and $f_i'$ are coprime,
  • By Bezout's, for all $n \geq A$ and all $i \neq j$, we have $\gcd(f_i(n),f_j(n))<A$,
  • By Dirichlet, there exists some prime $Q \equiv 1 \pmod{e_1\ldots e_k}$ such that $Q<A$ and $Q$ does not divide every value of $P(n)$.
Let the primes (less than $A$) be $r_1,\ldots,r_a$ (this includes $Q$) and let $g_1,\ldots,g_a$ be nonnegative integers such that for all $i$, $r_i^{g_i}$ divides $P(n)$ for all $n$, but $r_i^{g_i+1}$ doesn't always divide $P(n)$. Therefore, for each $i$, we can impose a modulo-$r_i^{g_i+1}$ condition on $m$ to ensure that $\nu_{r_i}(P(m))=g_i$.

Now let $q_1,\ldots,q_b$ be the primes in $[A,B)$. Since none of these divide every value of $P(n)$, for each $i$ we can impose a modulo-$q_i$ condition on $m$ to ensure that $q_i \nmid P(m)$.

Finally let $p_1,\ldots,p_c$ be the primes in $[B,C)$ which divide some value of $P(n)$. For any $i,j$, we have $p_i \mid f_j(n) \implies p_i \nmid f_j'(n)$, so for Hensel lifting-related reasons, for any $k$ we can impose a modulo-$p_i^{k+1}$ condition on $m$ so that $\nu_{p_i}(f_i(m))=k$. Therefore, for each $i$, pick some $j$ such that $p_i$ divides some value of $f_j(n)$, and impose a modulo-$p_i^{\frac{Q-1}{e_j}+1}$ condition on $m$ such that $\nu_{p_i}(f_j(m))=\tfrac{Q-1}{e_j}$. Since $p_i$ won't divide $f_h(n)$ for any $h \neq j$ since $p_i>A$, it follows that $\nu_{p_i}(P(m))=Q-1$.

Observe that we can evidently find some $m$ satisfying these conditions which is $O((q_1\ldots q_b)(p_1\ldots p_c)^Q)$ but also arbitrarily large (by taking $B$ large and selecting the second-smallest possibility for $m$), since the product of $r_i^{g_i+1}$ is a constant. Then $P(m)$ is $O((q_1\ldots q_b)^d(p_1\ldots p_c)^{Qd})$. Separate $P(m)$ into the product of two "parts": one consisting of the product of primes less than $C$, which equals
$$\left(\prod_{i=1}^a r_i^{g_i}\right)\left(\prod_{i=1}^c p_i^{Q-1}\right),$$and the other, which is the product of primes at least $C$. Hence the "other part" of $P(m)$ is $O((q_1\ldots q_b)^d(p_1\ldots p_c)^{Q(d-1)+1})$. Now consider the contribution of each part to $\gcd(P(m),\tau(P(m)))$: specifically, for the first part, the contribution from $\tau(r_i^{g_i})=g_i+1$ is bounded for each $i$, so we can ignore it, and the contribution from $\tau(p_i^{Q-1})=Q$ is zero, since $Q \nmid P(m)$. Therefore, if $\gcd(P(m),\tau(P(m)))$ is unbounded as $m$ grows large (by making $B$ large), we will need some prime $p \geq C$ such that $\tau(p^{\nu_p(P(m))})$ contributes arbitrarily large quantities to $\gcd(P(m),\tau(P(m)))$. This means that we must be able to find some $p$ with $p_i \mid \nu_p(P(m))$ for some $i$, since if $\nu_p(P(m))$ has prime factors in $[A,B)$ they don't matter, and the contribution of any primes smaller than $A$ is bounded by the product of $r_i^{g_i}$ (i.e. the "$<A$ part" of $P(m)$), which is finite.

On the other hand, if we take $C \approx B\sqrt{\log B}$, then by PNT the the "other part" of $P(m)$ is $O\left(B^{\frac{B}{\log B}d}(B\sqrt{\log B})^{\frac{B}{\sqrt{\log B}}(Q(d-1)+1)}\right)$, since $\frac{B\sqrt{\log B}}{\log(B\sqrt{\log B})} \leq \frac{B\sqrt{\log B}}{\log B}$, but if $p_i \mid \nu_p(P(m))$ for some $i$, then we need $P(m) \geq p^{p_i}$ which is $\Omega((B\sqrt{\log B})^B)$, which dominates as $B \to \infty$: contradiction. $\blacksquare$


Edit: Wait shoot $C$ (and everything related to it) isn't necessary lmao. If we drop everything related to $C$ then $P(m)$ will be $O(B^{\frac{B}{\log B}d})$, but we will need some prime $p \geq B$ such that $\nu_p(P(m))$ is divisible by a prime at least $B$, hence we need $P(m) \geq B^B$, which is a contradiction. This is the same as the other solutions
This post has been edited 3 times. Last edited by IAmTheHazard, Aug 19, 2023, 12:17 AM
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Krish230905
109 posts
#12
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I don't know whether this is correct or not. Can someone please check.

We claim that the answer is no.
Let $\deg P(n)=t$ Then, $$P(n)=\sum_{i=1}^t a_i x^i$$Evaluate $P(\alpha)$ for $n=p_1p_2\dots p_k$ where $p_1,p_2, \dots p_k$ are the smallest $n$ primes. For large enough $k$, we have that $P(\alpha) < (a_t+1)\alpha^t$ . Also, note that $P(n)$ doesn't have any prime divisors $< p_k$ except finitely many. Hence, $p^{kn} > P(\alpha)$ for any $p \vert P(n)$. From this, we can conclude that $\nu_p(P(n)) < kn$ for any $p$. We denote $\nu_p(P(n))$ by $e$

$\tau (P(n))$ is given by $\prod_{i=1}^n (e_i+1)$ where $e_i$ are the coefficients of the primes dividing $P(n)$. So, $\tau (P(n))$ is not divisible by any $p$ if $p>kn$ due to the fact proved earlier. So, $P(n)$ has prime factors $p>p_k$ (except finitely many) and $\tau (P(n))$ has prime factors $p<kn$. This forces that $\gcd(P(n), \tau (P(n)))$ is small for infinitely many $n$.
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YaoAOPS
1541 posts
#13
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Fix a polynomial $P = c_0x^d + c_1x^{d-1} + \dots + c_{b}x^b$ with positive leading coefficient, as the result is immediately no for negative coefficients. Let $M = \operatorname{lcm}(c_0, c_1, \dots, c_d)$ and $m = \gcd(c_0, c_1, \dots, c_d)$. over nonzero coefficients. Let $c_b$ be the minimal nonzero coefficient.
FTSOC assume that $a_n$ steadily increases with $n$.

Define by CRT \begin{align*} &N \equiv a_p \pmod{p}, d < p < n \\ &N \equiv 0 \pmod{p^{\nu_p(M) + 1}}, p \le d \end{align*}Choose $a_p$ such that $p \nmid \frac{P(a_p)}{m}$, which is possible by Lagrange's theorem.

Take $N$ sufficiently large such that $P(N) < n^n$ which is possible by the prime number theorem and primorial growth rate.

Claim: For $p < n$ and some fixed $K$ independent of $N$, either $b = 0$ and $\nu_p(P(N))$ is fixed over $N$, or $b \ge 1$ and $\nu_p(P(N)) = b \cdot \nu_p(N) = b \cdot (\nu_p(M) + 1)$.
Proof. Follows by construction. $\blacksquare$
Hence, since primes $p \le d$ are bounded in $P(N)$, primes $d < p < n$ aren't present in $P(N)$, and primes $p > n$ aren't present in $\tau(P(N))$, the result follows.
This post has been edited 10 times. Last edited by YaoAOPS, Aug 22, 2023, 6:56 PM
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dgrozev
2463 posts
#14 • 3 Y
Y by PRMOisTheHardestExam, YaoAOPS, MS_asdfgzxcvb
The answer is NO. The plan is the same as in #2 with different realization. We will show that for infinitely many $ n\in\mathbb{N}$ the prime factorization of $ P(n)$ consists of primes with small exponents. We prove it by counting the number of $ n$'s for which $ p^s\mid P(n)$ for a prime $ p$ and sufficiently large $ s$. Here the Hensel's lemma will do the job. This allows us to show existence of a bounded subsequence of $ (a_n)$.

Proposition 1. For each polynomial $ P(x)$ with integer coefficients there exists a constant $ M>0$ such that for infinitely many $ n\in\mathbb{N}$ all the exponents of the prime factoring of $ P(n)$ are less than $ M$.

Let us first see how this claim helps. Indeed, suppose
$$ \displaystyle P(n)=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k} \text{ and }  \alpha_i< M,i=1,2,\dots,k.$$Then $ \tau(P(n))=(\alpha_1+1)\cdots(\alpha_k+1)$ and the prime factors of it are less or equal to $ M$. Each of them can participate in the factoring of $ P(n)$ with degree at most $ M$, hence $ a_n\le M^M$ and it holds for infinitely many $ n$.

The proof of Proposition 1. is based on an interesting and important fact (Korea 2016 TST ), so we formulate it as a standalone claim.

Proposition 2. Let $ P(x)$ be an irreducible polynomial with integer coefficients and $ p$ be a prime number. Then there exists a constant $ M>0$ such that for sufficiently large $ m$ the number of the solutions of the congruence
$$ \displaystyle P(x)\equiv 0\pmod {p^m}$$is exactly $ M$.

Proof (of Proposition 2). The proof is based on Hensel's lemma. First, let us see that for sufficiently large $ N$ if
$$ \displaystyle P(x_0)\equiv 0\pmod{p^N}$$then
$$ \displaystyle P'(x_0)\not\equiv 0\pmod{p^N}.$$Indeed, since $ P$ is irreducible, $ P(x)$ and $ P'(x)$ are coprime as polynomials, hence
$$ \displaystyle a(x)P(x)+b(x)P'(x)=c$$where $ a(x),b(x)\in \mathbb{Z}[X], c\in\mathbb{Z}, c\neq 0.$ We take $ N$ so that $ 2^N>c.$ So, if $ p^N\mid P(x_0)$ it follows $ p^N\nmid P'(x_0).$
Now, let $ x_1,x_2,\dots,x_M$ be the roots of
$$ \displaystyle P(x)\equiv 0\pmod{p^N}.$$It follows by Hensel's lemma that for any $ m\ge N$ each of $ x_i, i=1,2,\dots,x_M$ is "lifted" uniquely to a root $ x'_i$ modulo $ p^m$ with $ x'_i\equiv x_i\pmod {p^N}$ and these are the only roots of $ P(x)\equiv\pmod {p^m}$. $ \square$

Remark. Note that we actually proved as a bonus that there is a constant $ N$ which depends only on the polynomial $ P$ such that for any prime $ p$ and any $ m>N$ the number of solutions of
$$ \displaystyle P(x)\equiv 0\pmod{p^m}$$is at most $ p^N$. We are ready to prove Proposition 1.

Proof (of Proposition 1). Let us first consider the case when $ P(x)$ is an irreducible polynomial. Let $ p_1,p_2,\dots,p_k$ be the first $ k$ prime numbers, where $ k\in \mathbb{N}$. According to the remark after the Proposition 1.2, for any $ s\in\mathbb{N}, s\ge N$ each congruence
$$ \displaystyle P(x)\equiv 0 \pmod{p^s}\,,\,i=1,2,\dots, k$$has at most $ p^N$ solutions. Let us now count the numbers of $ n$'s $ 1\le n\le p_k$ for which $ P(n)$ is multiple of $ p_i^s$ for some $ i, 1\le i\le k$. This number is at most $ \varepsilon\cdot p_k$ where
\begin{align*}\varepsilon&:= \frac{p_1^N}{p_1^s}+\frac{p_2^N}{p_2^s}+\dots+\frac{p_k^N}{p_k^s} \\
    &= \frac{1}{p_1^{s-N}}+\frac{1}{p_2^{s-N}}+\dots+\frac{1}{p_k^{s-N}}\\
    &< \sum_{j=1}^{\infty}\frac{1}{p_j^{s-N}}\qquad(1).
\end{align*}Since $ N$ depends only on the polynomial $ P(x)$, we can take $ M$ sufficiently large such that $ \varepsilon<1/2$ for $ s\ge M$ and all $ k\in \mathbb{N}$. It means that there is $ n\in[p_k/2, p_k]$ such that $ P(n)$ is not multiple of $ p_i^M, i=1,2,\dots, k$. But in this case $ P(n)$ is also not multiple of any $ p^M$ for a prime $ p>p_k$ since $ P(n)<n^M<p^M$ if we choose $ M$ be larger than $ \deg(P)$ and $ p_k$ is large enough.
Thus, we found $ n\ge p_k/2$ that satisfies the requirements of Proposition 1 for some $ M>0$ (not depending on $ k$). We can consecutively choose $ k$ bigger and bigger thus finding infinitely many $ n$ satisfying the statement of Proposition 1.

Let us now consider the case when $ P(x)$ is reducible polynomial with integer coefficients. It can be represented as
$$ \displaystyle P(x)=P_1(x)P_2(x)\cdots P_{\ell}(x)$$where $ P_i(x), i=1,2,\dots,\ell$ are irreducible polynomials with integer coefficients. We apply the same argument to each of these polynomials, but with $ \displaystyle \varepsilon<\frac{1}{2\ell},$ and conclude that there is $ n\in [p_k/2,p_k]$ such that $ p_j^M\nmid P_i(n), j=1,2,\dots,k;i=1,2,\dots,\ell$. Hence $ p_j^{\ell M}\nmid P(n)$.
This post has been edited 1 time. Last edited by dgrozev, Sep 1, 2023, 7:37 PM
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bin_sherlo
733 posts
#16
Y by
Hope this is true :blush:
There does not exist such polynomial. Suppose that the leading coefficient of $P$ is positive. Let $\deg P=d$. Call a prime special iff it divides $P(x)$ for some integer $x$. Let $p_1<p_2<\dots<p_m<\dots $ be all special primes. By Chinese Remainder Theorem we can pick $P(t)=p_1^{a_1}\dots p_m^{a_m}r_1^{b_1}\dots r_x^{b_x}$ where $t,m$ are sufficiently large. Suppose that $x\geq C$ implies $(P(x),\tau(P(x)))>P(t)$. For a sufficiently large $k>C$ that we will choose later, pick
\[a\equiv t(mod \ p_i^{a_i+1}) \forall i\leq m, \ \ a\equiv t(mod \ r_i^{b_i+1}),\ \ a\equiv w(mod \ \frac{p_{m+1}\dots p_k}{r_1\dots r_x})\]Where $p_k>P(w)>0$. Let $a$ be the second smallest positive integer holding these modulo restrictions. This yields $a<2P(t).rad P(t)p_{m+1}\dots p_k$.
$P(t)|P(a)$ and any prime divisor of $\frac{P(a)}{P(t)}$ is larger than $p_k$. Notice that $a>p_k>C$ hence $(P(a),\tau(P(a)))>P(t)$. Let $P(a)=p_1^{a_1}\dots p_m^{a_m}r_1^{b_1}\dots r_x^{b_x}p_{k+1}^{c_{k+1}}\dots p_l^{c_l}$ where $c_i$ are nonnegative. $k$ is sufficiently large such that $p_k>(a_1+1)\dots (b_x+1)$ hence $(p_i, (a_j+1)(b_j+1))=1$ for $i\geq k+1$. Since $(x,yz)\leq z(x,y)$ we have
\begin{align*}
P(t)< (P(a),(\tau(P(a)) &= (p_1^{a_1}\dots r_x^{b_x}\dots p_l^{c_l},(a_1+1)\dots (b_x+1)\dots (c_l+1)) \\
&\leq p_1^{a_1}\dots p_m^{a_m}r_1^{b_1}\dots r_x^{b_x}(p_{k+1}^{c_{k+1}}\dots p_l^{c_l},(a_1+1)\dots (b_x+1)\dots (c_l+1)) \\
& =P(t) (p_{k+1}^{c_{k+1}}\dots p_l^{c_l},(c_{k+1}+1)\dots (c_l+1))
\end{align*}Thus, $(p_{k+1}^{c_{k+1}}\dots p_l^{c_l},(c_{k+1}+1)\dots (c_l+1))>1$ which implies $p_i|c_j+1$ for some $i,j\geq k+1$.
\[(2P(t)^2)^{d+1}.(p_k)^{(k-m)(d+1)}>(2P(t)^2p_{m+1}\dots p_k)^{d+1}>a^{d+1}>P(a)>P(t)p_{k+1}^{p_k}>p_k^{p_k}\]Since $(2P(t)^2)^{d+1}=\ell $ is a constant, this requires $\ell>(p_k)^{p_k-(k-m)(d+1)}$ or $p_k<(k-m)(d+1)$ however this is not true since primes have density zero as desired.$\blacksquare$
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