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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
interesting geometry config (3/3)
Royal_mhyasd   0
29 minutes ago
Let $\triangle ABC$ be an acute triangle, $H$ its orthocenter and $E$ the center of its nine point circle. Let $P$ be a point on the parallel through $C$ to $AB$ such that $\angle CPH = |\angle BAC-\angle ABC|$ and $P$ and $A$ are on different sides of $BC$ and $Q$ a point on the parallel through $B$ to $AC$ such that $\angle BQH = |\angle BAC - \angle ACB|$ and $C$ and $Q$ are on different sides of $AB$. If $B'$ and $C'$ are the reflections of $H$ over $AC$ and $AB$ respectively, $S$ and $T$ are the intersections of $B'Q$ and $C'P$ respectively with the circumcircle of $\triangle ABC$, prove that the intersection of lines $CT$ and $BS$ lies on $HE$.

final problem for this "points on parallels forming strange angles with the orthocenter" config, for now. personally i think its pretty cool :D
0 replies
Royal_mhyasd
29 minutes ago
0 replies
A perverse one
darij grinberg   7
N an hour ago by ezpotd
Source: German TST 2004, IMO ShortList 2003, number problem 2
Each positive integer $a$ undergoes the following procedure in order to obtain the number $d = d\left(a\right)$:

(i) move the last digit of $a$ to the first position to obtain the numb er $b$;
(ii) square $b$ to obtain the number $c$;
(iii) move the first digit of $c$ to the end to obtain the number $d$.

(All the numbers in the problem are considered to be represented in base $10$.) For example, for $a=2003$, we get $b=3200$, $c=10240000$, and $d = 02400001 = 2400001 = d(2003)$.)

Find all numbers $a$ for which $d\left( a\right) =a^2$.

Proposed by Zoran Sunic, USA
7 replies
darij grinberg
May 18, 2004
ezpotd
an hour ago
a_ia_{i+1}a_{i+2}a_{i+3}=i(mod p)
Aryan-23   23
N an hour ago by Jupiterballs
Source: IMO SL 2020 N1
Given a positive integer $k$ show that there exists a prime $p$ such that one can choose distinct integers $a_1,a_2\cdots, a_{k+3} \in \{1, 2, \cdots ,p-1\}$ such that p divides $a_ia_{i+1}a_{i+2}a_{i+3}-i$ for all $i= 1, 2, \cdots, k$.


South Africa
23 replies
Aryan-23
Jul 20, 2021
Jupiterballs
an hour ago
Eventually constant sequence with condition
PerfectPlayer   4
N an hour ago by kujyi
Source: Turkey TST 2025 Day 3 P8
A positive real number sequence $a_1, a_2, a_3,\dots $ and a positive integer \(s\) is given.
Let $f_n(0) = \frac{a_n+\dots+a_1}{n}$ and for each $0<k<n$
\[f_n(k)=\frac{a_n+\dots+a_{k+1}}{n-k}-\frac{a_k+\dots+a_1}{k}\]Then for every integer $n\geq s,$ the condition
\[a_{n+1}=\max_{0\leq k<n}(f_n(k))\]is satisfied. Prove that this sequence must be eventually constant.
4 replies
PerfectPlayer
Mar 18, 2025
kujyi
an hour ago
Pentagon with given diameter, ratio desired
bin_sherlo   3
N an hour ago by tugra_ozbey_eratli
Source: Türkiye 2025 JBMO TST P7
$ABCDE$ is a pentagon whose vertices lie on circle $\omega$ where $\angle DAB=90^{\circ}$. Let $EB$ and $AC$ intersect at $F$, $EC$ meet $BD$ at $G$. $M$ is the midpoint of arc $AB$ on $\omega$, not containing $C$. If $FG\parallel DE\parallel CM$ holds, then what is the value of $\frac{|GE|}{|GD|}$?
3 replies
bin_sherlo
May 11, 2025
tugra_ozbey_eratli
an hour ago
Number Theory
fasttrust_12-mn   15
N an hour ago by Pal702004
Source: Pan African Mathematics Olympiad P1
Find all positive intgers $a,b$ and $c$ such that $\frac{a+b}{a+c}=\frac{b+c}{b+a}$ and $ab+bc+ca$ is a prime number
15 replies
fasttrust_12-mn
Aug 15, 2024
Pal702004
an hour ago
pairs (m, n) such that a fractional expression is an integer
cielblue   2
N an hour ago by cielblue
Find all pairs $(m,\ n)$ of positive integers such that $\frac{m^3-mn+1}{m^2+mn+2}$ is an integer.
2 replies
cielblue
May 24, 2025
cielblue
an hour ago
interesting geo config (2/3)
Royal_mhyasd   3
N 2 hours ago by Royal_mhyasd
Source: own
Let $\triangle ABC$ be an acute triangle and $H$ its orthocenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = |\angle ABC-\angle ACB|$. Define $Q$ and $R$ as points on the parallels through $B$ to $AC$ and through $C$ to $AB$ similarly. If $P,Q,R$ are positioned around the sides of $\triangle ABC$ as in the given configuration, prove that $P,Q,R$ are collinear.
3 replies
Royal_mhyasd
Yesterday at 11:36 PM
Royal_mhyasd
2 hours ago
interesting geo config (1\3)
Royal_mhyasd   2
N 2 hours ago by Royal_mhyasd
Source: own
Let $\triangle ABC$ be an acute triangle with $AC > AB$, $H$ its orthocenter and $O$ it's circumcenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = \angle ABC - \angle ACB$ and $P$ and $C$ are on different sides of $AB$. Denote by $S$ the intersection of the circumcircle of $\triangle ABC$ and $PA'$, where $A'$ is the reflection of $H$ over $BC$, $M$ the midpoint of $PH$, $Q$ the intersection of $OA$ and the parallel through $M$ to $AS$, $R$ the intersection of $MS$ and the perpendicular through $O$ to $PS$ and $N$ a point on $AS$ such that $NT \parallel PS$, where $T$ is the midpoint of $HS$. Prove that $Q, N, R$ lie on a line.

fiy it's 2am and i'm bored so i decided to look further into this interesting config that i had already made some observations on, maybe this problem is trivial from some theorem so if that's the case then i'm sorry lol :P i'll probably post 2 more problems related to it soon, i'd say they're easier than this though
2 replies
Royal_mhyasd
Yesterday at 11:18 PM
Royal_mhyasd
2 hours ago
Find all sequences satisfying two conditions
orl   35
N 2 hours ago by wangyanliluke
Source: IMO Shortlist 2007, C1, AIMO 2008, TST 1, P1
Let $ n > 1$ be an integer. Find all sequences $ a_1, a_2, \ldots a_{n^2 + n}$ satisfying the following conditions:
\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]

\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]
Author: Dusan Dukic, Serbia
35 replies
orl
Jul 13, 2008
wangyanliluke
2 hours ago
Gcd of N and its coprime pair sum
EeEeRUT   20
N 2 hours ago by Adywastaken
Source: EGMO 2025 P1
For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$for all $1 \leqslant i \leqslant m-1$

Here $\gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$. Integers $a$ and $b$ are coprime if $\gcd(a, b) = 1$.

Proposed by Paulius Aleknavičius, Lithuania
20 replies
EeEeRUT
Apr 16, 2025
Adywastaken
2 hours ago
geometry problem with many circumcircles
Melid   0
2 hours ago
Source: own
In scalene triangle $ABC$, which doesn't have right angle, let $O$ be its circumcenter. Circle $BOC$ intersects $AB$ and $AC$ at $A_{1}$ and $A_{2}$ for the second time, respectively. Similarly, circle $COA$ intersects $BC$ and $BA$ at $B_{1}$ and $B_{2}$, and circle $AOB$ intersects $CA$ and $CB$ at $C_{1}$ and $C_{2}$ for the second time, respectively. Let $O_{1}$ and $O_{2}$ be circumcenters of triangle $A_{1}B_{1}C_{1}$ and $A_{2}B_{2}C_{2}$, respectively. Prove that $O, O_{1}, O_{2}$ are collinear.
0 replies
Melid
2 hours ago
0 replies
Rootiful sets
InternetPerson10   38
N 2 hours ago by cursed_tangent1434
Source: IMO 2019 SL N3
We say that a set $S$ of integers is rootiful if, for any positive integer $n$ and any $a_0, a_1, \cdots, a_n \in S$, all integer roots of the polynomial $a_0+a_1x+\cdots+a_nx^n$ are also in $S$. Find all rootiful sets of integers that contain all numbers of the form $2^a - 2^b$ for positive integers $a$ and $b$.
38 replies
InternetPerson10
Sep 22, 2020
cursed_tangent1434
2 hours ago
weird conditions in geo
Davdav1232   2
N 3 hours ago by teoira
Source: Israel TST 7 2025 p1
Let \( \triangle ABC \) be an isosceles triangle with \( AB = AC \). Let \( D \) be a point on \( AC \). Let \( L \) be a point inside the triangle such that \( \angle CLD = 90^\circ \) and
\[
CL \cdot BD = BL \cdot CD.
\]Prove that the circumcenter of triangle \( \triangle BDL \) lies on line \( AB \).
2 replies
Davdav1232
May 8, 2025
teoira
3 hours ago
IMO Shortlist 2009 - Problem N6
April   15
N Apr 25, 2025 by luutrongphuc
Let $k$ be a positive integer. Show that if there exists a sequence $a_0,a_1,\ldots$ of integers satisfying the condition \[a_n=\frac{a_{n-1}+n^k}{n}\text{ for all } n\geq 1,\]then $k-2$ is divisible by $3$.

Proposed by Okan Tekman, Turkey
15 replies
April
Jul 5, 2010
luutrongphuc
Apr 25, 2025
IMO Shortlist 2009 - Problem N6
G H J
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April
1270 posts
#1 • 2 Y
Y by Mathuzb, Adventure10
Let $k$ be a positive integer. Show that if there exists a sequence $a_0,a_1,\ldots$ of integers satisfying the condition \[a_n=\frac{a_{n-1}+n^k}{n}\text{ for all } n\geq 1,\]then $k-2$ is divisible by $3$.

Proposed by Okan Tekman, Turkey
This post has been edited 1 time. Last edited by djmathman, Dec 13, 2015, 5:19 PM
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daniel73
253 posts
#2 • 4 Y
Y by mad, Aubery, Adventure10, Mango247
This may help in finding a solution:

Note that we can always find a polynomial $P_k(x)$ of degree $k-1$ and with integer coefficients such that $xP_k(x)=P_k(x-1)+x^k+r_k$, where $r_k$ is some residual constant to be determined; just express $P_k(x)=c_{k-1}x^{k-1}+c_{k-2}x^{k-2}+\dots+c_1x+c_0$, and work the coefficients down from $c_{k-1}$ to $c_0$, finding the coefficient $c_m$ by comparing the terms of degree $m+1$ at both sides of the equation. Clearly $c_{k-1}=1$, and all other coefficients are a linear combination with integral weights of higher-order coefficients, hence the polynomial coefficients are all integral. The residual constant $r_k$ is the one that makes the constant term in the RHS equal to zero, and is also clearly integral.

Note now that a sequence as described in the problem statement exists for $k$ iff $r_k=0$; indeed, if $r_k=0$, take $a_n=P_k(n)$ for $n\geq0$, and we got the right sequence, whereas is the sequence exists, define $d_n=P_k(n)-a_n$, or $d_n=\frac{d_{n-1}+r_k}{n}$. After trivial induction, we find that $d_n=\frac{d_0}{n!}+r_k\frac{0!+1!+\dots+(n-1)!}{n!}$. Take now $n$ large enough that $\left|\frac{d_0}{n!}\right|<\frac{1}{2}$, and $|r_k|\frac{0!+1!+\dots+(n-1)!}{n!}<\frac{1}{2}$ (*), and since $d_n$ is clearly integral, then $d_n=0$, and since the same applies to $n+1$ because it is even larger, then $d_n=d_{n+1}=0$, yielding $r_k=0$.

(*) It can be shown by induction that for $n\geq4$, $\frac{0!+1!+\dots+(n-1)!}{n!}<\frac{1}{\sqrt{n}}$, or it suffices to take $n>4r_k^2$.

Once we have all this, it suffices to show that if $r_k=0$, then $k\equiv2\pmod 3$, which can probably obtained as follows: note that $P_1(x)=1$ and $r_1=-1$, $P_2(x)=x+1$ and $r_2=0$, and $P_3(x)=x^2+x-1$ and $r_3=1$, but after this $r_k$ increases rapidly in absolute value, with parity odd,even,odd,odd,even,odd,odd,even,odd,... or we may postulate that $r_k$ is even iff $k\equiv2\pmod3$; if we manage to prove that, then we are done...

Somebody cares to finish it?

Note: I have worked $r_k$ for $k=2,5,8,11$, but it is only zero for $k=2$, yielding sequence $a_n=n+1$, clearly satisfying the condition in the problem statement, hence I'd settle just for proving the parity of $r_k$, since it is clearly false that $r_k=0$ for all $k\equiv2\pmod3$...
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Zhero
2043 posts
#3 • 10 Y
Y by mad, littlemushroom, Aubery, Mathuzb, Aryan-23, Adventure10, and 4 other users
Lemma 1: If there exist integers $k$ and $c$ such that $n! | (1! + 2! + \cdots + (n-1)!) k + c$, then $k = c = 0$.
Proof: Suppose for the sake of contradiction that $k \neq 0$. We may assume without loss of generality that $k > 0$. It is obvious that for sufficiently large $n$,
\begin{align*}
k(1! + 2! + \cdots + (n-1)!) + c
&< 1 \cdot 1! + 2 \cdot 2! + \cdots + (n-1)! \cdot (n-1)! \\
&= (2! - 1!) + (3! - 2!) + \cdots + (n! - (n-1)!) \\
&= n! - 1 < n!
\end{align*}
which is a contradiction.

Lemma 2: $n(n+k)! = (n+k+1)! - (k+1)(n+k)!$.
Proof:
\begin{align*}
n(n+k)!
&= (n+k)(n+k)! - k(n+k)! \\
&= (n+k+1)! - (n+k)! - k(n+k)! \\
&= (n+k+1)! - (k+1)(n+k)!.
\end{align*}

Lemma 3: Let $\left \{ \begin{matrix} n \\ k \end{matrix} \right \}$ denote the number of ways to partition $n$ objects into $k$ groups (a Stirling number of the second kind.) Then
\[n^{k} n! = \sum_{i=0}^{k} \left\{\begin{matrix} k+1 \\ i+1 \end{matrix}\right\} (-1)^{i+k} (n+i)!. \]
Proof: We will use induction and lemma 2. The base case, $k=0$, is trivial. Suppose the result is true for all integers less than some $k > 0$ now. Then
\begin{align*}
n^{k+1} n!
&= \sum_{i=0}^{k} \left\{\begin{matrix} k+1 \\ i+1 \end{matrix}\right\} (-1)^{i+k} (-1)^{i+k} n (n+i)! \\
&= \sum_{i=0}^{k} \left\{\begin{matrix} k+1 \\ i+1 \end{matrix}\right\} (-1)^{i+k} (-1)^{i+k} ((n+i+1)! - (i+1)(n+i)!) \\
&= \left( (n+k+1)! + \sum_{i=1}^{k} \left\{\begin{matrix} k+1 \\ i \end{matrix}\right \} -(-1)^{i+k}(n+i)! \right) - \left((-1)^k n! +  \sum_{i=1}^{k} \left\{\begin{matrix} k+1 \\ i+1 \end{matrix}\right \} (-1)^{i+k} (i+1)(n+i)!\right) \\
&= (n+k+1)! - (-1)^k n! + \sum_{i=1}^k (-1)^{i+(k+1)} (n+i)! \left ( \left \{ \begin{matrix} k+1 \\ i \end{matrix}  \right \} + (i+1) \left \{ \begin{matrix} k+1 \\ i+1 \end{matrix} \right \}  \right ) \\
&= (n+k+1)! - (-1)^k n! + \sum_{i=1}^k (-1)^{i+(k+1)}(n+i)! \left \{ \begin{matrix} k+2 \\ i+1 \end{matrix} \right \} \\
&= \sum_{i=0}^{k+1} \left \{ \begin{matrix} (k+1)+1 \\ i+1 \end{matrix} \right \}  (-1)^{i+(k+1)}(n+i)!,
\end{align*}
as desired.


Let $a_0 = c$. It can easily be shown by induction that $a_n = \frac{c + 1^{k-1} 1! + 2^{k-1} 2! + \cdots + n^{k-1} n!}{n!}$. It is clear that $a_n$ is an integer if and only if $n! | c + 1^{k-1} 1! + 2^{k-1} 2! + \cdots + (n-1)^{k-1} (n-1)!$, so we wish to show that if $n! | c + 1^{k-1} 1! + 2^{k-1} 2! + \cdots + (n-1)^{k-1} (n-1)!$ for all integers $n$, then $3 | k - 2$.

Applying lemma 4, we find that there is a constant $C_1$ and a function $f_k(n) : \mathbb{Z} \to \mathbb{Z}$ for which
\[ 1^{k-1} 1! + 2^{k-1} 2! + \cdots + (n-1)^{k-1} (n-1)! = C_1 + \left(1! + 2! + \cdots + (n-1)!\right) \left( \sum_{i=0}^{k-1} \left\{\begin{matrix} k \\ i+1 \end{matrix}\right\} (-1)^{i+k-1} \right) + f(n)(n!).\]
(Essentially, when we add these sums, almost all of the terms "telescope" to a sum of $1! + 2! + \cdots + (n-1)!$ multiplied by an alternating sum of Stirling numbers. The terms past $n!$ together form the $f(n)(n!)$ term in our sum and the beginning terms that were not handled, added to the $c = a_0$ yield the constant $C_1$.)

By lemma 1, in order for $n!$ to divide each term of the $1^{k-1} 1! + 2^{k-1} 2! + \cdots + (n-1)^{k-1} (n-1)!$, we must have
\[ \sum_{i=0}^{k-1} \left\{\begin{matrix} k \\ i+1 \end{matrix}\right\} (-1)^{i+k-1} = 0.\]

In order for the sum to be 0, it must be even. We claim that the sum is even if and only if $3 | k - 2$. Note that
\begin{align*}
\sum_{i=0}^{k-1} \left\{\begin{matrix} k \\ i+1 \end{matrix}\right\} (-1)^{i+k-1}
&\equiv \sum_{i=0}^{k-1} \left\{\begin{matrix} k \\ i+1 \end{matrix}\right\} \\
&= \sum_{i=0}^k \left\{\begin{matrix} k \\ i \end{matrix}\right\} \\
&= B_k \pmod{2},
\end{align*}
where $B_{k}$ is the $k$th Bell number. The Bell numbers satisfy $B_0 = B_1 = 1$ and Touchard's congruence; in the case of $p=2$, we have $B_{n+2} \equiv B_{n+1} \equiv B_n \pmod{2}$. It is easily seen that $(B_{3k}, B_{3k+1}, B_{3k+2}) \equiv (1, 1, 0) \pmod{2}$, i.e., the Bell numbers are even only if $3 | k - 2$, which completes our proof.

Comment: Alternating sums of Stirling numbers are called Complementary Bell Numbers, or Uppuluri-Carpenter numbers. A result proven here seems to imply that there actually exist at most two $k$ for which the alternating sum is 0, that is, for which every term in our sequence is a positive integer (one of which is $k=2$).
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math154
4302 posts
#4 • 5 Y
Y by mad, Aubery, Adventure10, Mango247, and 1 other user
Suppose that the $a_n$ are all integers. Let $S_k(n)=\sum_{i=1}^{n}i^k (i-1)!$, so $n!a_n-a_0=S_k(n)$. Then $S_1(n)\equiv S_0(n)-1\pmod{n!}$ and for $k\ge1$ (to avoid $0^0$), the binomial theorem yields
\begin{align*}
\sum_{j=0}^{k} (-1)^{k-j}\binom{k}{j}S_j(n) = \sum_{i=1}^{n}(i-1)^k (i-1)! = \sum_{i=2}^{n}(i-1)^{k+1}(i-2)!
&= \sum_{i=1}^{n-1}i^{k+1}(i-1)! \\
&\equiv \sum_{i=1}^{n}i^{k+1}(i-1)! = S_{k+1}(n) \pmod{n!}.
\end{align*}Now define $p_0=1$, $q_0=0$, $p_1=1$, $q_1=-1$, and
\[p_{k+1} = \sum_{j=0}^{k} (-1)^{k-j}\binom{k}{j} p_j,\; q_{k+1} = \sum_{j=0}^{k} (-1)^{k-j}\binom{k}{j} q_j\]for $k\ge1$ (these sequences are independent of $n$) so that $-a_0\equiv S_k(n) \equiv p_k S_0(n) + q_k \pmod{n!}$ for all $n\ge1$.

We can prove by strong induction that $p_k$ is even iff $3|k-2$. Indeed,
\[p_{k+1} \equiv \sum_{j=0}^{k}\binom{k}{j}p_j \pmod{2}\]for all $k\ge1$, so by the inductive hypothesis,
\[p_{k+1} \equiv 2^k - \sum_{3|j-2}\binom{k}{j} \equiv \omega^{2k+1}+\omega^{k+2} \equiv 1-[3|k-1] \pmod{2}\]by a third roots of unity filter.

Now fix $k$ not congruent to $2\pmod{3}$ and assume for the sake of contradiction that the $a_n$ are all integers. From our induction we have $p_k$ odd and therefore nonzero, so for sufficiently large $n$, $n!$ divides
\[|p_k S_0(n) + (a_0+q_k)| \in (0,n!),\]a contradiction (see the first lemma in the previous post for a proof of this bound).
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DrMath
2130 posts
#6 • 3 Y
Y by huricane, CyclicISLscelesTrapezoid, Adventure10
We start with the observations that:
1) $a_n=\dfrac{a_0+\sum_{j=1}^n j^k\cdot (j-1)!}{n!}$.
2) There always exists a polynomial $p$ with integer coefficients such that $xp(x)-p(x-1)=x^k-r_k$ where $r_k$ is some constant dependent on $k$. This is easy to show by repeatedly determining the coefficients of the polynomial.
3) Let $s_k(n)=\sum_{j=1}^n j^k(j-1)!$. Then for some fixed $t_k$, $s_k(n)=p(n)\cdot n!+r_k\cdot s_0 (n)+t_k$. This is trivial to show with induction, where $t_k$ is just defined by taking $n=1$.

Thus, the problem becomes finding all $k$ such that for some $a_0$, $n!\mid \left(p(n)\cdot n!+r_k s_0 (n)+t_k+a_0\right)$. It's clear that for this to happen, $r_k$ must be $0$ by a size argument. In other words, we want to show that if $xp(x)-p(x-1)=x^k$ then $k\equiv 2\pmod{3}$.

Working in $\mathbb{Z}[x]/(x^6-1)$ gives us $p(x)\equiv ax^5+bx^4+cx^3+dx^2+ex+f$. Take $\zeta$ to be a $6$th root of unity. Note that $\zeta - 1 = \zeta ^2$. Thus $\zeta p(\zeta) - p(\zeta ^2 ) = \zeta ^k$. It follows that $\zeta (a-2b-c+d+f)+(a+2b-c-d-f)=\zeta ^k$.

It's easy to see that $a-2b-c+d+f$ and $a+2b-c-d-f$ have the same parity. Thus, it is impossible for one of them to be 1 and the other to be 0, ruling out $k\equiv 0, 1\pmod{3}$, therefore implying $k\equiv 2\pmod{3}$ as desired.
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trumpeter
3332 posts
#7 • 3 Y
Y by AlastorMoody, Adventure10, Mango247
April wrote:
...Show that if there exists a consequence...

Hmm... nobody's mentioned this yet (just a technicality) but I think in the OP this should be "sequence" instead of "consequence".

Fixed ~dj
This post has been edited 1 time. Last edited by djmathman, Dec 13, 2015, 5:20 PM
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Pathological
578 posts
#8 • 4 Y
Y by stroller, mijail, Adventure10, Mango247
Finally solved this :).

Solution
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stroller
894 posts
#9 • 1 Y
Y by Adventure10
Solution from ~2 years ago when I still knew what "Stirling Numbers" are... (not that this knowledge is required, but it's helpful to motivate the solution below)

First, a general formula for the sequence $(a_i)$:

$$a_n = \frac{a_0 + \sum_{m=1}^n m^k(m-1)!}{n!}.$$Note that there are integer coefficients $s_{k,0},...,s_{k,k-1}$ such that $m^k = \sum_{i=0}^{k-1} s_{k,i}\prod_{j=0}^i(m+j)$, which means that
$$a_n = \frac{1}{n!}(a_0 + \sum_{m=1}^n \sum_{i=0}^{k-1}s_{k,i}(m+i)!).$$Consider the double sum mod $n!$. $$\sum_m \sum_i s_{k,i}(m+i)! \equiv \sum_{c=1}^{k-1} c! \sum_{i=0}^{c-1} s_{k,i} + \sum_{c=k}^n c!\sum_{i=0}^{k-1} s_{k,i}.$$Thus we have, for some $a_0' \in \mathbb Z$, that
$$a_n^* := \frac{1}{n!}(a_0' + \sum_{c =k}^{n-1} c! \left(\sum_{i=0}^{k-1}s_{k,i}\right)) \in \mathbb Z \forall n \text{ large}.$$The denominator will be larger than the numerator in absolute value for all $n$ large, hence $a_n^*$ is an integer for all $n$ if and only if $a_0' = \sum_{i=0}^{k-1} s_{k,i} = 0$.

Note that the sequence $s_{k,i}$ satisfies the relation $s_{k,i} = s_{k-1,i-1} - (i+1)s_{k-1,i}, s_{k,j} = 0 \quad \forall \ \ j \ne 0,...,k-1$, and the initial condition $s_{1,0} = 1$. Let $B_k = \sum_{i=0}^{k-1} s_{k,i}$. We'll prove that $B_k\equiv 0 \pmod 2$ if and only if $k \equiv 2 \pmod 3$, which will finish the problem.

We have $B_k= \sum_{i=0}^{k-1} \left(s_{k-1,i-1} - (i+1)s_{k-1,i}\right) \equiv B_{k-1} + \sum_{t \ge 0} s_{k-1, 2t}$
$$\implies B_k - B_{k-1} \equiv \sum_{t\ge 0} s_{k-1,2t} \equiv \sum_{t\ge 0} s_{k-2, 2t-1} - (2t+1)s_{k-2,2t} \equiv \sum_{t\ge 0} s_{k-2,2t-1} + s_{k-2,2t} = B_{k-2}.$$Moreover, we also have $B_1 = 1, B_2 = 0$. We can compute from the above recurrence mod 2 that $B_3 = 1, B_4 = 1 = B_1, B_5 = 0 = B_2$, from here the result is immediate.
This post has been edited 1 time. Last edited by stroller, Feb 19, 2020, 3:17 AM
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ayan.nmath
643 posts
#10
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April wrote:
Let $k$ be a positive integer. Show that if there exists a sequence $a_0,a_1,\ldots$ of integers satisfying the condition \[a_n=\frac{a_{n-1}+n^k}{n}\text{ for all } n\geq 1,\]then $k-2$ is divisible by $3$.

Proposed by Okan Tekman, Turkey

Solution. Define $S_m(n)=1^m\cdot 1!+2^m\cdot 2!+\cdots+n^m\cdot n!.$ Note that $n!\cdot a_{n+1}=(n-1)!\cdot a_n+n^k\cdot (n-1)!,$ hence
$$a_{n+1}=\frac{1}{n!}(a+S_t(n))$$where $t=k-1$ and $a=a_0.$

Claim 1. For all $t\ge 1$ and $n\ge 2$ it follows that
$$S_t(n-1)\equiv (-1)^tS_0(n-1)+\sum_{i=1}^t\binom{t}{i}(-1)^{t-i}(S_{i-1}(n-1)-1)\pmod{n!}.$$Proof. By computing,
\begin{align*}
    S_t(n) = \sum_{m=1}^n m^t\cdot m! &= \sum_{m=1}^n((m+1)-1)^t\cdot m!\\
    &= \sum_{m=1}^n\left(\sum_{i=0}^{t}(m+1)^i(-1)^{t-i}\binom{t}{i}\right)m!\\
    &= \sum_{i=0}^{t}\sum_{m=1}^n(m+1)^i(-1)^{t-i}\binom{t}{i}m!\\
    &= \sum_{i=0}^{t}(-1)^{t-i}\binom{t}{i}\sum_{m=1}^n(m+1)^im!\\
    &= (-1)^tS_0(n)+\sum_{i=0}^{t}(-1)^{t-i}\binom{t}{i}(S_{i-1}(n+1)-1).
\end{align*}Now consider modulo $n!$ to get our claim.$~~\square$

Define the sequences $\{a_n\}_{n=1}^{\infty}$ and $\{b_n\}_{n=1}^{\infty}$ as
$$\begin{cases}a_0 &= 1\\ a_k&=(-1)^k+\sum_{i=1}^k(-1)^{k-i}\binom{k}{i}a_{i-1}~~\forall~k\ge 1\end{cases}$$$$\begin{cases}b_0 &= 0\\ b_k&=\sum_{i=1}^k(-1)^{k-i}\binom{k}{i}(b_{i-1}-1)~~\forall~k\ge 1\end{cases}$$
Claim 2. For all $k\ge 0$ and $n\ge 1,$
$$S_k(n-1)\equiv a_k S_0(n-1)+b_k\pmod{n!}.$$Proof. Induct on $k$ and use claim 1. $~~\square$

Coming back to our original problem,

We are given that $S_t(n-1)\equiv -a\pmod{n!}$ holds for all positive integers $n.$ Using our second claim, we have that $$\frac 1{n!}(a_t S_0(n-1) +b_t+a)$$is an integer for all $n\ge 1.$ Using Stolz-Cesaro theorem,
$$\lim_{n\to\infty}\frac{S_0(n-1)}{n!}=\lim_{n\to\infty}\frac{1!+2!+\cdots+(n-1)!}{n!}=\lim_{n\to\infty}\frac{n!}{(n+1)!-n!}=0.$$Therefore, for all sufficiently large $n,$ it follows that $a_t S_0(n-1) +b_t+a=0,$ which forces $a_t=0.$

Claim 3. $a_n$ is even if and only if $n\equiv 1\pmod 3.$
Proof. Easy. Omitted. $~~\square$

Our last claim finishes the problem since $t=k-1.~~\blacksquare$
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william122
1576 posts
#11
Y by
Obviously $k\neq 1$ so suppose $k\ge 2$. Note that we can write $a_n$ as $\frac{a_0+0!\cdot 1^k+1!\cdot 2^k+2!\cdot 3^k+\ldots +(n-1)!n^k}{n!}$, so it suffices to have $$a_0\equiv -\sum_{i=0}^{n-1}i!\cdot(i+1)^k\equiv -\sum_{i=1}^{n-1}i^{k-1}i!\pmod{n!}$$We will use $S(m,n)$ to denote $\sum_{i=0}^{m-1} i^n\cdot i!\pmod{m!}$. Also, for convenience, we denote $P_m=S(m,0)$
Claim 1: $\frac{P_m}{m!}$ approaches $0$ as $m$ grows large.
Proof: As we have $\sum_{i=1}^{m-1}i!<m!$, this means that we want $\lim_{m\to\infty}\frac{0!+1!+\ldots+(m-1)!}{m!}=0$. To prove, this, it suffices to show $\frac{P_m}{m!}<\frac{2020}{\sqrt{m}}$, which is evident by induction since it's true for $1,2$ and $$P_{m+1}=\frac{P_m+1}{m+1}<\frac{\frac{2020}{\sqrt{m}}+1}{m+1}\le\frac{2021}{m+1}<\frac{2020}{\sqrt{m+1}}$$for $m>2$.

Claim 2: We have $S(m,n)\equiv S(m+1,n-1)-\sum_{i=0}^{n-1}S(m,i)\cdot\binom{n}{i}$ for $n>1$
Proof: This is because $S(m+1,n-1)\equiv \sum_{i=1}^m i^{n-1}i!\equiv \sum_{i=1}^m i^n(i-1)!\equiv\sum_{i=0}^{m-1}(i+1)^ni!\pmod{(m+1)!}$, so $$S(m+1,n-1)-S(m,n)\equiv\sum_{i=0}^{m-1}\sum_{j=0}^{n-1}\binom{n}{j}i^ji!\equiv\sum_{j=0}^{n-1}\binom{n}{j}\sum_{i=0}^{m-1}i^ji!\equiv\sum_{j=0}^{n-1}\binom{n}{j}S(m,j)$$as desired.

Now, with Claim 2, we can show that $S(m,n)$ for fixed $n$ is a linear equation in $P_m$ through induction. If it is true for numbers $<n$, then $\sum_{i=0}^{n-1}S(m,i)\binom{n}{i}$, and $S(m+1,n-1)$ is one in $P_{m+1}\equiv P_m\pmod{m!}$, as desired. Thus, for the original question, we want fixed constants $A>0,B$ such that $AS_m+B\equiv 0\pmod{n!}$ for all $m$. However, if $A\neq 0$, this is impossible by Claim 1, since $AS_m+B$ will be $<n!$ for large enough $m$.

Hence, we need $A=0$, and it suffices to show that $S(m,n)$ can only be a constant if $n\equiv 1\pmod 3$. In fact, we can show the statement that the coefficient of $P_m$ is even if and only if $n\equiv 1\pmod{3}$. For this, we once again use strong induction. If we denote the coefficient of $S_m$ as $A_n$, our base cases of $0,1$ yield $(A_0,A_1)\equiv(1,0)\pmod 2$, as desired. For the inductive step, note that $$A_n\equiv A_{n-1}-\sum_{i=0}^{n-1}A_i\binom{n}{i}\equiv A_{n-1}+\sum_{\stackrel{i\not\equiv 1\pmod 3}{i\le n-1}}\binom{n}{i}\equiv\sum_{i\equiv 1\pmod 3}\binom{n}{i}+\begin{cases}1&\text{ if }n\equiv 1\pmod 3\\0&\text{ otherwise }\end{cases}\pmod 2$$By Roots of Unity Filter, $$\sum_{i\equiv 1\pmod 3}\binom{n}{i}\equiv \omega^2(1+\omega)^n+\omega(1+\omega^2)^n\equiv \omega^{2n+2}+\omega^{n+1}\equiv\begin{cases}0&\text{ if }n\equiv2\pmod 3\\1&\text{ otherwise }\end{cases}\pmod 2$$Putting these two together, $A_n\equiv \begin{cases}0&\text{ if }n\equiv1\pmod 3\\1&\text{ otherwise }\end{cases}\pmod 2$, as desired.
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Aryan-23
558 posts
#14 • 2 Y
Y by GioOrnikapa, Mango247
Hopefully correct :stretcher:

Solution
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IAmTheHazard
5005 posts
#15 • 1 Y
Y by centslordm
We can clearly write
$$a_n=\frac{a_0+0!\cdot 1^k+1!\cdot 2^k+\cdots+(n-1)!\cdot n^k}{n!}=\frac{a_0+1!\cdot 1^{k-1}+2!\cdot 2^{k-1}+\cdots+n!\cdot n^{k-1}}{n!}.$$Let $p=k-1$; I will show that $p \equiv 1 \pmod{3}$. Let $S(n,p)=1!\cdot 1^p+\cdots+n!\cdot n^p$. If this above expression is an integer for all $n \geq 1$, then $S(n,p) \equiv S(n-1,p)$ must be congruent to a fixed integer modulo $n!$. The idea is to write $S(n,p)$ in terms of $S(n,q)$, where $q<p$. We first need the following lemma.

Lemma: $\lim_{n \to \infty} \frac{S(n-1,0)}{n!}=0$.
Proof: It suffices to show that $\frac{S(n-1,0)}{n!} \leq \frac{10^{10}}{\sqrt{n}}$ for all $n \geq 1$. This is true for $n=1$, and then for $n \geq 2$ we have
$$S(n-2,0) \leq \frac{10^{10}(n-1)!}{\sqrt{n-1}} \implies \frac{S(n-1,0)}{n!}=\frac{S(n-2,0)+(n-1)!}{n!}\leq \frac{\frac{10^{10}}{\sqrt{n-1}}+1}{n},$$whic is clearly at most $\frac{10^{10}}{\sqrt{n}}$. $\blacksquare$

We now set up the key recurrence relation.

Claim: We have $S(n,p)\equiv S(n,p-1)-1-\sum_{q=0}^{p-1} \binom{p}{q}S(n,q) \pmod{(n+1)!}$.
Proof: Write
$$S(n+1,p-1)=\sum_{i=1}^{n+1} i!\cdot i^{p-1}=\sum_{i=1}^{n+1} (i-1)!\cdot i^p=\sum_{i=0}^n i!\cdot (i+1)^p=\sum_{i=0}^n i!\sum_{q=0}^p\binom{p}{q}i^q=\sum_{q=0}^p \binom{p}{q}\sum_{i=0}^n i!\cdot i^q=1+\sum_{q=0}^p \binom{p}{q}S(n,q),$$where the $1$ comes from the fact that for $q>0$, $\sum_{i=0}^n i!\cdot i^q=\sum_{i=1}^n i!\cdot i^q=S(n,q)$, but for $q=0$ we have an extra $1$ on the LHS. Hence $S(n,p)=S(n+1,p-1)-1-\sum_{q=0}^{p-1} \binom{p}{q}S(n,q)$. Since $S(n+1,p-1) \equiv S(n,p-1) \pmod{(n+1)!}$, we obtain the desired result. $\blacksquare$

Now, by our lemma, and the fact that $S(n-1,0)$ is unbounded, $S(n-1,0)$ cannot be congruent to a fixed integer modulo $n!$. On the other hand, it is both well-known and easily proven by induction that $S(n-1,1) \equiv -1 \pmod{n!}$. From our recursion, we can express $S(n-1,p)$ as a linear combination of $S(n-1,0)$ and $S(n-1,1)$ (working modulo $n!$) whose coefficients depend only on $p$, so if $S(n-1,p)$ is congruent to a fixed integer modulo $n!$ then the coefficient of $S(n-1,0)$ must be zero. I will actually show that if the coefficient of $S(n,0)$ in $S(n,p)$ is divisible by $2$ only when $p \equiv 1 \pmod{3}$.

Let $a_p \in \{0,1\}$ be the coefficient of $S(n,0)$ in $S(n,p)$ when taken modulo $2$, so $a_0=1$, $a_1=0$, and $a_p\equiv a_{p-1}+\sum_{q=0}^{p-1} \binom{p}{q}a_q \pmod{2}$ for all $p \geq 2$. We now use strong induction, with the bases cases done already. By inductive hypothesis, it is sufficient to show that $\sum_{\substack{0 \leq q \leq p-1\\q \not \equiv 1 \pmod{3}}} \binom{p}{q}$ is even only when $p \equiv 0 \pmod{3}$, or equivalently $\sum_{\substack{0 \leq q \leq p\\q \equiv 1 \pmod{3}}} \binom{p}{q}$ is even only when $p \equiv 2 \pmod{3}$, since $\sum_{0 \leq q \leq p} \binom{p}{q}=2^p \equiv 0 \pmod{2}$.

When $p \equiv 0 \pmod{3}$, we have $\sum_{\substack{0 \leq q \leq p\\q \equiv 1 \pmod{3}}}\binom{p}{q}=\sum_{\substack{0 \leq q \leq p\\q \equiv 2 \pmod{3}}} \binom{p}{q}$, hence it suffices to show that $\sum_{\substack{0 \leq q \leq p\\3 \mid q}} \binom{p}{q}$ is not divisible by $4$. Letting $\omega=e^{2\pi i/3}$, this is equivalent to $(1+\omega)^p+(1-\omega)^p$ never being divisible by $4$ when $3 \mid p$, which is true since this quantity is evidently always $\pm 2$ since $1\pm \omega$ are primitive sixth roots of unity.

When $p \equiv 1 \pmod{3}$, we have $\sum_{\substack{0 \leq q \leq p\\q \equiv 1 \pmod{3}}}\binom{p}{q}=\sum_{\substack{0 \leq q \leq p\\3 \mid q}} \binom{p}{q}$, hence defining $\omega$ as before, it is equivalent to show that $(1+\omega)^p+(1-\omega)^p$ is odd when $p \equiv 1 \pmod{3}$. It is clear that this quantity is always $\pm 1$ for the same reason as above.

When $p \equiv 2 \pmod{3}$, we can pair every $\binom{p}{q}$ with $\binom{p}{p-q}$ in the summation, except for $\binom{p}{p/2}$ if it appears. Therefore, if $\binom{p}{p/2}$ doesn't appear, the summation is clearly even, otherwise we just need to show that $\binom{p}{p/2}$ is even which is just true by Lucas/Kummer.

Therefore, if $S(n,p)$ is congruent to a fixed integer modulo $(n+1)!$ for all $n$, $p \equiv 1 \pmod{3}$, hence $k \equiv 2 \pmod{3}$, which is the desired result. $\blacksquare$


Edit: I have realized that I also need to handle p=2 mod 3 because we’re doing strong induction but you basically do it the same way. I am sick rn though so I will edit it in a few days later ok fixed
This post has been edited 6 times. Last edited by IAmTheHazard, Aug 7, 2023, 9:00 PM
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LightChaser
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#16
Y by
daniel73 wrote:
This may help in finding a solution:

Note that we can always find a polynomial $P_k(x)$ of degree $k-1$ and with integer coefficients such that $xP_k(x)=P_k(x-1)+x^k+r_k$, where $r_k$ is some residual constant to be determined; just express $P_k(x)=c_{k-1}x^{k-1}+c_{k-2}x^{k-2}+\dots+c_1x+c_0$, and work the coefficients down from $c_{k-1}$ to $c_0$, finding the coefficient $c_m$ by comparing the terms of degree $m+1$ at both sides of the equation. Clearly $c_{k-1}=1$, and all other coefficients are a linear combination with integral weights of higher-order coefficients, hence the polynomial coefficients are all integral. The residual constant $r_k$ is the one that makes the constant term in the RHS equal to zero, and is also clearly integral.

Note now that a sequence as described in the problem statement exists for $k$ iff $r_k=0$; indeed, if $r_k=0$, take $a_n=P_k(n)$ for $n\geq0$, and we got the right sequence, whereas is the sequence exists, define $d_n=P_k(n)-a_n$, or $d_n=\frac{d_{n-1}+r_k}{n}$. After trivial induction, we find that $d_n=\frac{d_0}{n!}+r_k\frac{0!+1!+\dots+(n-1)!}{n!}$. Take now $n$ large enough that $\left|\frac{d_0}{n!}\right|<\frac{1}{2}$, and $|r_k|\frac{0!+1!+\dots+(n-1)!}{n!}<\frac{1}{2}$ (*), and since $d_n$ is clearly integral, then $d_n=0$, and since the same applies to $n+1$ because it is even larger, then $d_n=d_{n+1}=0$, yielding $r_k=0$.

(*) It can be shown by induction that for $n\geq4$, $\frac{0!+1!+\dots+(n-1)!}{n!}<\frac{1}{\sqrt{n}}$, or it suffices to take $n>4r_k^2$.

Once we have all this, it suffices to show that if $r_k=0$, then $k\equiv2\pmod 3$, which can probably obtained as follows: note that $P_1(x)=1$ and $r_1=-1$, $P_2(x)=x+1$ and $r_2=0$, and $P_3(x)=x^2+x-1$ and $r_3=1$, but after this $r_k$ increases rapidly in absolute value, with parity odd,even,odd,odd,even,odd,odd,even,odd,... or we may postulate that $r_k$ is even iff $k\equiv2\pmod3$; if we manage to prove that, then we are done...

Somebody cares to finish it?

Note: I have worked $r_k$ for $k=2,5,8,11$, but it is only zero for $k=2$, yielding sequence $a_n=n+1$, clearly satisfying the condition in the problem statement, hence I'd settle just for proving the parity of $r_k$, since it is clearly false that $r_k=0$ for all $k\equiv2\pmod3$...
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vsamc
3789 posts
#17 • 1 Y
Y by megarnie
Solution
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HamstPan38825
8872 posts
#18
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This is not a number theory problem. Also how one motivates the second step would be much appreciated (I got carried there :stretcher:)

First Step: We find an integer polynomial $P$ such that there is a sequence $\{b_n\} = \{a_n - P(n)\}$ that satisfies $b_n = \frac{b_{n-1} + \ell}n$ for some constant $\ell$. This is simply a matter of equating coefficients: setting $P(n) = n^{k-1} + a_{k-2} n^{k-2} + \cdots + a_0$, we must have \[a_{k-2} n^{k-1} + a_{k-3} n^{k-2} + \cdots + a_0 n = (n-1)^{k-1} + a_{k-2} (n-1)^{k-2} + \cdots + a_0+ r.\]This is a linear system where $a_i$ is written in terms of linear combinations of $a_j$ for $j \geq i$ (in particular $a_{k-2} = 1$), so it is given by a nonsingular $k \times k$ matrix. Thus such coefficients $a_0, a_1, \dots, a_{k-2}, r$ exist and are unique.

Second Step: Observe that $a_n$ is an integer if and only if $b_n$ is an integer. If $\ell \neq 0$, then notice that for sufficiently large $n$, $b_n$ will not be an integer; thus we must have $\ell = 0$. In other words, our polynomial $P$ satisfies \[xP(x) = x^k + P(x-1).\]In fact, for $\omega$ a third root of unity, we have
\begin{align*}
-\omega P(-\omega) &= P(\omega^2) + (-\omega)^k \\
-\omega^2 P(-\omega^2) &= P(\omega) + (-\omega^2)^k.
\end{align*}Now some black magic: taking this equation modulo $2$ and adding the first equation to $\omega$ times the second, we need $\omega^k + \omega^{2k + 1} \equiv 0$ which can only happen when $k \equiv 2 \pmod 3$, as needed.

Remark: The last step still feels like black magic. A lot of more *natural* things one would do with the cube root of unity (filter to get $2$ mod $3$ exponent coefficients, etc) don't work, but this somehow restricts to $k \equiv 2 \pmod 3$.
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luutrongphuc
58 posts
#19
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Pathological wrote:
Finally solved this :).

Solution
why do you think to consider modulo $x^4+x^2+1$
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