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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.

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[*]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
Three circles are concurrent
Twoisaprime   23
N 3 minutes ago by Curious_Droid
Source: RMM 2025 P5
Let triangle $ABC$ be an acute triangle with $AB<AC$ and let $H$ and $O$ be its orthocenter and circumcenter, respectively. Let $\Gamma$ be the circle $BOC$. The line $AO$ and the circle of radius $AO$ centered at $A$ cross $\Gamma$ at $A’$ and $F$, respectively. Prove that $\Gamma$ , the circle on diameter $AA’$ and circle $AFH$ are concurrent.
Proposed by Romania, Radu-Andrew Lecoiu
23 replies
Twoisaprime
Feb 13, 2025
Curious_Droid
3 minutes ago
|a_i/a_j - a_k/a_l| <= C
mathwizard888   32
N 13 minutes ago by ezpotd
Source: 2016 IMO Shortlist A2
Find the smallest constant $C > 0$ for which the following statement holds: among any five positive real numbers $a_1,a_2,a_3,a_4,a_5$ (not necessarily distinct), one can always choose distinct subscripts $i,j,k,l$ such that
\[ \left| \frac{a_i}{a_j} - \frac {a_k}{a_l} \right| \le C. \]
32 replies
mathwizard888
Jul 19, 2017
ezpotd
13 minutes ago
Two lines meeting on circumcircle
Zhero   54
N 35 minutes ago by Ilikeminecraft
Source: ELMO Shortlist 2010, G4; also ELMO #6
Let $ABC$ be a triangle with circumcircle $\omega$, incenter $I$, and $A$-excenter $I_A$. Let the incircle and the $A$-excircle hit $BC$ at $D$ and $E$, respectively, and let $M$ be the midpoint of arc $BC$ without $A$. Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$. If $TI$ intersects $\omega$ again at $S$, prove that $SI_A$ and $ME$ meet on $\omega$.

Amol Aggarwal.
54 replies
Zhero
Jul 5, 2012
Ilikeminecraft
35 minutes ago
Help me this problem. Thank you
illybest   3
N an hour ago by jasperE3
Find f: R->R such that
f( xy + f(z) ) = (( xf(y) + yf(x) )/2) + z
3 replies
illybest
Today at 11:05 AM
jasperE3
an hour ago
line JK of intersection points of 2 lines passes through the midpoint of BC
parmenides51   4
N an hour ago by reni_wee
Source: Rioplatense Olympiad 2018 level 3 p4
Let $ABC$ be an acute triangle with $AC> AB$. be $\Gamma$ the circumcircle circumscribed to the triangle $ABC$ and $D$ the midpoint of the smallest arc $BC$ of this circle. Let $E$ and $F$ points of the segments $AB$ and $AC$ respectively such that $AE = AF$. Let $P \neq A$ be the second intersection point of the circumcircle circumscribed to $AEF$ with $\Gamma$. Let $G$ and $H$ be the intersections of lines $PE$ and $PF$ with $\Gamma$ other than $P$, respectively. Let $J$ and $K$ be the intersection points of lines $DG$ and $DH$ with lines $AB$ and $AC$ respectively. Show that the $JK$ line passes through the midpoint of $BC$
4 replies
parmenides51
Dec 11, 2018
reni_wee
an hour ago
AGM Problem(Turkey JBMO TST 2025)
HeshTarg   3
N an hour ago by ehuseyinyigit
Source: Turkey JBMO TST Problem 6
Given that $x, y, z > 1$ are real numbers, find the smallest possible value of the expression:
$\frac{x^3 + 1}{(y-1)(z+1)} + \frac{y^3 + 1}{(z-1)(x+1)} + \frac{z^3 + 1}{(x-1)(y+1)}$
3 replies
HeshTarg
2 hours ago
ehuseyinyigit
an hour ago
Shortest number theory you might've seen in your life
AlperenINAN   8
N 2 hours ago by HeshTarg
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
8 replies
AlperenINAN
Yesterday at 7:51 PM
HeshTarg
2 hours ago
Squeezing Between Perfect Squares and Modular Arithmetic(JBMO TST Turkey 2025)
HeshTarg   3
N 2 hours ago by BrilliantScorpion85
Source: Turkey JBMO TST Problem 4
If $p$ and $q$ are primes and $pq(p+1)(q+1)+1$ is a perfect square, prove that $pq+1$ is a perfect square.
3 replies
HeshTarg
3 hours ago
BrilliantScorpion85
2 hours ago
A bit too easy for P2(Turkey 2025 JBMO TST)
HeshTarg   0
2 hours ago
Source: Turkey 2025 JBMO TST P2
Let $n$ be a positive integer. Aslı and Zehra are playing a game on an $n \times n$ chessboard. Initially, $10n^2$ stones are placed on the squares of the board. In each move, Aslı chooses a row or a column; Zehra chooses a row or a column. The number of stones in each square of the chosen row or column must change such that the difference between the number of stones in a square with the most stones and a square with the fewest stones in that same row or column is at most 1. For which values of $n$ can Aslı guarantee that after a finite number of moves, all squares on the board will have an equal number of stones, regardless of the initial distribution?
0 replies
HeshTarg
2 hours ago
0 replies
D1030 : An inequalitie
Dattier   0
2 hours ago
Source: les dattes à Dattier
Let $0<a<b<c<d$ reals, and $n \in \mathbb N^*$.

Is it true that $a^n(b-a)+b^n(c-b)+c^n(d-c) \leq \dfrac {d^{n+1}}{n+1}$ ?
0 replies
Dattier
2 hours ago
0 replies
Long and wacky inequality
Royal_mhyasd   0
2 hours ago
Source: Me
Let $x, y, z$ be positive real numbers such that $x^2 + y^2 + z^2 = 12$. Find the minimum value of the following sum :
$$\sum_{cyc}\frac{(x^3+2y)^3}{3x^2yz - 16z - 8yz + 6x^2z}$$knowing that the denominators are positive real numbers.
0 replies
Royal_mhyasd
2 hours ago
0 replies
Vietnamese national Olympiad 2007, problem 4
hien   16
N 3 hours ago by de-Kirschbaum
Given a regular 2007-gon. Find the minimal number $k$ such that: Among every $k$ vertexes of the polygon, there always exists 4 vertexes forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.
16 replies
hien
Feb 8, 2007
de-Kirschbaum
3 hours ago
2n^2+4n-1 and 3n+4 have common powers
bin_sherlo   4
N 3 hours ago by CM1910
Source: Türkiye 2025 JBMO TST P5
Find all positive integers $n$ such that a positive integer power of $2n^2+4n-1$ equals to a positive integer power of $3n+4$.
4 replies
bin_sherlo
Yesterday at 7:13 PM
CM1910
3 hours ago
An interesting geometry
k.vasilev   19
N 3 hours ago by Ilikeminecraft
Source: All-Russian Olympiad 2019 grade 10 problem 4
Let $ABC$ be an acute-angled triangle with $AC<BC.$ A circle passes through $A$ and $B$ and crosses the segments $AC$ and $BC$ again at $A_1$ and $B_1$ respectively. The circumcircles of $A_1B_1C$ and $ABC$ meet each other at points $P$ and $C.$ The segments $AB_1$ and $A_1B$ intersect at $S.$ Let $Q$ and $R$ be the reflections of $S$ in the lines $CA$ and $CB$ respectively. Prove that the points $P,$ $Q,$ $R,$ and $C$ are concyclic.
19 replies
k.vasilev
Apr 23, 2019
Ilikeminecraft
3 hours ago
Polynomials in Z[x]
BartSimpsons   16
N Apr 22, 2025 by bin_sherlo
Source: European Mathematical Cup 2017 Problem 4
Find all polynomials $P$ with integer coefficients such that $P (0)\ne  0$ and $$P^n(m)\cdot P^m(n)$$is a square of an integer for all nonnegative integers $n, m$.

Remark: For a nonnegative integer $k$ and an integer $n$, $P^k(n)$ is defined as follows: $P^k(n) = n$ if $k = 0$ and $P^k(n)=P(P(^{k-1}(n))$ if $k >0$.

Proposed by Adrian Beker.
16 replies
BartSimpsons
Dec 27, 2017
bin_sherlo
Apr 22, 2025
Polynomials in Z[x]
G H J
G H BBookmark kLocked kLocked NReply
Source: European Mathematical Cup 2017 Problem 4
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BartSimpsons
159 posts
#1 • 4 Y
Y by son7, Kanep, Adventure10, Mango247
Find all polynomials $P$ with integer coefficients such that $P (0)\ne  0$ and $$P^n(m)\cdot P^m(n)$$is a square of an integer for all nonnegative integers $n, m$.

Remark: For a nonnegative integer $k$ and an integer $n$, $P^k(n)$ is defined as follows: $P^k(n) = n$ if $k = 0$ and $P^k(n)=P(P(^{k-1}(n))$ if $k >0$.

Proposed by Adrian Beker.
This post has been edited 1 time. Last edited by BartSimpsons, Dec 27, 2017, 12:26 PM
Reason: added source
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talkon
276 posts
#2 • 7 Y
Y by kwanglee123456, MF163, mijail, Kanep, oVlad, hakN, Adventure10
Choose a prime $p\nmid P(0)$ and plug in $(m,n)=(p,0)$ to get $p\cdot P^p(0)$ is a perfect square, so $p\mid P^p(0)$.
Consider the orbit $$0\to P(0)\to P(P(0))\to\cdots$$in $\bmod\ p$. Clearly it is periodic, and we know that its period is not $1$, but must divides $p$, so it is precisely $p$. Therefore every $p$ consecutive numbers in the orbit of $0$ forms a complete residue system $\bmod\ p$. Now consider any $t\in\mathbb Z$. We know that $P^k(0)\equiv t\pmod{p}$ for exactly one value of $0\leqslant k\leqslant p-1$. Therefore,
$$\{t,P(t),P(P(t)),\ldots,P^{p-1}(t)\}\equiv \{P^k(0),P^{k+1}(0),\ldots, P^{k+p-1}(0)\}\pmod{p}$$is also a CRS $\bmod\ p$

If $P(n)$ is not of the form $n+c$, then by Schur, $\{P(n)-n\mid n\in\mathbb Z\}$ has infinitely many prime divisors, so for some prime $q\nmid P(0)$ and $n_0\in\mathbb Z$, $q\mid P(n_0)-n_0$, contradicting the previous paragraph.
Therefore $P(n) = n+c$ for some constant $c$. The problem now reduces to finding all $c\in\mathbb Z$ such that
$$(m+cn)(n+cm)$$is always a perfect square, and this is easy:
- If $c<0$, when $(m,n)=(1,1-c)$ we have $(m+cn)(n+cm) = 1+(1-c)c<0$ which is obviously not a perfect square.
- If $c=0$ then it's obvious that $c$ doesn't work, and if $c=1$ it's obvious that $c$ works.
- Finally, if $c\geqslant 2$, choose a prime $p>c^2$ and plug in $(m,n)=(p-c,1)$ to get $(m+cn)(n+cm)=p(1+pc-c^2)$ which is divisible by $p$ but not $p^2$, and so is not a perfect square.

Therefore, the only such $P$ is $\boxed{P(n)=n+1}$
This post has been edited 1 time. Last edited by talkon, Dec 27, 2017, 4:29 PM
Reason: formatting
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kwanglee123456
50 posts
#3 • 2 Y
Y by Adventure10, Mango247
Does this problem help ?
https://artofproblemsolving.com/community/c6h6448p22328
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monkey322
3 posts
#5 • 3 Y
Y by Daniil02, Adventure10, Mango247
using equivalence relations it becomes 2010 IMO #3
hint: let f(n)=P^{n}(0) and a~b iff ab is a square.
This post has been edited 1 time. Last edited by monkey322, Mar 22, 2018, 2:12 PM
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william122
1576 posts
#7 • 1 Y
Y by Adventure10
Note that, if $m=0$, we get $P^n(0)n$ is a perfect square, and $n=1$ gives $P(0)$ is a square. Considering the sequence $a_i=P^i(0)$, note that $a_{i}-a_{i-1}|a_{i+1}-a_i$ since $P$ is an integer polynomial. This implies that $P(0)|a_i$. Also, $a_{i+1}\equiv P(0)\pmod {a_i}$, since $a_{i+1}=P(a_i)$. Now, I will inductively show that $a_i=iP(0)$. Clearly, the base case is true, so suppose $a_k=kP(0)$. Note that $a_{k+1}-a_k$ must be of the form $(1+nk)P(0)$ for $n\ge 0$. Note that as all future partial differences are divisible by this, $a_i\equiv kP(0)\pmod{(1+nk)P(0)}\forall i>k$. So, this means that $\gcd(a_i/P(0),(1+nk))=1\forall i>k$. Now, if $n>1$, consider a squarefree factor, $f>1$, of $1+nk$, and a very large squarefree multiple of $f$, $F$. Of course, we want $P^F(0)F$ to be a perfect square. However, as $P(0)$ is a square, we need $F|\frac{a_F}{P(0)}$, which is a contradiction, since $f\not\arrowvert \frac{a_F}{P(0)}$. Therefore, we must have $n=0$ and $a_{k+1}=(k+1)P(0)$, as desired. Now, as $P(x)-(x+P(0))$ has infinite roots, located at all $a_i$, we must have that $P(x)=x+P(0)$.

Denoting $P(0)$ as $c^2$, $(n+mc^2)(m+nc^2)$ is always a perfect square. Now, if $c>1$, choose $(m,n)$ with the following properties: $n=c^2n'$, $m+n'$ is squarefree, $\gcd(m,n)=1$, and $\gcd(m+n',c^4-1)=1$. Then, this becomes $c^2(m+n')(m+n'c^4)$. So, we must have $m+n'|m+n'c^4$. However, $\gcd(m+n',m+n'c^4)=\gcd(m+n',n'(c^4-1))=1$, which is a contradiction. Therefore, $c=1$ and $P(x)=x+1$ is the only solution.
This post has been edited 2 times. Last edited by william122, Sep 8, 2019, 7:52 PM
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math_pi_rate
1218 posts
#8 • 1 Y
Y by amar_04
Nice! Here's my solution (which is heavily dependent on induction): Define the sequence $(x_j)_{j \geq 0}$ with $x_0=0$ and $x_j=P(x_{j-1})$ for all $j \in \mathbb{N}$. Then the given condition states that $nx_n$ is a perfect square for all $n \in \mathbb{N}$. We first prove some claims which will help us strengthen our foothold over the sequence $(x_j)_{j \geq 0}$. Throughout we'll be using the deep fact that $a-b \mid P^k(a)-P^k(b)$ for all non-negative integers $a,b,k$.

CLAIM 1 $x_i \mid x_{iz}$ for all $i,z \in \mathbb{N}$.

Proof of Claim 1

CLAIM 2 For all $n \in \mathbb{N}$, we have $n \mid x_n$.

Proof of Claim 2

CLAIM 3 Let $x_1=P(0)=A^2$ for some $A \in \mathbb{N}$ (Since $P(0)$ is a non-zero square). Then $x_i-x_{i-1}=A^2$ for all $i \in \mathbb{N}$.

Proof of Claim 3

Return to the problem at hand. By Claim 3, we get that the equation $P(x)-x-A^2=0$ has infinitely many roots, namely $x_0,x_1,x_2 \dots$ (which are all distinct, since our claim ascertains that this sequence is an increasing sequence). Thus, we must have $P(x)=x+A^2$. Then $P^m(x)=x+mA^2$ for all $m \geq 0$. Suppose $A>1$, and choose a prime $p \mid A$. Then putting $(m,n)=(A^2,p-1)$ in the given conditions, we get that $pA^2(A^4+p-1)$ is a perfect square. But, $p \mid A$ gives that $\gcd(A^4+p-1,p)=1$. Then $$\nu_p(pA^2(A^4+p-1))=\nu_p(pA^2)=2\nu_p(A)+1$$which means that an odd power of $p$ divides $pA^2(A^4+p-1)$, contradicting the fact that this is a square. Thus, we must have $A=1$, and so we get the desired polynomial as $P(x)=x+1$. This clearly works, and so we're done. $\blacksquare$
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Physicsknight
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#9
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Consider $P\ne\text{constant}.$ Since $P\equiv C$ is easy to notice.
If
\begin{align*}
n=0\to mp^m(0)&=x_m^{2}\quad\forall m\left(x_m\in\mathbb N\right)\\
\implies p\mid p^p(0)&=x_p^2\quad\forall p\in\text{(prime)}\\
\implies p\mid p^p(0)\quad\forall p\in\text{(prime)}\\
\end{align*}Take $p\nmid p(0)$
$\{0,p(0),\cdots,p^p(0)\}\implies\exists 0\leqslant i<j\leqslant p, p^i(0)\equiv p^j(0)[p]$ by $\text{Dirichlet's principle}$
Let $a_k=p^k(0)\implies\exists  T\in\mathbb Z_p\leqslant p: a_{k+T}\equiv a_k[\pmod {p}]\forall k\ge 0$
Since $p\mid p^p(0)\implies T\mid P\implies\begin{cases} T&=1\\ T&=P\end{cases}$
$T=1\implies a_i\equiv a_{i+1}\equiv\hdots\equiv a_p\equiv a_{p+1}[p]\implies a_{p+1}=P(a_p)\equiv P(0)[\pmod {p}]\implies p\mid p(0)(\text{constant})\implies T=P$
For any $a\in\mathbb Z, P^k(0)\equiv a\pmod{p}$ for exactly one $k, 0\le k<P\implies\{a,P(a),P(P(a)),\hdots,P(a)\}\equiv\{P^k(0),P^{k+1}(0),\hdots,P^{k+p-1}(0)\}(\pmod {p}).$
This is a complete residue system $\pmod{p}.$
If $P(x)-x=Q(x)$ is a non-constant polynomial.
By $\text{Schur's theorem}$ $\{P(n)-n: n\in\mathbb Z^+\}$ has infinitely many prime.
Take $Q\mid P(0),n_0\in\mathbb Z, Q\mid P(n_0)-n_0.$
Note, $Q$ is prime this contradict to the above conditions when $q=p.$
Therefore $P(x)=x+c$ for some constant $c\in\mathbb Z.$
$(m+cn)(n+cm)$ is a square $\forall m,n\in\mathbb N.$
case 1
$c<0: m=1, n=1-c\implies(m+cn)(m+cm)=1+c-c^2<0$
case 2
$c=0: (m+cn)(n+cm)=mn$ not possible.
case 3
$c=1$ is easy to check
case 4
$\exists p\in\mathbb P : p >c^2+1$
$m=p-c,n=1\implies (m+cn)(n+cm)=p(1+pc-c^2): p\nmid p^2\implies\text{Contradiction}$
Hence, $c=1\implies P(x)\equiv x+1$
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IndoMathXdZ
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#10 • 4 Y
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What a nice problem, took the wrong approach several times before trying "arrow" :what?:
From the problem, we have $p \cdot P^p(0)$ is a square for any prime number $p$, forcing $p \mid P^p(0)$ for all primes $p$.

Take a prime number $p \nmid P(0)$.
Claim 01. We have $\{ P(0), P^2(0), \dots, P^p(0) \}$ being complete residues modulo $p$.
Proof. Consider
\[ 0 \mapsto P(0) \mapsto P(P(0)) \mapsto \dots \mapsto P^p(0) \]Let $a$ be the smallest positive integer such that $P^a(0) \equiv 0 \ ( \text{mod} \ p)$. We must have $a \mid p$ by the definition of $a$. Thus, $a = 1$ or $a = p$. However, we have taken $p \nmid P(0)$, so $a = p$. If there exists two positive integers $1 \le a < b \le p - 1$ such that $P^a(0) \equiv P^b(0) \ (\text{mod} \ p)$. This forces $P^{p} (0) \equiv P^{p - b + a} (0) \ (\text{mod} \ p)$, forcing $k = p - b + a < p$ to satisfy $P^k(0) \equiv 0 \ (\text{mod} \ p)$, a contradiction. Thus, all of them must be distinct residues, resulting in a complete residue modulo $p$.
Claim 02. This can be extended slightly to $\{ P^k(0), \dots, P^{k + p - 1}(0) \}$ is a complete residue modulo $p$.
Proof. Since $P^p(0) \equiv 0 \ (\text{mod} \ p)$, then since $P \in \mathbb{Z}[x]$, then $P^{a + p}(0) = P^a (P^p (0)) \equiv P^a (0) \ (\text{mod} \ p)$. Now, just notice that $\{ k, k + 1, \dots, k + p - 1 \} \equiv \{ 1, 2, \dots, p \}$ in $\mathbb{Z}_p$, and we are done.
Why do we consider this?
Take any integer $t \in \mathbb{Z}$.
Claim 03. $\{ t, P(t), \dots, P^{p - 1}(t) \}$ forms a complete residue modulo $p$.
Proof. Since $\{ P(0), P^2(0), \dots, P^p(0) \}$ forms a complete residue modulo $p$. Then, we can take $1 \le k \le p - 1$ such that $P^k(0) \equiv t \ (\text{mod} \ p)$. Thus,
\[ \{ t, P(t), \dots, P^{p - 1}(t) \} \equiv \{ P^k(0), P^{k + 1}(0), \dots, P^{k + p - 1}(0) \} (\text{mod} \ p) \]
Main Claim. $P(n) - n$ is a constant.
Proof. Now, we know that for any prime number $p \nmid P(0)$ and any integer $t \in \mathbb{Z}$, then $p \nmid P(t) - t$ (since $\{ t, P(t) \}$ are distinct residues modulo $p$ for any $t \in \mathbb{Z}$.)
Therefore, we conclude that if $\mathcal{P}$ is the set of primes dividing $P(n) - n$ for all $n \in \mathbb{N}$, and $\mathcal{Q}$ is the set of primes dividing $P(0)$, then
\[ \mathcal{P} \subseteq \mathcal{Q} \Rightarrow |\mathcal{P}| \le |\mathcal{Q}| \]which proves that $|\mathcal{P}|$ is finite.
However, by Schur Theorem, unless $P(n) - n$ is a constant, $|\mathcal{P}|$ is infinite, which is a contradiction.
Thus, we conclude that $P(n) = n + b$ for some integer $b$.
We have $P^m(n) \cdot P^n(m) = (n + mb)(m + nb)$ is a square for all $n,m \in \mathbb{N}_0$.
Take $n = 0, m = 1$ gives us $b$ being a perfect square.
Take $m = 4, n = 1$, and we have $4b^2 + 17b + 4$ being a square. However, since
\[ (2b + 2)^2 = 4b^2 + 8b + 4 < 4b^2 + 17b + 4 < 4b^2 + 20b + 25 = (2b+5)^2 \]We have $4b^2 + 17b + 4 = (2b + 3)^2 \rightarrow b = 1$ or $4b^2 + 17b + 4 = (2b + 4)^2 \rightarrow b = 12$, which is not a square.
Therefore, the only solution is $\boxed{P(x) = x + 1}$ which works since
\[ P^m(n) \cdot P^n(m) = (m + n)^2 \]is a square for all nonnegative $m,n$.
Comments: 3 hour play and pitfalls
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MatBoy-123
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#11
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BartSimpsons wrote:
Find all polynomials $P$ with integer coefficients such that $P (0)\ne  0$ and $$P^n(m)\cdot P^m(n)$$is a square of an integer for all nonnegative integers $n, m$.

Remark: For a nonnegative integer $k$ and an integer $n$, $P^k(n)$ is defined as follows: $P^k(n) = n$ if $k = 0$ and $P^k(n)=P(P(^{k-1}(n))$ if $k >0$.

Proposed by Adrian Beker.

Nice Problem !!
First note that $p \mid P^{p} (0)$ , suppose there exist a $x <p$ such that $ P^{x} (0) \equiv 0 mod p$ , as $p-x >0$ , choose a minimum such $x$.

Now ,$P^{x}(0) = P^{p-x}(P^{p}(0)) \equiv P^{p-x}(0) \equiv 0  mod p$ , so in general $P^{p-kx}(0) \equiv 0 modp$.

But we can choose $k > \frac{p}{x} - 1 > 0$ , so $p-kx <x$ , a contradiction to the minimality of $x$. , so using this we get that $P^t (0) \equiv 0 modp$ if and only if $t  \equiv 0 mod p$ , for any prime $p$, so $P^{1} (0) = P(0) = 1$.

But now $P^{p}(0) = P^{p-1}(1)$ , but from the statement $\nu_p(P^{p}(0))$ is odd , and as $ P^{p-1}(1). P(p-1)$ is perfect square this implies $P(p-1) \equiv P(-1) \equiv 0 mod p$ , but this is true for infinitely many primes $p$ , so $P(-1) = 0$ .

So we can write $P(x) = (x+1)^{r}. g(x)$ where $g(-1) \neq 0$ , suppose $g$ is not constant $P^{p}(0)= P(P^{p-1}(0)) = (P^{p-1}(0) +1)^{r} .g( P^{p-1}(0)) = p^{odd}$ , so $P^{p-1}(0)  = p^{t} - 1$ , for some natural $t$ , and $ g(p^{t} - 1) \equiv 0 mod p$ , but this is true for infinite $p,s$ , so $g(-1) = 0$ , a contradiction , so $g$ is indeed constant , this implies $P(x) =c(x+1)^{r}$ , but as $P(0) = c = 1$ , so $P(x) = (x+1)^{r}$ , from here we can easily prove $r = 1$ , (I will post it later as it just involved putting some values , to prove that $r = 1$) , Hence $P \equiv x+1$ $\blacksquare$
This post has been edited 1 time. Last edited by MatBoy-123, Sep 22, 2021, 12:14 PM
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guptaamitu1
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Answer is $P(x) \equiv x+1$, which clearly works.

We will be showing that cycle of each negative number contains $0$ which would then be enough to imply $P(x) \equiv x+1$.
Fix any $r \in \mathbb Z_{>0}$, to prove $0 \in -r,f(-r),f(f(-r)),\ldots$.
$$m=0 \implies P^n(0) \cdot n \in S \qquad \qquad (1) $$Claim: $0,P(0),P(P(0)),\ldots$ is not a cycle.

Proof: Assume on the contrary the sequence is eventually periodic. Pick $k,c$ (with $c \ge 1$) such that $P^k(0) = x \ne 0$ and $$ x = P^k(0) = P^{k+c} (0) = P^{k+2c}(0) = \cdots $$Then $(1)$ gives
$$ x \cdot k, x \cdot (k+c), x \cdot (k+2c), \ldots \in S $$But this is an easy contradiction, say by using the fact that $S$ has density $0$ or making $\nu_p(k+yc)$ odd for prime $p \nmid x$. $\square$

Our Claim particularly gives $P^n(0) \ne 0 ~ \forall ~ n \ge 1$.
$$ m = P^r(0) \implies  P^{n+r}(0) \cdot P^{P^r(0)}(n) \in S  \qquad \qquad (2)$$Combining $(1),(2)$ along with our Claim gives
$$ Q(n) := (n+r) \cdot P^{P^r(0)} (n) \in S ~ ~ \forall ~ n \ge 0 \qquad \qquad (3)$$This forces $Q(n)$ be a square of polynomial with integer coefficients (see the proof in post #26 to Iran TST 2008/8). Then
$$ n+r \mid Q(n) \implies (n+r)^2 \mid Q(n) \implies n+r \implies P^{P^r (0)} (n) \implies P^{P^r(0)}(-r) =0 $$So we have proven $0$ is in the cycle of $-r$. We are ready to finish.

Claim $P$ must be linear and its leading coefficient must be $\pm 1$.

Proof: Assume contrary. Choose a constant $\mu > 0$ such that $|P(x)| \ge |x|$ whenever $|x| \ge \mu$. But then cycle of $-\mu$ does not contain any $0$, which is a contradiction. $\square$

Now write $P(x) \equiv x+c$ with $c \in \mathbb Z \setminus \{0\}$. Again, exploiting $0$ is in cycle of each negative integer we obtain $c=1$, as desired. $\blacksquare$
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megarnie
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#13
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The only solution is $\boxed{P(x) = x + 1}$, which works. Now we prove that nothing else works.

Setting $m = 0$, we have \[P^n(0) \cdot n\]is a perfect square for nonnegative integers $n$. Therefore $p\mid P^p(0) $ for any prime $p$.

For any prime $p$ not dividing $P(0)$, the period of $0,P(0), P(P(0)), \ldots, $ modulo $p$ divides $p$, so it must be equal to $p$. Therefore, $\{0,P(0), P(P(0)), \ldots, P^{p-1}(0) \}$ forms a complete residue set modulo $p$. This in fact implies that $p$ cannot divide $P(n) - n$ for any integer $n$ (if $P(n)\equiv n\pmod p$, take $P^k(0) \equiv n\pmod p$ for $k\le p-1$). Since $P(0)$ has finitely many prime divisors, $P(n) - n$ is constant by Schur. Let $P(n) = n+c$. We have \[(m + nc)(n + cm)\]is a perfect square. From setting $m = 0$, we see that $c$ is a perfect square. From setting $m = 1$ and $n = 4$, we have that $(4c + 1)(c + 4) = 4c^2 + 17c + 4$ is a perfect square. It is clearly strictly in between $(2c + 2)^2$ and $(2c + 5)^2$, so it must be equal to $(2c+3)^2$ or $(2c+4)^2$. This gives the only possibilities $c = 1$ and $12$, but $12$ isn't a perfect square, so $c = 1$, as desired.
This post has been edited 4 times. Last edited by megarnie, Aug 10, 2023, 11:37 AM
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atdaotlohbh
186 posts
#14
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Substitute $m=0$:
$P^n(0)n$ is a perfect square. Note that if $P$ is constant, then $cn$ is a perfect square, which is false when $n$ is a big prime

Now let $p$ be a prime. Because $P^p(0) \vdots p$, the length of the cycle of $0,P(0),P^2(0),\ldots$ mod $p$ should divide $p$, so it is $1$ or $p$. If it is $1$, then $p | P(0)$, so only finitely many $p$. For all other $p$ all residues $P(0),P^2(0),\ldots ,P^p(0)$ are different, and thus $Q(x)=P(x+1)-P(x) \not \vdots p$ for all $x$. Because $Q$ is divisible only by finitely many primes, it should be constant, and so $P$ is linear, say $P(x)=ax+b$. As $P(0)$ is a perfect square, $b=t^2$. Now let $P^k(0)=t^2a_k$. Then $a_1=1$ and $a_{k+1}=a*a_k+1$. And $k*a_k$ is a perfect square for all $k$. Suppose $q | a$, where $q$ is a prime. Then $a_q \equiv 1$ (mod $q$), and so $v_q(qa_q)=1$, contradicting the fact that it is perfect square. So $a$ is either one or negative one, as our values should be positive, it is $1$.
Now the original equation tells us that $(m+nt^2)(n+mt^2)$ is always a perfect square. Take $m=p-nt^2$, where $p$ is a large prime. Then $n-nt^4 \vdots p$ for any large $p$, and thus $t^2=1$. So $P(x)=x+1$, which works because $P^n(m) \dot P^m(n)=(m+n)(m+n)=(m+n)^2$
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MathLuis
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Clearly no $P$ constant works so suppose $\text{deg} P \ge 1$, now let $P(m,n)$ the assertion.
$P(p,0)$ gives $p \mid P^p(0)$ for all $p$ primes. Now we focus on all $p>\text{max}(P(0), 1434)$ and consider the orbit $0 \to P(0) \to P(P(0)) \to \cdots$ in $\pmod p$.
It's clear that since we found a cycle we can consider minimal $k$ for which $p \mid P^k(0)$, but this means $k \mid p$ so since $p>P(0)$ we end up having $k=p$ and as a result $p$ is the lenght of the minimal cycle, now if there was a sub cycle in the graph we would never reach $0$ therefore $0,P(0), \cdots P^{p-1}(0)$ is a complete residue system $\pmod p$.
Select $P^{k}(0) \equiv \ell \pmod p$ and now check that due to the cycle $\ell, P(\ell), \cdots, P^{p-1}(\ell)$ is also a complete residue system $\pmod p$. Suppose FTSOC $P(x)-x$ isn't constant, then by Schur theorem we get that for some $p>\text{max}(P(0), 1434)$ we can find $x_0$ such that $p \mid P(x_0)-x_0$ but this means the orbit of $x_0$ has a cycle of lenght $1$, contradicting minimality of the cycle with lenght $p$.
Therefore $P(x)=x+c$ for all integers $x$, now $P(p,0)$ gives $c$ is a perfect square and the problem becomes $(m+cn)(n+cm)$ perfect square for all non-negative integers $m,n$. Trivially $c=0$ fails, and if $c<0$ then fix $m$ and set $n>-cm$ and $m+cn<0$ for large $n$ which gives that something $<0$ is a percect square, contradiction!. Therefore $c$ is a positive integer, now assume FTSOC $c \ne 1$.
Note that by dirchlet we can get some large prime $p=m+cn$ by fixing some $m$ such that $\gcd(m,c)=1$ and setting very large $n$, now we have $p \mid n+cm$ which means $n+cm \ge m+cn$ but clearly since $c \ge 2$ we get RHS is larger than LHS at some point, contradiction!.
Therefore $c=1$ which means $(m+n)(m+n)$ is a perfect square. This is completely true, therefore all polynomials $P \in \mathbb Z[x]$ that work are $P(x)=x+1$ thus we are done :cool:.
This post has been edited 2 times. Last edited by MathLuis, Jul 12, 2024, 1:26 AM
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YaoAOPS
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#16
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Claim: $P$ has positive leading coefficient.
Proof. Suppose not, then for sufficiently large $|x|$ it follows that $|P(x)| > |x|$, and that $x$ and $P(x)$ have negative signs.
Taking sufficiently large $n$ and $m$ of opposite parity $\pmod{2}$ gives the result. $\blacksquare$

Claim: $\nu_p(P^n(m))$ is unbounded for all primes $p$ and nonnegative $n, m$.
Proof. FTSOC suppose that $\nu_p(P^n(0))$ is bounded above by some integer $N$.
It then follows that if $a \equiv b \pmod{p^N}$, that then $P(a) \equiv P(b) \pmod{p^N}$.
Note that $P^n(0) \cdot n$ is always a perfect square. Consider the cyclic chain taken $\pmod{p^N}$ of \[ 0, P(m), P(P(m)), \dots \]It follows that $\nu_p$ of the chain is eventually cyclic with some period $T$.
However, this implies that $P^{kT}(0)$ and $P^{kpT}(0)$ have the same $\nu_p$ for sufficiently large $k$, contradiction. $\blacksquare$

Claim: $P(x) = x + 1$.
Proof. Note that $\gcd(P(x), x) \mid P(0)$ for all $x$. Then take sufficiently large $x$ such that $\gcd(x, P(0)) = 1$ and $P(x) - x > 0$. If any prime $p \mid P(x) - x$, then $p \nmid x$, so $\nu_p(P^n(x))$ is always $0$, contradiction. $\blacksquare$
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Saucepan_man02
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#17
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Let $a_n = P^(n)(0)$. Plugging $m=0$ gives $na_n$ to be a perfect square for all $n \ge 0$.

Claim: For large enough prime $p$, we have $a_n \pmod p$ to be purely periodic with period $p$.
Proof: Let $p > \max(0, |P(0)|)$ be a large enough prime.
Note that sequence $a_n \pmod p$ is eventually periodic, as $a_j \equiv a_i \pmod p$ for some $i, j $ (due to PHP). Thus, letting $t = j-i$ implies $a_{k+st} \equiv a_k$ for all $k \ge i$. Let $T$ denote the period (minimum) of sequence $a_n$.
Note that: $p a_p$ is a perfect square, which implies $p | a_p$. Notice that: $$a_{n+p} = P^{(n)}(a_p) \equiv P^{(n)}(0) \equiv a_n \pmod p$$which implies $a_n$ is purely periodic. Thus: $T|p$ which implies $T = 1$ or $T=p$.

If $T=1$, then $a_n \pmod p$ will be constant eventually. Notice that: $a_{np} \equiv 0 \pmod p$ which implies that, it should equal to $0$ eventually. But: $P(a_p) \equiv P(0) \pmod p$ and $P(a_p) \equiv a_p \equiv 0 \pmod p$, thus contradiction that $p|P(0)$.
Therefore $T=p$.

Claim: $\deg(P) \le 1$
Proof:FTSOC, assume $\deg(P) \ge 2$.
Thus, note that we must have: $$\{ P(0), P(1), \cdots, P(p-1) \} = \{ 0, 1, \cdots, p-1 \}$$since $a_n$ is purely periodic modulo $p$.
Let $Q(x) = P(x+1)-P(x)$ be of $\deg P -1$ polynomial with $b_n = Q(n)$. Thus, due to Schurs theorem, infinitely many primes $q$ divide $b_n$ which implies $P(n+1) \equiv P(n) \pmod q$. Taking $q$ to be a prime which is much larger than $p$, it shows that $P(n+1) = P(n)$ which contradicts that $\{ P(0), P(1), \cdots, P(p-1) \} = \{ 0, 1, \cdots, p-1 \}$.

Therefore, $P(x)=ax+b$ for some $a, b$ integers.
Consider $G(x)=P(x)-x$. Then, by Schur's theorem, $P(n) \equiv n \pmod p$ for infinitely many primes $p$ but this contradicts that $a_n \pmod p$ has period of length $p$, for large enough primes $p$, unless $G$ is constant.

Thus: $P(x) = x+c$ and plugging back (followed by a NT-bash), we conclude with $c=1$.
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quantam13
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Sketch
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Reason: missed one detail
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bin_sherlo
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This is a similar one. Note that $P$ is non-constant since $qP^q(0)\in \mathbb{Z}^2$ and $P(0)\neq 0$ where $q|P^q(0)$. Answer is $P(x)=x+1$ which indeed fits.
If $r|P^d(0)$ is the smallest positive integer $d$ satisfying the condition and $d<q$, then $r|P^{d+k}(0)-P^k(0)$ and since $r|P^r(0)$ we get $d|r$. If $r\not | P(0)$, then $d=q$ which implies $\{P(0),P^2(0),\dots,P^{r-1}(0)P^r(0)\}$ is the complete residue system on modulo $r$.
Let $Q(x)=P(x)-x$. Suppose that $Q$ is non-constant. By Schur, we can pick a sufficiently large prime $p|Q(x)$ for some sufficiently large posiitve integer $x$. Since $p\not |P^n(0)$ for $p\not | n$ we have
\[1=(\frac{P^x(m)P^m(x)}{p})=(\frac{P^x(m)x}{p})\implies (\frac{x}{p})=(\frac{P^x(m)}{p})\overset{m=P^{n-x}(0)}{\implies} (\frac{x}{p})=(\frac{P^n(0)}{p})=(\frac{n}{p})\]However, this is impossible since we can choose $n$ such that $(\frac{n}{p})\neq (\frac{x}{p})$ where $p$ is sufficiently large. Thus, $Q$ is constant which implies $P(x)=x+c$. We see that $(m+nc)(n+mc)\in \mathbb{Z}^2$. Pick $n=q$ to get that $(\frac{c}{q})=1$ for all sufficiently large primes. This yields $c$ is a perfect square. Let $c=t^2$. If $n=t^2$, then $(m+t^4)(t^2+mt^2)\in \mathbb{Z}^2$ or $(m+t^4)(m+1)\in \mathbb{Z}^2$. If $t^4\neq 1$, we pick a sufficiently large prime $p^2|m+1-p$ where $p\not | m+t^4$ but we observe that this is impossible. Thus, $t^2=1$ or $P(x)=x+1$ as desired.$\blacksquare$
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