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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
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
A problem
kwin   0
5 minutes ago
Let x, y, z > 0 and $ x^2+y^2+z^2=3 $ . Find max:
$$P=xy(x^2+y^2-\frac{1}{z})+yz(y^2+z^2-\frac{1}{x})+zx(x^2+z^2-\frac{1}{y})$$
0 replies
kwin
5 minutes ago
0 replies
tangent line to 3 circles of center A, B, C and all pass through orthocenter
parmenides51   1
N 7 minutes ago by ilovemath0402
Source: 2018 Saudi Arabia BMO TST I p4
Let $ABC$ be an acute, non isosceles with $I$ is its incenter. Denote $D, E$ as tangent points of $(I)$ on $AB,AC$, respectively. The median segments respect to vertex $A$ of triangles $ABE$ and $ACD$ meet$ (I)$ at$ P,Q,$ respectively. Take points $M, N$ on the line $DE$ such that $AM \perp BE$ and $AN \perp  C D$ respectively.
a) Prove that $A$ lies on the radical axis of $(MIP)$ and $(NIQ)$.
b) Suppose that the orthocenter $H$ of triangle $ABC$ lies on $(I)$. Prove that there exists a line which is tangent to three circles of center $A, B, C$ and all pass through $H$.
1 reply
parmenides51
Jul 25, 2020
ilovemath0402
7 minutes ago
geometry party
pnf   0
10 minutes ago
pnf
10 minutes ago
0 replies
n Tans
MithsApprentice   17
N 30 minutes ago by AshAuktober
Source: USAMO 1998
Let $a_0,a_1,\cdots ,a_n$ be numbers from the interval $(0,\pi/2)$ such that \[ \tan (a_0-\frac{\pi}{4})+ \tan (a_1-\frac{\pi}{4})+\cdots +\tan (a_n-\frac{\pi}{4})\geq n-1.  \] Prove that \[ \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}.  \]
17 replies
MithsApprentice
Oct 9, 2005
AshAuktober
30 minutes ago
Roots of unity problem
Ferid.---.   10
N 39 minutes ago by Avron
Source: Polish MO 2019 P4
Let $n,k,l$ be positive integers.Define injective function $f$ from $\{1,2,\dots,n\}$ to itself such that $f(i)-i\in \{k,-l\}$.Prove that $k+l$ divides $n$.
10 replies
Ferid.---.
May 19, 2019
Avron
39 minutes ago
Coloring
demmy   6
N an hour ago by Kaimiaku
Source: Thailand TST 2015
What is the maximum number of squares in an $8 \times 8$ board that can be colored so that for each square in the board, at most one square adjacent to it is colored.
6 replies
demmy
Dec 2, 2023
Kaimiaku
an hour ago
Nice and easy FE on R+
sttsmet   21
N an hour ago by bo18
Source: EMC 2024 Problem 4, Seniors
Find all functions $ f: \mathbb{R}^{+} \to \mathbb{R}^{+}$ such that $f(x+yf(x)) = xf(1+y)$
for all x, y positive reals.
21 replies
sttsmet
Dec 23, 2024
bo18
an hour ago
max power of 2 that divides \lceil(1+\sqrt{3})^{2n}\rceil for pos. integer n
parmenides51   2
N an hour ago by Inspector_Maygray
Source: Gulf Mathematical Olympiad GMO 2017 p4
1 - Prove that $55 < (1+\sqrt{3})^4 < 56$ .

2 - Find the largest power of $2$ that divides $\lceil(1+\sqrt{3})^{2n}\rceil$ for the positive integer $n$
2 replies
parmenides51
Aug 23, 2019
Inspector_Maygray
an hour ago
Points in general position
AshAuktober   1
N an hour ago by Rdgm
Source: 2025 Nepal ptst p1 of 4
Shining tells Prajit a positive integer $n \ge 2025$. Prajit then tries to place n points such that no four points are concyclic and no $3$ points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?

(Prajit Adhikari, Nepal and Shining Sun, USA)
1 reply
AshAuktober
Yesterday at 2:15 PM
Rdgm
an hour ago
GMO 2017 #1
m2121   3
N an hour ago by Inspector_Maygray
Source: GMO 2017
1- Find a pair $(m,n)$ of positive integers such that $K = |2^m-3^n|$ in all of this cases :

$a) K=5$
$b) K=11$
$c) K=19$

2-Is there a pair $(m,n)$ of positive integers such that : $$|2^m-3^n| = 2017$$3-Every prime number less than $41$ can be represented in the form $|2^m-3^n|$ by taking an Appropriate pair $(m,n)$
of positive integers. Prove that the number $41$ cannot be represented in the form $|2^m-3^n|$ where $m$ and $n$ are positive integers

4-Note that $2^5+3^2=41$ . The number $53$ is the least prime number that cannot be represented as a sum or an difference of a power of $2$ and a power of $3$ . Prove that the number $53$ cannot be represented in any of the forms $2^m-3^n$ , $3^n-2^m$ , $2^m-3^n$ where $m$ and $n$ are positive integers
3 replies
m2121
Sep 28, 2017
Inspector_Maygray
an hour ago
2^a + 3^b + 1 = 6^c
togrulhamidli2011   0
an hour ago
Find all positive integers (a, b, c) such that:

\[
2^a + 3^b + 1 = 6^c
\]
0 replies
togrulhamidli2011
an hour ago
0 replies
Inequality stroke
giangtruong13   0
an hour ago
Let $a,b,c$ be real positive numbers such that: $a+b+c=abc-2$. Prove that $$\sum \frac{1}{\sqrt{ab}} \leq \frac{3}{2} $$
0 replies
1 viewing
giangtruong13
an hour ago
0 replies
Help to prove an inequality
JK1603JK   2
N 2 hours ago by whwlqkd
Source: unknown
If a,b,c\ge 0: ab+bc+ca=1 then prove \frac{a\left(b+c+2\right)}{bc+2a}+\frac{b\left(c+a+2\right)}{ca+2b}+\frac{c\left(a+b+2\right)}{ab+2c}\ge 3
* Please help me convert it to latex form. Thank you.
2 replies
JK1603JK
2 hours ago
whwlqkd
2 hours ago
Perfect Squares, Infinite Integers and Integers
steven_zhang123   0
2 hours ago
Source: China TST 2001 Quiz 5 P1
For which integer \( h \), are there infinitely many positive integers \( n \) such that \( \lfloor \sqrt{h^2 + 1} \cdot n \rfloor \) is a perfect square? (Here \( \lfloor x \rfloor \) denotes the integer part of the real number \( x \)?
0 replies
steven_zhang123
2 hours ago
0 replies
Polynomial produces perfect powers
TheUltimate123   21
N Yesterday at 8:18 PM 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
Yesterday at 8:18 PM
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
4999 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
811 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
248 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
661 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
3357 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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