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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
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0 replies
1 viewing
jlacosta
Mar 2, 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
Number theory - Iran
soroush.MG   32
N 2 minutes ago by Nobitasolvesproblems1979
Source: Iran MO 2017 - 2nd Round - P1
a) Prove that there doesn't exist sequence $a_1,a_2,a_3,... \in \mathbb{N}$ such that: $\forall i<j: gcd(a_i+j,a_j+i)=1$

b) Let $p$ be an odd prime number. Prove that there exist sequence $a_1,a_2,a_3,... \in \mathbb{N}$ such that: $\forall i<j: p \not | gcd(a_i+j,a_j+i)$
32 replies
soroush.MG
Apr 20, 2017
Nobitasolvesproblems1979
2 minutes ago
Inspired by my own results
sqing   2
N 27 minutes ago by cazanova19921
Source: Own
Let $ a,b,c\geq \frac{1}{2}  . $ Prove that
$$ (a+1)(b+2)(c +1)-15 abc\leq \frac{15}{4}$$$$ (a+1)(b+3)(c +1)-21abc\leq \frac{21}{4}$$$$(a+2)(b+1)(c +2)-25a b c \leq \frac{25}{4}$$$$ (a+2)(b+3)(c +2)-35a b c \leq  \frac{35}{2}$$$$    (a+3)(b+1)(c +3)-49a b c \leq  \frac{49}{4}$$$$ (a+3)(b+2)(c +3)-49a b c \leq \frac{49}{2}$$
2 replies
sqing
2 hours ago
cazanova19921
27 minutes ago
Line through incenter tangent to a circle
Kayak   32
N 27 minutes ago by L13832
Source: Indian TST D1 P1
In an acute angled triangle $ABC$ with $AB < AC$, let $I$ denote the incenter and $M$ the midpoint of side $BC$. The line through $A$ perpendicular to $AI$ intersects the tangent from $M$ to the incircle (different from line $BC$) at a point $P$> Show that $AI$ is tangent to the circumcircle of triangle $MIP$.

Proposed by Tejaswi Navilarekallu
32 replies
Kayak
Jul 17, 2019
L13832
27 minutes ago
D1015 : A strange EF for polynomials
Dattier   3
N 34 minutes ago by Dattier
Source: les dattes à Dattier
Find all $P \in \mathbb R[x,y]$ with $P \not\in \mathbb R[x] \cup \mathbb R[y]$ and $\forall g,f$ homeomorphismes of $\mathbb R$, $P(f,g)$ is an homoemorphisme too.
3 replies
Dattier
Mar 16, 2025
Dattier
34 minutes ago
Turkey EGMO TST 2017 P6
nimueh   4
N an hour ago by Nobitasolvesproblems1979
Source: Turkey EGMO TST 2017 P6
Find all pairs of prime numbers $(p,q)$, such that $\frac{(2p^2-1)^q+1}{p+q}$ and $\frac{(2q^2-1)^p+1}{p+q}$ are both integers.
4 replies
nimueh
Jun 1, 2017
Nobitasolvesproblems1979
an hour ago
An inequality
JK1603JK   4
N an hour ago by Quantum-Phantom
Source: unknown
Let a,b,c>=0: ab+bc+ca=3 then maximize P=\frac{a^2b+b^2c+c^2a+9}{a+b+c}+\frac{abc}{2}.
4 replies
1 viewing
JK1603JK
Yesterday at 10:28 AM
Quantum-Phantom
an hour ago
Inspired by Abelkonkurransen 2025
sqing   1
N an hour ago by kiyoras_2001
Source: Own
Let $ a,b,c $ be real numbers such that $  a^2+4b^2+16c^2= abc. $ Prove that $$\frac{1}{a}+\frac{1}{2b}+\frac{1}{4c}\geq -\frac{1}{16}$$Let $ a,b,c $ be real numbers such that $ 4a^2+9b^2+16c^2= abc. $ Prove that $$ \frac{1}{2a}+\frac{1}{3b}+\frac{1}{4c}\geq -\frac{1}{48}$$
1 reply
sqing
Yesterday at 1:06 PM
kiyoras_2001
an hour ago
Inspired by Titu Andreescu
sqing   0
an hour ago
Source: Own
Let $ a,b,c>0 $ and $ a+b+c\geq 3abc . $ Prove that
$$a^2+b^2+c^2+1\geq \frac{4}{3}(ab+bc+ca) $$
0 replies
sqing
an hour ago
0 replies
Geometry challenging question
srnjbr   0
2 hours ago
Given a triangle ABC. A1, B1 and C1 are the points of contact of the inner circumcircle of the triangle with the sides BC, AC and AB respectively. The point of contact of AA1 with B1C1 and the circumcircle are called L and Q respectively. M is the midpoint of B1C1. The point of intersection of lines BC and B1C1 is called T. P is the foot of the perpendicular drawn to AT from point L. Show that points A1, M, Q and P lie on a circle.
0 replies
srnjbr
2 hours ago
0 replies
Plane normal to vector
RenheMiResembleRice   0
2 hours ago
Source: Bian Wei
Solve the attached
0 replies
RenheMiResembleRice
2 hours ago
0 replies
Complex numbers should be easy
RenheMiResembleRice   1
N 2 hours ago by RenheMiResembleRice
Source: Wenjing Kong
I cant do the last part. :(
1 reply
RenheMiResembleRice
2 hours ago
RenheMiResembleRice
2 hours ago
Strange NT
magicarrow   20
N 2 hours ago by Yuvi01
Source: Romanian Masters in Mathematics 2020, Problem 6
For each integer $n \geq 2$, let $F(n)$ denote the greatest prime factor of $n$. A strange pair is a pair of distinct primes $p$ and $q$ such that there is no integer $n \geq 2$ for which $F(n)F(n+1)=pq$.

Prove that there exist infinitely many strange pairs.
20 replies
magicarrow
Mar 1, 2020
Yuvi01
2 hours ago
D1010 : How it is possible ?
Dattier   13
N 2 hours ago by Dattier
Source: les dattes à Dattier
Is it true that$$\forall n \in \mathbb N^*, (24^n \times B \mod A) \mod 2 = 0 $$?

A=1728400904217815186787639216753921417860004366580219212750904
024377969478249664644267971025952530803647043121025959018172048
336953969062151534282052863307398281681465366665810775710867856
720572225880311472925624694183944650261079955759251769111321319
421445397848518597584590900951222557860592579005088853698315463
815905425095325508106272375728975

B=2275643401548081847207782760491442295266487354750527085289354
965376765188468052271190172787064418854789322484305145310707614
546573398182642923893780527037224143380886260467760991228567577
953725945090125797351518670892779468968705801340068681556238850
340398780828104506916965606659768601942798676554332768254089685
307970609932846902
13 replies
Dattier
Mar 10, 2025
Dattier
2 hours ago
IMO problem 1
iandrei   76
N 3 hours ago by ihategeo_1969
Source: IMO ShortList 2003, combinatorics problem 1
Let $A$ be a $101$-element subset of the set $S=\{1,2,\ldots,1000000\}$. Prove that there exist numbers $t_1$, $t_2, \ldots, t_{100}$ in $S$ such that the sets \[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100  \] are pairwise disjoint.
76 replies
iandrei
Jul 14, 2003
ihategeo_1969
3 hours ago
Polynomial produces perfect powers
TheUltimate123   21
N Mar 15, 2025 by pi271828
Source: ELMO 2023/1
Let \(m\) be a positive integer. Find, in terms of \(m\), all polynomials \(P(x)\) with integer coefficients such that for every integer \(n\), there exists an integer \(k\) such that \(P(k)=n^m\).

Proposed by Raymond Feng
21 replies
TheUltimate123
Jun 26, 2023
pi271828
Mar 15, 2025
Polynomial produces perfect powers
G H J
G H BBookmark kLocked kLocked NReply
Source: ELMO 2023/1
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TheUltimate123
1739 posts
#1 • 4 Y
Y by GeoKing, ImSh95, Stuart111, Rounak_iitr
Let \(m\) be a positive integer. Find, in terms of \(m\), all polynomials \(P(x)\) with integer coefficients such that for every integer \(n\), there exists an integer \(k\) such that \(P(k)=n^m\).

Proposed by Raymond Feng
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EpicNumberTheory
250 posts
#2 • 3 Y
Y by Nuterrow, GeoKing, ImSh95
We claim that all polynomials $P(x) = (\pm x+ c)^d$ work where $c$ is an arbitrary integer, and $d$ is any positive divisor of $m$. If $P(x) = (x+c)^d$ then picking $x = n^{\frac{m}{d}}-c$ gives $P(x) = n^m$ thus these works. If $P(x) = (-x+c)^d$ then picking $x = -n^{\frac{m}{d}}+c$ gives $P(x) = n^m$ thus these work as well. Now we show that these are the only solutions.

Firstly letting $n=0$ implies that $P$ has an integer root. Let it be $-b$. Then $P(x) = P_{d-1}(x)\cdot (x+b)$ (Let $P_j(x)$ denote $\frac{P(x)}{(x+b)^{d-j}}$ in the induction). Now we inductively show that $(x+b) \mid P_i(x)$ if $i \geq 1$. Let $p$ be any prime. Then now to show that $x+b$ divides $P_i(x)$ let $P_i(x) = (x+b)Q(x) + R(x)$. But letting $n =p$ we get that $P(k)$ is a prime power, thus so is $k+b$ and $P_i(k)$. If $i>1$ then letting $p$ be large shows that $|k|$ is large and $| P_i(k) | > k+b$ thus $k+b$ divides $P_i(k)$. So $R(k) = 0$. Considering all large primes $p$, there exist infinitely many zeros of $R(x)$. Thus $R \equiv 0$ and thus $ak+b \mid P_i(x)$ completing the induction hypothesis when $i > 1$. If $i =1$ then let $P_1(x) = rx+s$. Now by the same argument $k+b$ and $rk+s$ are both perfect powers of $p$. If $|rk+s| \geq |k+b|$ for all large $k$ then the same argument shows that $k+b$ divides $rk+s$. Suppose that $|k+b| \geq |rk+s|$ for all large $k$. Then the same argument shows that $rk+s$ divides $k+b$. If $|r| > 1$ then the former holds ($|rk+s| \geq | k +b|$ for large $k$). Thus $|r| = 1$. If $r =1$ then $k+s \mid k + b \implies k +s \mid b-s$. Letting $k$ large shows $b=s$. If $r = -1$ then $-k+s$ divides $k+b$ and so $-k+s$ divides $b+s$, and letting $k$ large gives $b=-s$. Thus in all cases we get that $P(x) = z(x+b)^d$ where $d$ is degree of $P$. Now to show that $z = \pm 1$, suppose for the sake of contradiction that $z$ has a prime divisor $p$. Then let $n = q \neq p$. Thus $p \mid z \mid q^m$, a contradiction! So $z = \pm 1$.

It remains to show that $d$ is a divisor of $m$. Again let $p$ be a prime and let $n = p$. Then $|(k+b)^d|$ is a power of prime $p$ , thus $|k+b|$ is also a power of $p$. Letting $|k+b| = p^{\ell} \implies p^m = P(x) = p^{d \ell}$ (As $p^m>0$ we need not care about sign here). Thus $d$ divides $m$.

Now if $d$ is even and $z = -1$ then $P \leq 0$ always, but its equal to positive integers too (Set $n = 1$ for example). So $z = 1$. Thus $P(x) = (x+b)^d$ if $d$ is even. Note that $(x+b)^d = (-1)^d(x+b)^d = (-x-b)^d = (-x+(-b))^d$.
If $d$ is odd then $z = \pm 1$ is fine: $P(x) = \pm(x+b)^d = (\pm x + (\pm b))^d$
Thus we can say that the general solution is $P(x) = (\pm x + b)^d$ where $d$ is a positive divisor of $m$ , and $b$ is any arbitrary integer.
This post has been edited 1 time. Last edited by EpicNumberTheory, Jun 26, 2023, 5:59 AM
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blackbluecar
302 posts
#3 • 8 Y
Y by Nuterrow, PRMOisTheHardestExam, LoloChen, ImSh95, GeoKing, Rounak_iitr, XbenX, Math_legendno12
The answer is $P(x)=(\pm x -k)^\alpha$ where $k \in \mathbb{Z}$ and $\alpha \in \mathbb{N}$ where $\alpha$ divides $m$.

Note that there exists $k$ where $P(k)=0^m=0$. So, wlog, shift $P$ so that $P(0)=0$. Thus, we see that $d$ divides $P(d)$ for all positive integers $d$. Note that for any prime $q$ there exists an $a$ for which $P(a)=q^m$ but recall that $a$ divides $P(a)$. So, it must follow that $a = \pm q^\ell$ for some $0 \leq \ell \leq m$. Thus, by infinite PHP, there exists a positive integer $\alpha \leq m$ and an infinite sequence of primes $q_1,q_2, \ldots$ for which either $P(q_i^\alpha)=q_i^m$ or $P(-q_i^\alpha)=q_i^m$. Assume wlog that $P(q_i^\alpha)=q_i^m$

Claim: if $P$ has degree $t$ then $\alpha t=m$

Let $Q(x)=P(x^\alpha)$. Note that $Q(a)=a^m$ has infinite solutions $a \in \mathbb{R}$, so it follows that $Q(x)=x^m$ so the degree of $Q$ is both $m$ and $\alpha t$ as desired. $\square$

Thus, since $\alpha t = m$ it follows that $\alpha$ divides $m$. Thus, undoing all of the wlogs, we are left with $P(x)=(\pm x - k)^\alpha$ as desired.
This post has been edited 4 times. Last edited by blackbluecar, Feb 29, 2024, 1:01 PM
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Mogmog8
1080 posts
#4 • 2 Y
Y by centslordm, ImSh95
Letting $n=0$ we see $P(k_0)=0$ for some integer $k_0$. Note if $P(x)$ works, then $P(x+c)$ works for any integer $c$, so $Q(x)=P(x+k_0)$ works. Note $Q(0)=P(k_0)=0$. Hence, $x\mid Q(x)$.

Claim: If we can write $Q(x)=x^\ell R(x)$ for a polynomial $R$ with integer coefficients an integer $\ell\ge 1$, then $Q(x)=\pm x^\ell$ (both may not work) or we can write $Q(x)=x^{\ell+1} R_1(x)$ for a polynomial $R_1$ with integer coefficients.
Proof. Consider a prime $p$. Note that there exists $k$ such that $k^\ell R(k)=Q(k)=p^m$. Hence, $k\mid p^m$ so let $k=\pm p^t$. Then, $R(p^t)=\pm p^{m-\ell t}$. Letting $R(x)=a_sx^x+\dots +a_1 x+a_0$, we see \[a_s p^{ts}+a_{s-1}p^{t(s-1)}+\dots+p^ta_1+a_0\]Hence, we must have one or more of the following true: (i) $t=0$ (ii) $m=\ell t$ (iii) $p\mid a_0$.

Varying $p$, we see (i) is true at most once as it implies $Q(1)=p^m$, which occurs for at most one $p$. Suppose $p\mid a_0$ is true finitely many times as $p$ varies. Then, $m=\ell t$ infinity many tries; that is, $R(p^t)=\pm 1$ infinity many times. Hence, either $R(x)-1$ or $R(x)+1$ has infinite roots so $R(x)=\pm 1$ so $Q(x)=\pm x^\ell$. $\blacksquare$

Let $\deg Q=d$. We claim by induction that for positive integers $z\le d$, we can write $Q(x)=x^zR_z(x)$ where $R_z$ is a polynomial with integer coefficients. For the base case $z=1$, we proved earlier that $x\mid Q(x)$. For the inductive step, if $z=y$ where $y+1\le d$ holds, by our claim $Q(x)=x^{y+1}R_{y+1}(x)$ or $Z(x)=\pm x^y$. The latter is false by degree comparison. Hence, our induction is complete so we can let $Z(x)=x^dR_d(x)$. Applying our claim, we have $Q(x)=\pm x^d$ or $Q(x)=x^{d+1}R_{d+1}(x)$. Again, the latter case is false by degrees.

Suppose FTSOC that $d\mid m$. Then, letting $n=2$ we have $\pm k^d=2^m$ for some $k$. Note $k=\pm 2^a$ so $\pm 2^{ad}=2^m$ so $ad=m$, contradiction. Hence, $Q(x)=\pm x^d$ where $d\mid m$. Note if $d$ is odd both plus and minus work but if $d$ is even only plus works (or else $Q$ can't cover positive values). Hence, $Q(x)=(\pm x)^d$ so $P(x)=(\pm x+c)^d$. This works if we let $x=n^{m/d}\mp c$. $\square$
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Orestis_Lignos
555 posts
#5 • 1 Y
Y by ImSh95
We claim that the only solutions are polynomials of the form $P(x)=(u-x)^{m/t}$ or $P(x)=(x-u)^{m/t}$, with $u$ an arbitrary integer and $t$ a positive divisor of $m$. The problem naturally splits into two parts.

Part 1: All these polynomials work. Indeed, for $P(x)=(u-x)^{m/t}$ we may take $k=u-n^t$, and for $P(x)=(x-u)^{m/t}$ we may take $x=u+n^t$.

Part 2: All polynomials that work are of this form. For $n=0$ in the given condition, there exists a $u$ such that $P(u)=0$. Now, for any prime $p$, let $p'$ be (one of) the corresponding integer such that $P(p')=p^m$. Then,

$p'-u \mid P(p')-P(u)=p^m,$

hence $p'-u=\pm p^t$ for some $t \leq m$. Therefore, there exists an $\epsilon \in \{-1, 1 \}$ and an integer $t \leq m$, such that $p'+\epsilon p^t=u$ for infinitely many primes $p$. Therefore,

$P(u-\epsilon p^t)=p^m$ for infinitely many $p$, which implies that $P(u-\epsilon x^t)=x^m$ for all $x$. Comparing the degrees of the two polynomials implies that $t \mid m$.

If $\epsilon=1$, then $P(u-x^t)=x^m$ for all $x$, which implies that $P(x)=(u-x)^{m/t}$ for infinitely many $x$, and so

$P(x)=(u-x)^{m/t}$ for all $x$.

If $\epsilon=-1,$ then $P(u+x^t)=x^m$ for all $x$, and we similarly obtain that

$P(x)=(x-u)^{m/t}$ for all $x$.
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P2nisic
406 posts
#6 • 2 Y
Y by GeoKing, ImSh95
Let $f:Z->Z$ such that $P(f(n))=n^m$ then: $f(x)-f(y)|P(f(x))-P(f(y))=x^m-y^m$

Take now $y=0$ and $x=r=prime$ we get that $f(r)-c|r^m\Rightarrow f(r)=\pm r^d +c$ where $f(0)=c$.
So since $d=0,1,2,3,..m$ there will be infinity primes which will have the same $d$ then we have:
$P(a*r^d+c)=r^m$ for infinity primes which means that $P(a*x^d+c)=x^m$ for all $x$ (becayse if we let $R(x)=P(a*r^d+c)-r^m$ then it has infinity roots so $R(x)=0$ )with $a=1,-1$

Let now $P(x)=(x-c)^dQ(x)$ with $Q(c)\neq 0$ then we have:
$x^m=P(a*x^b-c)=a^dx^{bd}Q(ax^b-c)\Rightarrow x^{m-bd}=a^dQ(ax^b-c)$ (1)
If $m-bd\geqslant 1$ then for $x=0$ we have $Q(c)=0$ contradiction
Obvioysly $m-bd\geqslant 0$ otherwise $deg(Q(x))$ will be negative wrong
So $m=bd$ and (1) became $1=a^d*Q(ax^b-c)$ so $Q(ax^b-c)=+-1$ which gives $P(x)=+-(x-c)^d$
we have to pay a little attention to the parity of $d$ because for an even number it gives only positive or only negative.
Esily we get the sollution:
If $m=even=bd$ then $P(x)=+-(x-c)^d$ if $d=odd$ and $P(x)=+(x-c)^d$ if $d=even$
If $m=odd=bd$ then $P(x)=+-(x-c)^d$.
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MrOreoJuice
594 posts
#7 • 5 Y
Y by GeoKing, PRMOisTheHardestExam, SatisfiedMagma, ImSh95, EndipifromBMT
Answer
Solution
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PRMOisTheHardestExam
409 posts
#9 • 1 Y
Y by ImSh95
just 1 idea:
let $P(f(x))=x$ then $f(x)-f(y) \mid x^m - y^m$.. if $m$ is odd then $f$ has a solution by https://artofproblemsolving.com/community/c6h488541p2737651
then it is easy to get $P(x)$
but if $m$ is not odd then we cannot use this solution :(
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VicKmath7
1385 posts
#10 • 1 Y
Y by ImSh95
Algebraic solution
This post has been edited 9 times. Last edited by VicKmath7, Jul 19, 2023, 8:29 AM
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IAmTheHazard
5000 posts
#11 • 8 Y
Y by GeoKing, PRMOisTheHardestExam, ImSh95, Aryan-23, centslordm, megarnie, khina, ohiorizzler1434
Anyone want to post their density solution? I know you're out there :ninja:
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a_n
162 posts
#12 • 1 Y
Y by ImSh95
Well, I'm finally getting to latexing my elmo solutions now...

I claim the only solution(s) is: $P(x)=(\pm 1)^d \times (x+c)^d$ where $d$ is any natural number that divides $m$ and $c$ is an arbitrary integer.

Clearly if $P(x)$ works so does $P(x+c)$ so WLOG assume $P(0)=0$.

Also note that $P(\infty) \to -\infty$ implies $P(- \infty) \to \infty$ so WLOG assume $P(x)$ has a positive leading coefficient.

Let $P(k_n)=n^m$. Let $\mathbb{P}$ be the set of primes.

And now as we know $a-b \mid P(a)-P(b)$, so $P(k_p) \mid p^m \  \forall \ p \in \mathbb{P}$, but since primes are cool this implies that $P(k_p)=p^{t_p}$ for some $t_p \in [1,2 \cdots m]$ for all $p$.

And now here is the main step, since primes are infinite, and $t_p$ can be zero at max once (because $P(1)$ can only take one value), this implies that there exists an infinite subset $\mathbb{Q}$ of $\mathbb{P}$ such that all the $t_q$ ($q \in \mathbb{Q})$ share some common value $t \in [1,2 \cdots m]$

And now our work is basically done, let $P(x)= \sum_{i=0}^d a_i x^i$.

And now think only in terms of large enough $q \in \mathbb{Q}$. Since we have established that $P(q^t)=q^m$,

$q^{td-1} < \sum_{i=0}^d a_i q^{it} < q^{td+1}$ but the middle thing is $P(q^t)=q^m$, so $q^m$ is a power of $q$ between $q^{td-1}$ and $q^{td+1}$ so we must have $m=td$

And so now we have $\sum_{i=0}^{d-1} a_i q^it + (a_d -1)q^td=0$ for all large enough $q$ and the only polynomial that is zero infinitely often is the zero polynomial so this implies $P(x)=x^d$ for some $d \mid m$ (because $td=m$) and the rest of the solution set comes from our initial two assumptions.

Really cool problem ;)

What is submitted is attached below
Attachments:
p1.pdf (102kb)
This post has been edited 1 time. Last edited by a_n, Jun 30, 2023, 1:30 PM
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a_n
162 posts
#13 • 1 Y
Y by ImSh95
IAmTheHazard wrote:
Anyone want to post their density solution? I know you're out there :ninja:

I really want to see this, could you post it yourself please?
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TheUltimate123
1739 posts
#14 • 3 Y
Y by PRMOisTheHardestExam, Rounak_iitr, ike.chen
For any positive \(d\mid n\), the polynomials \(P(x)=(x+a)^d\) and \(P(x)=(-x+a)^d\) work. We show they are the only solutions.

First, there exists \(k_0\) so that \(P(k_0)=0^m\). Instead consider the polynomial \(Q(x)=P(x-k_0)\), which has 0 as a root and but still contains every \(m\)th power in its range. We will show that \(Q(x)\) is of the form either \(x^d\) or \((-x)^d\) for some \(d\mid n\).

For each prime \(p\), there is some \(k_p\) with \(Q(k_p)=p^m\). Since 0 is a root of \(Q\), we have \(x\mid Q(x)\) for every \(x\). In particular, \(k_p\mid P(k_p)=p^m\). This implies that \[k_p\in\{1,p,p^2,\ldots,p^m,-1,-p,-p^2,\ldots,-p^m\}\quad\text{for all }p.\]
By the Pigeonhole principle, one of the following must occur:
  • There is some \(0\le r\le m\) such that \(k_p=p^r\) for infinitely many \(p\). In particular, \(Q(x^r)=x^m\) infinitely often, implying it holds for all real \(x\), so \(Q(x)=x^{m/r}\) for all \(x\).
  • There is some \(0\le r\le m\) such that \(k_p=-p^r\) for infinitely many \(p\). In particular, \(Q(-x^r)=x^m\) infinitely often, implying it holds for all real \(x\), so \(Q(x)=(-x)^{m/r}\) for all \(x\).
This completes the proof.
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GrantStar
813 posts
#15 • 2 Y
Y by PRMOisTheHardestExam, ike.chen
The answers are $P(x)=(\pm x-k)^n$ for any integer $k$ and $n\mid n$. These work as $P\left(\pm x^{\frac mn} +k\right)=x^m.$

Notice that if $P(x)$ is a solution, so is $P(\pm x +k)$ for integer $k$. There exists a $j$ with $p(j)=0^m=0,$ thus WLOG shift $P$ so $P(0)=0.$

Now turn this into NT! For each prime $p$, let $x_p$ be a number with $P(x_p)=p^m.$ There might be multiple $\ell$ with $P(\ell)=x^m$ so choose the one with the smallest absolute value. Then, notice that $x_p=x_p-0\mid P(x_p)-P(0)=p^m$ so thus $x_p=\pm 1, \pm p, \pm p^2, \dots, \pm p^m$ for each prime. Then, let $f(p)=v_p(x_p)=1,2,\dots, m.$ As there are infinitely many primes but finitely many choices of $f(p),$ be pigeonhole f(p) takes the same value $y$ infinitely many times. Considering $p_1(x)=P(x)^y\pm x^m$, we see $P_1$ is a polynomial with infinitely many zeroes so $P_1(x)=0$ and $P(x)^y=\pm x^m$ or $P(x)=\pm x^{\frac my}$ for $y\mid m$. Thus $P(x)=\pm x^j$ for $j\mid m$, and we get the other results by shifting $P(x)$ to $P(\pm x+k)$.
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GoldenBoy03
22 posts
#16 • 1 Y
Y by PRMOisTheHardestExam
Seemingly similar though very different problem appeared on Kazakhstan National Olympiad $2021$. Apparently, if $n$ is a positive integer and $P(x) \in R[x]$ is such that for any positive integer $k$, there exists positive integer $l$ such that $P(l) = m^n$, then $P(x) = (ax + b)^k$ for some real numbers $a, b$ and positive integer $k$.
This post has been edited 1 time. Last edited by GoldenBoy03, Jul 24, 2023, 7:46 PM
Reason: Typo
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sdninajanlari
25 posts
#17 • 1 Y
Y by bin_sherlo
An alternative solution.

Take $P(X)=A.R_1(X)^{\alpha _1}...R_s(X)^{\alpha _s}$ where $R_i \in \mathbb Z[X]$ is irreducible over $Z[X]$ for $i=1,...,s$; $A$ is an integer constant; $Q \in \mathbb Z[X]$. Since $(R_i,R_j)=1$, there exist $T_{ij}, T_{ji} \in \mathbb Z[X]$ such that $R_i(X).T_{ij}(X)+R_j(X).T_{ji}(X)=c_{ij}$ where $c \in \mathbb Z^+$. Take a prime $p$ s.t. it is larger than all $c_{ij}$'s and for an $r$ such that at least one of $R_i(r)$'s is $\pm 1$, $p^m>P(r)$. Hence, if $s \geq 2$, then we have contradiction ($p|R_i(k)$ for any $i$.). Then $s=0,1$. If $s=0$, then $P(X)=A$, but this results in contradiction obviously. If $s=1$, we have $P(X)=A.R(X)^{\alpha}$. By taking a prime $p>|A|$, we get that $A=\pm 1$. We may suppose that $A=1$ (if $\alpha$ is even, it is obvious; otherwise, multiply $R$ by $-1$). Using the methods mentioned above, we conclude that $P(x)=(\pm x +c)^\alpha$ for an $\alpha | m$ and an integer $c$.
This post has been edited 1 time. Last edited by sdninajanlari, Nov 25, 2023, 3:48 PM
Reason: typo again
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cosmos1999
5 posts
#18
Y by
We claim that the only sols are $(\pm x+c)^d$ with $c$ an integer and $d$ a positive divisor of $m$

We know that there exists a number whose image with $P$ is $0$, shift $P$ so that it's $0$ and replace $P(x)$ by $P(-x)$ if necessary so that $P$ is infinitely positive, take a prime $p$, then if $P(k)=p^m$ then $k|P(k)-P(0)=p^m$ so $k$ is a power of $p$ which we set to be $p^k,$ clearly for large enough $p$ we must have $k\le m$ so by infinite PHP(since there is a finite number of $k$s) we get a value of $k$ which we call $d$ that's repeated for infinitely many primes $p$ thus $P(p^d)=p^m$ for infinitely many primes $p$ implying $P(x)=x^d$ for all $x,$ checking we find that $d\mid m$ which gives us our sols
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TheHazard
93 posts
#19 • 1 Y
Y by PRMOisTheHardestExam
This can be generalized to having this hold for a set of any integers $n_1 < n_2 < \dots$ such that $\frac{n_{N+k}}{n_N}$ approaches $1$ (sub-exponential) that are dense in residues and contain infinitely many $d$th powers for each $d$.

Let $P$ have degree $d$. Note that $P(k)$ contains terms of the form $n^{md}$ for each $d$.
Define $x_i = k_i$ for $x_1, \dots, x_N$ such that $P(x_i) = (n_{N+i})^{md}$ for arbitrarily large $N$ such that $\frac{n_{N+n}}{n_{N+1}} < 2^{-1434md}$. We can verify that $\frac{1}{2} < \frac{P(x_i)}{P(x_i)} < 2$ and that $\frac{P(x_i)}{P(x_j)}$ is a $d$-th power of a rational.

As such, by generalized Shortlist 2016 N8 it follows that $P(x) = c (ax + b)^d$ for integers $a, b$ and rational $c$. It is well-known that $n_i$ is divisible by infinitely many primes, so taking large prime $p$ such that $p$ doesn't appear in the factorization of $c$, it must follow that $d \mid m$, and that $c$ is a $d$th power. It thus follows that we can re-express $P(x) = (ax + b)^d$ due to being integral. By density it follows that $a = \pm 1$.
This post has been edited 1 time. Last edited by TheHazard, Jan 19, 2024, 6:49 PM
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Pyramix
419 posts
#20 • 1 Y
Y by PRMOisTheHardestExam
There exists some $k$ for which $P(k)=0$. Then, shift by $k$ so that we have $P(0)=0$. So, we have $a\mid P(a)$ for every $a$. So, for every prime $p$ we have $a\mid P(a)=p^{m}$, which means $a=p^l$ with $0\leq l\leq m$. Since there are infinitely many primes, one exponent $l_0$ occurs infinitely many times, i.e. there are infinitely primes for which $P(p^l) = p^m$.
Let $Q(x) = P(x^l)$. Then, there are infinitely many solutions to $Q(x) = x^m$, which means $Q(x)$ and $x^m$ always agree. So, $Q(x) = x^m$ for every real $x$. Hence, $\frac ml$ is an integer and $P(x) = x^{\frac ml}$ or $P(x) = -x^{\frac ml}$ (if $\frac ml$ is odd.)
Re-shifting $x$ in $P$, we see that all the valid solutions are
\[P(x)\in \left\{(x-k)^{\frac ml}, \ \ \  \text{and} \ \ \ (k-x)^{\frac ml} \ \ \ \text{given } l\mid m\text{ and }k\in\mathbb{N}\right\}\]Indeed, these solutions work.
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L13832
250 posts
#21
Y by
Claim: If $P(x)$ is a solution then so is $P(\pm x+k)$.
Proof: There exists a $c$ such that $P(c)=0^m=0$, now shift $P$ such that $P(0)=0$.
Let $a_p$ be a number such that $P(a_p)=p^m$, since $P(0)=0$ we get $a_p\mid p^m$.
So possible values of $a_p$ are
$\{\pm1,\pm p,\pm p^2, \dots, \pm p^m\}$.
By Pigeonhole Principle $\exists$ $0\leq t\leq m$ such that $a_p=\pm p^t$ for infinitely many $t$ and $P_1(x)=P(x)^t+\pm x^m$, we also have $P_1(x)=0$.
So $P(x)^t=\pm x^m$ .
We finally get our desired answers to be $(\pm x+k)^i, i \mid m$ by shifting $P(x) \to P(\pm x+k)$.
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bin_sherlo
662 posts
#22 • 1 Y
Y by tiny_brain123
After writing down the solution, I realised that this one is really similar to sdninajanlari's solution at #17. How couldn't I know this? :D
Answer is $P(x)=(\pm x+b)^r$ where $b$ is any integer and $r$ is a positive divisor of $m$.
Lemma: If $R(x)$ and $S(x)$ are coprime polynomials in integers, then there exists a constant $D$ with $(R(x),S(x))\leq D$.
Proof: By Bezout, there exists $A,B$ such that $A(x)R(x)+B(x)S(x)=C$ where $C$ is a constant. Hence $p^k|R(x),S(x)$ implies $p^k|C$ which is bounded.$\square$
Let $P(x)=Q_1(x)^{\alpha_1}\dots Q_l^{\alpha_l}(x)$ with $Q_i(x),Q_j(x)$ are coprime. By the lemma, we see that $(Q_i(x),Q_j(x))\leq D$ for some constant $C$ for each $1\leq i\leq j\leq l$. By the condition, there exists some $k$ for a prime $p\gg D$ where $Q_1(k)^{\alpha_1}\dots Q_l^{\alpha_l}(k)=P(k)=p^m$ and since $p$ is sufficiently large, $p$ cannot divide both $Q_i(k)$ and $Q_j(k)$ at the same time thus, $l-1$ polynomials among $Q_i'$s must be constant (because $l-1$ of them are $\pm 1$ infinitely many times) which yields $P(x)=c.Q(x)^{\alpha}$ and $|c|=1$ since $c|n$ for each positive integer $n$. The polynomial $Q(x)$ can be divisible by each prime so by Chebotarev we get that $Q$ is linear. Let $Q(x)=ax+b$ and $P(x)=\pm (ax+b)^{\alpha}$. We have $\pm (ax+b)^{\alpha}=p^m$ for a prime and $ax+b=p^t$ for some $t$. This gives $t.\alpha=m$ hence $\alpha|m$. Let $m=\alpha .r$ so there exists $x$ such that $\pm (ax+b)=n^r$. We see that $b=0$ yields $a=\pm 1$ which is a solution. Suppose that $b\neq 0$. We see that $a\neq 0$. If $|a|>1$, then multiples of $a$ cannot be in the image of $P$ which contradicts with the condition. So $a=\pm 1$. We get the desired conclusion.$\blacksquare$
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pi271828
3363 posts
#23
Y by
The answer is $P(x) = (x+k)^d$ and $P(x) = (-x+k)^d$ where $d \mid m$, which clearly works. Note that from setting $n = 0$, $P(x)$ must have an integer root, so we shift such that $P(0) = 0$. The following claim finishes the problem.

Claim: $P(x)$ must be of the form $x^{\ell}$ or $-x^{\ell}$ where $\ell \mid m$.

Proof. Set $n$ to be a prime $p$, such that $P(a_p) = p^m$. Clearly this implies $a_p \mid p^m$, or in other words \begin{align*} a_p \in \{-p^{m}, -p^{m-1}, \dots, p^{m-1}, p^{m} \} = S\end{align*}There are infinitely many primes and $|S|$ is finite, so by pigeonhole, there must be an integer $\ell \le m$ and integer $k \in \{0, 1\}$ such that $a_p = (-1)^k \cdot p^{\ell}$ for infinitely many primes $p$. This implies there are infinitely many $x$ such that $P(x) = (-1)^k \cdot x^{m/ \ell}$. The polynomial $Q(x) = P(x)^\ell - (-1)^{k \ell} x^m$ therefore has infinitely many roots, implying $P(x)^{\ell} = (-1)^{k\ell} x^m$, which means $\ell \mid m$. The claim readily follows. $\square$
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