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

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

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[*]May 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
ISI UGB 2025 P7
SomeonecoolLovesMaths   2
N 5 minutes ago by Primeniyazidayi
Source: ISI UGB 2025 P7
Consider a ball that moves inside an acute-angled triangle along a straight line, unit it hits the boundary, which is when it changes direction according to the mirror law, just like a ray of light (angle of incidence = angle of reflection). Prove that there exists a triangular periodic path for the ball, as pictured below.

IMAGE
2 replies
1 viewing
SomeonecoolLovesMaths
an hour ago
Primeniyazidayi
5 minutes ago
Knights NOT crowded on the chessboard
mshtand1   1
N 6 minutes ago by sarjinius
Source: Ukrainian Mathematical Olympiad 2025. Day 1, Problem 8.4
What is the maximum number of knights that can be placed on a chessboard of size \(8 \times 8\) such that any knight, after making 1 or 2 arbitrary moves, does not land on a square occupied by another knight?

Proposed by Bogdan Rublov
1 reply
mshtand1
Mar 13, 2025
sarjinius
6 minutes ago
ISI UGB 2025 P5
SomeonecoolLovesMaths   2
N 19 minutes ago by SomeonecoolLovesMaths
Source: ISI UGB 2025 P5
Let $a,b,c$ be nonzero real numbers such that $a+b+c \neq 0$. Assume that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c}$$Show that for any odd integer $k$, $$\frac{1}{a^k} + \frac{1}{b^k} + \frac{1}{c^k} = \frac{1}{a^k+b^k+c^k}.$$
2 replies
1 viewing
SomeonecoolLovesMaths
an hour ago
SomeonecoolLovesMaths
19 minutes ago
A bit tricky invariant with 98 numbers on the board.
Nuran2010   0
29 minutes ago
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
The numbers $\frac{50}{1},\frac{50}{2},...\frac{50}{97},\frac{50}{98}$ are written on the board.In each step,two random numbers $a$ and $b$ are chosen and deleted.Then,the number $2ab-a-b-1$ is written instead.What will be the number remained on the board after the last step.
0 replies
Nuran2010
29 minutes ago
0 replies
Looks like power mean, but it is not
Nuran2010   0
35 minutes ago
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
For $a,b,c$ positive real numbers satisfying $a^2+b^2+c^2 \geq 3$,show that:

$\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+\frac{a+b+c}{9} \geq \frac{4}{3}$.
0 replies
+1 w
Nuran2010
35 minutes ago
0 replies
Taking antipode on isosceles triangle's circumcenter
Nuran2010   0
39 minutes ago
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
In isosceles triangle, the condition $AB=AC>BC$ is satisfied. Point $D$ is taken on the circumcircle of $ABC$ such that $\angle CAD=90^{\circ}$.A line parallel to $AC$ which passes from $D$ intersects $AB$ and $BC$ respectively at $E$ and $F$.Show that circumcircle of $ADE$ passes from circumcenter of $DFC$.
0 replies
Nuran2010
39 minutes ago
0 replies
R to R, with x+f(xy)=f(1+f(y))x
NicoN9   3
N 40 minutes ago by EeEeRUT
Source: Own.
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that\[
x+f(xy)=f(1+f(y))x
\]for all $x, y\in \mathbb{R}$.
3 replies
NicoN9
4 hours ago
EeEeRUT
40 minutes ago
find angle
TBazar   7
N 44 minutes ago by TBazar
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
7 replies
TBazar
May 8, 2025
TBazar
44 minutes ago
ISI UGB 2025 P4
SomeonecoolLovesMaths   0
an hour ago
Source: ISI UGB 2025 P4
Let $S^1 = \{ z \in \mathbb{C} \mid |z| =1 \}$ be the unit circle in the complex plane. Let $f \colon S^1 \longrightarrow S^2$ be the map given by $f(z) = z^2$. We define $f^{(1)} \colon = f$ and $f^{(k+1)} \colon = f \circ f^{(k)}$ for $k \geq 1$. The smallest positive integer $n$ such that $f^{(n)}(z) = z$ is called the period of $z$. Determine the total number of points in $S^1$ of period $2025$.
(Hint : $2025 = 3^4 \times 5^2$)
0 replies
SomeonecoolLovesMaths
an hour ago
0 replies
ISI UGB 2025 P8
SomeonecoolLovesMaths   0
an hour ago
Source: ISI UGB 2025 P8
Let $n \geq 2$ and let $a_1 \leq a_2 \leq \cdots \leq a_n$ be positive integers such that $\sum_{i=1}^{n} a_i = \prod_{i=1}^{n} a_i$. Prove that $\sum_{i=1}^{n} a_i \leq 2n$ and determine when equality holds.
0 replies
SomeonecoolLovesMaths
an hour ago
0 replies
ISI UGB 2025 P6
SomeonecoolLovesMaths   0
an hour ago
Source: ISI UGB 2025 P6
Let $\mathbb{N}$ denote the set of natural numbers, and let $\left( a_i, b_i \right)$, $1 \leq i \leq 9$, be nine distinct tuples in $\mathbb{N} \times \mathbb{N}$. Show that there are three distinct elements in the set $\{ 2^{a_i} 3^{b_i} \colon 1 \leq i \leq 9 \}$ whose product is a perfect cube.
0 replies
SomeonecoolLovesMaths
an hour ago
0 replies
ISI UGB 2025 P2
SomeonecoolLovesMaths   0
an hour ago
Source: ISI UGB 2025 P2
If the interior angles of a triangle $ABC$ satisfy the equality, $$\sin ^2 A + \sin ^2 B + \sin^2  C = 2 \left( \cos ^2 A + \cos ^2 B + \cos ^2 C \right),$$prove that the triangle must have a right angle.
0 replies
SomeonecoolLovesMaths
an hour ago
0 replies
Six variables
Nguyenhuyen_AG   1
N 2 hours ago by TNKT
Let $a,\,b,\,c,\,x,\,y,\,z$ be six positive real numbers. Prove that
$$\frac{a}{b+c} \cdot \frac{y+z}{x} + \frac{b}{c+a} \cdot \frac{z+x}{y} + \frac{c}{a+b} \cdot \frac{x+y}{z} \geqslant 2+\sqrt{\frac{8abc}{(a+b)(b+c)(c+a)}}.$$
1 reply
Nguyenhuyen_AG
Today at 5:09 AM
TNKT
2 hours ago
Anything real in this system must be integer
Assassino9931   3
N 2 hours ago by Sardor_lil
Source: Al-Khwarizmi International Junior Olympiad 2025 P1
Determine the largest integer $c$ for which the following statement holds: there exists at least one triple $(x,y,z)$ of integers such that
\begin{align*} x^2 + 4(y + z) = y^2 + 4(z + x) = z^2 + 4(x + y) = c \end{align*}and all triples $(x,y,z)$ of real numbers, satisfying the equations, are such that $x,y,z$ are integers.

Marek Maruin, Slovakia
3 replies
Assassino9931
May 9, 2025
Sardor_lil
2 hours ago
IMO Shortlist 2011, Number Theory 3
orl   47
N Apr 25, 2025 by Ilikeminecraft
Source: IMO Shortlist 2011, Number Theory 3
Let $n \geq 1$ be an odd integer. Determine all functions $f$ from the set of integers to itself, such that for all integers $x$ and $y$ the difference $f(x)-f(y)$ divides $x^n-y^n.$

Proposed by Mihai Baluna, Romania
47 replies
orl
Jul 11, 2012
Ilikeminecraft
Apr 25, 2025
IMO Shortlist 2011, Number Theory 3
G H J
Source: IMO Shortlist 2011, Number Theory 3
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popop614
271 posts
#35
Y by
solved at 4 AM after dying on (read: unable to solve) other problems in DNY-euclid (how do i unbad...)

All functions of the form $f(x) = \pm x^{D} + c$ where $c \in \mathbb{Z}$ and $D \in \mathbb{N}$ and $D \mid n$ work. It is clear that these do work by well known identities.

Observe that if $g(x) = f(x) - f(0)$ satisfies the condition then $f(x)$ does as well; henceforth assume that $f(0) = 0.$

Note that
\[ f(1) - f(0) \mid 1, \]so we obtain that $f(1) = \pm 1$. Assume from now on that $f(1) = 1$, as the other is handled in exactly the same manner.

Now note that
\[ f(1) - f(-1) \mid 1 - (-1) \implies f(-1) - 1 \mid 2, \]and also
\[ f(-1) - f(0) \mid -1 \implies f(-1) \in \{-1, 1\}.\]In particular we must obtain that $f(-1) = -1$ as $0 \nmid 2$.

Let $p$ be an arbitrarily large prime. Notice that
\[ f(p) \mid p^n, \]so therefore for this value of $p$ we obtain that $f(p) = \pm p^d$ for some value $d$ and some choice of sign.

Now assume that $f(p) = -p^d$ for some integer $d$, and let $k$ and $r$ be integers such that $dk+r=n$ and $0 \le r < d$. Then observe that
\[ f(p) - f(-1) \mid p^n + 1 \]or
\[ p^d - 1 \mid p^n + 1 \implies p^d - 1 \mid p^r + 1.\]Clearly for $p$ sufficiently large we can't have this. Therefore the negative sign is out.

Now suppose that $f(p) = p^d$ for some integer $d \nmid n$, and let $k$ and $r$ be integers such that $dk+r=n$ and $0 < r < d$. Then observe that
\[ f(p) - f(1) \mid p^n - 1 \]or
\[ p^d - 1 \mid p^n - 1 \implies p^d - 1 \mid p^r - 1,\]again obviously impossible. Therefore for sufficiently large (realistically like $p>2$) we must have that $f(p) = p^d$ for some positive divisor $d$ of $n$.

let $D$ be such that there are infinitely many primes $p$ such that $f(p) = p^D$. This exists as $n$ has a finite number of divisors. Now,
\[ f(x) - p^D \mid x^n - p^n \implies f(x) - p^D \mid x^n - p^n - f(x)^{n/D} + p^n. \]
However, taking $p$ incredibly large, we obtain that $f(x)^{n/D} = x^n$, or $f(x) = x^D$ for said divisor $D$. Undoing all the WLOG stuff we get our result.
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andyxpandy99
365 posts
#36
Y by
The key idea is to realize that we can freely shift $f$ by a constant and that if $f$ works then $-f$ works. This implies that WLOG we can set $f(0) = 0$ and since $f(1) \mid 1$ take $f(1) = 1$. Now consider a particular prime $p$. Plug in $x = p$ and $y = 0$ to get $f(p) \mid p^n$. It follows that $f(p) = \pm p^m$ for some $m \leq n$. We will prove that $f(p) \neq -p^m$. Assume otherwise.

Plugging in $m = p$ and $n = 1$ now yields $$-p^m-1 \mid p^n-1$$which is equivalent to $$p^m+1 \mid p^n-1$$Since $p^m+1 \mid p^{2m}-1$ we have $$p^m+1 \mid p^{\gcd(2m,n)}-1 = p^{\gcd(m,n)}-1$$because $n$ is odd. If $m > 0$ then $p^m +1$ is clearly greater than $p^{\gcd(m,n)}-1 \leq p^m-1$. It follows that $p^{\gcd(m,n)} -1 = 0$ or $m = 0$. This means that $f(p) = -1$.

We will now prove that $f$ is injective. To see this, note that if $f(x) = f(y)$ then $x^n-y^n = 0$ so $x =y$ as desired. Plugging in $x = -1$ and $y = 0$ yields $f(-1) \mid -1$ but since $f$ is injective $f(-1) \neq f(1)$ so $f(-1) = -1$. It follows that $f(p) \neq f(-1) = -1$ so we arrive at a contradiction and $f(p) = p^m$ as desired.

If $f(p) = p^m$ plugging in $x = p$ and $y = 1$ yields $$p^m-1 \mid p^n-1$$which means $m \mid n$. So, for a particular $p$ we have $f(p) = p^m$ where $m$ is a factor of $n$. Since there are infinitely many primes and only a finite number of factors of $n$, there has to exist a $k \mid n$ such that $f(p) = p^k$ for infinitely many primes $p$, not just a particular value of $p$. Now note that $$p^k - f(y) \mid p^n-y^n$$and $$p^k-f(y) \mid p^n-f(y)^{\frac{n}{k}}$$implies $$p^k-f(y) \mid f(y)^{\frac{n}{k}}-y^n$$The RHS is now not dependent on $p$ and we are free to take a big enough $p$ such that we force $f(y)^{\frac{n}{k}} = y^n$ which yields $f(y) = y^k$. Recalling that we can freely shift and negate, our answer is thus $f(y) = \pm y^k + c$ for some constant $c$.
This post has been edited 1 time. Last edited by andyxpandy99, Aug 24, 2023, 3:26 PM
Reason: typo
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huashiliao2020
1292 posts
#38
Y by
It's obvious that if f is a sol then f+c is also a sol, so is -f, so WLOG f(0)=0, and henceforth ignore the +c and sign stuff; (1,0) gives f(1)=1, (p,0) for a prime p gives $f(p)=p^k$ (WLOG from f$\leftrightarrow$-f), (p,1) gives $p^k-1\mid p^n-1\implies k\mid n$. Now, since n has finite divisors but f is infinite, take $f(p)=p^k$ where there are infinite p that give the same k. Then, $f(x)-p^k\mid x^n-p^n-f(x)^{n/k}+p^n$; taking sufficiently large p (there are infinite of them), since the RHS doesn't depend on p, once the LHS>RHS we must have RHS=0, so $x^n=f(x)^{n/k}\iff f(x)=x^k$ for all x, as desired.
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YaoAOPS
1541 posts
#39
Y by
WLOG shift $f$ and invert such that $f(0) = 0$ and $f(1) = 1$.

Claim: For primes $p$, $f(p) = p^k$ for some not fixed integer $k \mid n$.
Proof. Note that $f(x) \mid x^n$ and $f(1) = 1$.
Let $f(p) = p^k$. Then $p^k - 1 \mid p^n - 1$ so $p^{\gcd(n,k)} - 1 = p^k - 1$, and thus $k \mid n$. $\blacksquare$

Claim: For all $n$, $f(y) = y^k$ for $y \le n$ and some fixed $k$.
Proof. Take a prime $q > n^{n^n}$. Then \[ q^k - f(y) \mid (q^n - y^n) - (q^n - f(y)^{\frac{n}{k}}) = f(y)^{\frac{n}{k}} - y^n \]so $f(y)^{\frac{n}{k}} = y^n$ and $f(y) = y^k$ for all $y \le n$. $\blacksquare$
Then $k$ is independent of $n$ by considering $f(2)$, so $f(x) = x^k$ for all $k$.
As such, $f(x) = \pm x^k + c$ for $k \mid n$ is the solution set.
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thdnder
198 posts
#40
Y by
There exists a positive integer $d$ dividing $n$, $c \in \mathbb{Z}$ and an $\epsilon \in \{1, -1\}$ such that $f(x) = \epsilon x^d + c$ for all $x \in \mathbb{Z}$.

If $f$ is a solution, then for all $c \in \mathbb{Z}$, $f - c$ is a solution. Thus we can assume that $f(0) = 0$. Then since $f(x) - f(0) \mid x^n - 0^n$, so $f(x) \mid x^n$. Thus for all prime $p$, we have $f(p) \mid p^n$. Since if $f$ is a solution, then $-f$ is also a solution, so we can assume $f(1) = 1$. Let $f(p) = \epsilon p^k$ for some $\epsilon \in \{-1, 1\}$, $k \le n$. Then $\epsilon p^k - 1 \mid p^n - 1$, thus $k \mid n$ and $\epsilon = 1$ for all prime $p$. Since there are infinitely many primes, so by pigeonhole principle, there are infinitely many primes $p$ such that $f(p) = p^d$ for some $d \mid n$.

Now take large enough prime $p$ such that $f(p) = p^d$. Then $f(x) - f(p) \mid x^n - p^n$, so $x^n - p^n \equiv x^n - f(x)^{\frac{n}{d}} (f(x) - p^d)$. Since $p$ is large enough, this forces $f(x) = x^d$. Thus $f(x) = \epsilon x^d + c$ for some constant $c$, $\epsilon \in \{1, -1\}$, $d \mid n$. So we're done. $\blacksquare$
This post has been edited 1 time. Last edited by thdnder, Oct 5, 2023, 8:43 AM
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HamstPan38825
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#41
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The answer is $f(x) = \varepsilon \cdot x^d + c$ for any $d \mid n$, $\varepsilon \in \{-1, 1\}$, and $c$ an integer. These obviously work.

To show that these are the only functions, note that if $f$ works, then $f+c$ works for any $c \in \mathbb Z$; thus, we may assume $f(0) = 0$. Furthermore, $f(1) - f(0) \mid 1$, and we can also assume $f(1) = 1$.

Now fix some prime $p$. As $f(p) \mid p^n$, set $f(p) = p^k$ for some $k$. Furthermore, because $f(p) - 1 \mid p^n - 1$, we have $k \mid n$.

I claim that $k$ is consistent across all $p$. To show this, assume that $f(q) = q^\ell$. Then as $p^k - q^\ell \mid p^n - q^n$, we have
$$p^k - q^{\ell} \mid p^{k+n-\ell} - q^n - p^n + q^n = p^n(p^{\ell - k} - 1).$$As the LHS is relatively prime to $p^n$, it follows for size reasons that $p^{\ell - k} = 1$, or $\ell = k$.

So we have $f(p) = p^d$ for some fixed $d$ across all primes. Then for any $n$, by setting $p$ big we have $$f(x) - p^d \mid x^d - f(x)$$implying $f(x) = x^d$ too.
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kamatadu
480 posts
#42
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I was absolutely delighted by this problem. Really loved this!! :love:

The solutions are $f(x) \equiv x^d + c$ and $f(x) \equiv -x^d + c$, where $d>0$ and $d \mid n$, $c\in\mathbb Z$.

Note that scaling the function by a constant does not effect the condition $f(x) - f(y) \mid x^n - y^n$. So WLOG assume that $f(0) = 0$. Also note that if $f$ works, then $-f$ works too. So multiplying the entire function by $-1$ does not change the condition either. We will use this fact later on.

Firstly note that $f$ is injective. Otherwise let $f(a) = f(b)$ for $a\neq b$. But then substituting $P(a,b)$ gives a contradiction as the divisor becomes undefined. Now note that $P(1,0) \implies f(1) \mid 1$ and $P(-1,0) \implies f(-1) \mid -1$. So we get that $f(1),f(-1) \in \left\{+1,-1\right\}$. Now using the injectivity, we get that one of them must equal $1$. So let, $f(a) = 1$.

Let $\left\{p_i\right\}$ be the sequence of primes.

$P(p_i,0) \implies f(p_i) \mid p_i^n \implies f(p_i) = p_i^{k_i}$ for some $1 \le k_i \le n$.

I claim that there are infinitely many $i$ such that $f(p_i) = p_i^{d_i}$ where $d_i$ is a positive divisor of $n$. Suppose on the contrary that there are finitely many such $i$. Thus there are infinitely many $j$ for which $f(p_j) = p_j^{k_j}$ where $k_j$ is not a divisor of $n$. By infinite PHP, we get a sequence of primes $\left\{q_i\right\}$ for which $f(q_i) = q_i^k$ where $k$ is fixed.

$P(q_i,a) \implies f(q_i) - 1 \mid q_i^n - a^n \implies q_i^k - 1 \mid q_i^n - a^n$.

Now let $n = ks + t$ where $0 < t < k$. Then we get that,
\[ q_i^k - 1 \mid q_i^n -a^n \equiv (q_i^k)^s \cdot q_i^t - a^n \equiv (1)^s \cdot q_i^t - a^n = q_i^t - a^n. \]
But then note that $a^n$ is just a constant. So after some sufficiently large $q_i$, we get that $q_i > a^n$. Thus we get that $q_i^k -1 \le q_i^t - a^n$ for all large enough $q_i$. We obviously have that $k > t$ and thus by taking a very large $q_i$, we get a contradiction.

Thus we must have had that there are infinitely many $i$ such that $f(p_i) = p_i^{d_i}$ where $d_i$ is a positive divisor of $n$. Now again by infinite PHP, we get that $f(p_i) = p_i^d$ where $d$ is a fixed divisor of $n$.

Now we fix some $u \in \mathbb Z$. Then note that $f(u) - f(p_i) \mid f(u)^{n/d} - f(p_i)^{n/d}= f(u)^{n/d} - p_i^n$.

Then we have that,
\[ P(u,p_i) \implies f(u) - f(p_i) \mid u^n - p_i^n \equiv (u^n - p_i^n) - (f(u)^{n/d} - p_i^n) = u^n - f(u)^{n/d} \implies f(u) - p_i^d \mid u^n - f(u)^{n/d}. \]
Now taking a sufficiently large $p_i$, we get that $u^n - f(u)^{n/d} \equiv 0$ that is $f(u) = u^d$. Now since $u$ was arbitrary, and we are done. :yoda:
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shendrew7
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#43
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We claim our only answers are $\boxed{f(x)=-x^a+b, ~ f(x)=-x^a+b}$, where $a \mid n$. Denote the assertion as $A(x,y)$.

Shifting tells us $b$ can be any integer, so we can assume WLOG $f(0)=0$. $A(1,0)$ gives us the $\pm$, so we can assume WLOG $f(1)=1$, from which $A(-1,0)$ forces $f(-1)=-1$.
  • Consider an arbitrarily large prime $p$. Then $A(p,0)$ and $A(p,1)$ implies $f(p)=p^a$, where $a \mid n$.
  • Consider a prime $q<p$. Then $A(p,q)$ says
    \[p^a-q^b = f(p)-f(q) \mid p^n-q^n, \quad p^a-q^b \mid p^n-q^{nb/a}.\]
    Hence the LHS must also divide $q^{nb/a}-q^n$. Since $p$ is arbitrarily large, we must have $\frac{nb}{a}=n$, or $a=b$, so $f(x)=x^a$ for all primes.
  • Fix an integer $x$, and let $n=ka+r$, where $0 \leq r \leq k-1$. Now $A(p,x)$ says
    \[p^a-f(x) = f(p)-f(x) \mid p^n-x^n = p^r \cdot f(p)^k-x^n,\]\[p^a-f(x) \mid p^r \left(f(p)^k-f(x)^k\right).\]
    Hence the LHS must also divide $p^r \cdot f(x)^k - x^n$, from which the size and infinite possibilities of $p$ forces this quantity to be 0 and $r=0$. Hence $f(x)=x^a$ for all $x$. $\blacksquare$
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megarnie
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#44 • 1 Y
Y by bjump
The answer is $x^d + C$ and $-x^d + C$ for any positive integer $d\mid n$ and integer constant $C$. These clearly work. Now we show they are the only solutions.

Let $P(x,y)$ denote the assertion that\[ f(x) - f(y) \mid x^n - y^n\]Since $f$ works iff $x + c$ works, we may WLOG that $f(0) = 0$. Now, since $f$ works iff $-f$ works, we may WLOG $f(1) \ge 0$. It suffices to show that $f(x) = x^d$ for some positive integer $d$ dividing $n$.

Claim: $f$ is injective
Proof: If $f(a) = f(b)$, then $P(a,b)$ gives $0\mid a^n - b^n$, so $a^n = b^n \implies a = b$. $\square$

$P(x,0): f(x) \mid x^n$

Now setting $x = 1$ here gives that $f(1) \mid 1$. Since $f(1) \ge 0$, $f(1) = 1$. Similarly, setting $x = -1$ gives $f(-1) \mid 1$. Since $f(1) \ne f(-1)$, we have $f(-1) = -1$.

For any prime $p$, $P(p,0)$ gives that $f(p) \mid p^n$, so $|f(p)|$ must be a power of $p$.

Claim: For any prime $p$, we have $f(p) > 0$.
Proof: Suppose otherwise. By injectivity, $f(p) \ne 0$. Let $f(p) = -p^k$ for some positive integer $k \le n$.

$P(p,1)$ gives that $p^k + 1 \mid p^n - 1$, so $p^k + 1 \mid p^{nk} - 1$. Since $n$ is odd, we also have $p^k + 1\mid p^{nk} + 1$, so $p^k + 1 \mid 2$, which is absurd. $\square$

Hence $f(p)$ is a power of $p$ for any prime $p$. Then by infinite pigeonhole there exists a positive integer $d \le n$ such that infinitely many primes $p$ satisfy $f(p) = p^d$.

For any such prime $p$, $P(p,1)$ gives $p^d - 1 \mid p^n - 1$. Now $p^n \equiv 1\pmod{p^d - 1}$, so $\frac{p^n}{p^{dk}}$ is also $1\pmod{p^d - 1}$, meaning that $p^d - 1 \mid p^{n - kd} - 1$ for any positive integer $k$. If $d \nmid n$, we could choose $k$ such that $1 \le a = n - kd \le d - 1$. We have $p^d  - 1 \mid p^a - 1$, which is a contradiction by size as $1 \le a < d$. Therefore, $d \mid n$.

Now, if $p$ is a prime with $f(p) = p^d$, then $P(x,p)$ gives $f(x) - p^d \mid x^n - p^n$. Hence $f(x) - p^d \mid x^{nd} - p^{nd}$.

Since $f(x) - p^d \mid f(x)^n - p^{nd}$, we have\[f(x) - p^d \mid x^{nd} - p^{nd} - (f(x)^n - p^{nd}) = x^{nd} - f(x)^n \]Taking $p$ sufficiently large gives that $x^{nd} = f(x)^n$, so $f(x) = x^d$, as desired.
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AshAuktober
1006 posts
#45
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Shift and invert $f$ so that $f(0) = 0, f(1) = 1$. Then some case-checking and bounding shows that for all primes $p$, $f(p) = p^a$ with $a \mid n$. Choose the $a$ that appears infinitely many times (which exists by infinite PHP).
Now for $y = p$ satisfying the condition, $f(x) - y^a \mid x^n - y^n \implies f(x) - y^a \mid x^n - f(x)^{\frac{n}{a}}$. As this means the quantity on the RHS has infinitely many divisors, we do indeed have $f(x) = x^a$ for fixed $a \mid n$. Un-transforming, the general function is $f(x) = cx^a + d,$ where $c \in \{-1, 1\}, d \in \mathbb{Z}, a \mid n$.
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Zsnim
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#46
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Note for the start of the solution, for simplicity, we only play around for the answer $f(x)=x^k+c$ for $k\mid n$, but the solution $f(x)=-x^k+c$ for $k\mid n$ can be gotten by completely analog means.

Claim:
$f(p)=p^k+f(0)$ for some $k\mid n$

Proof:
Consider the assertion $P(p,0)$ where $p$ is a prime, we get:
\[
f(p)-f(0) \mid p^n \implies f(p)=p^k+f(0)  \quad \text{for some $k\leq n$}
\]Now consider the assertions $P(p,q)$ where $p$ and $q$ are prime:
\[
f(p)-f(q) \mid p^n-q^n \iff p^k-q^l \mid p^n - q^n
\]Notice that this is an expression on which we can use the Euclidean algorithm and get something beneficial:
\[
p^k-q^l \mid p^n - q^n-(p^n-q^{l}\cdot p^{n-k}) \implies p^k-q^l \mid p^{n-k} - q^{n-l}
\]WLOG assume that $k>l$, which means that the power of $p$ is going to be the first one to dip under its respective power (basically, what I mean is that after appyling Euclidean algorithm we will get that the power of $p$ is smaller then $k$). So after applying the Euclidean algorithm a couple of times, we will get
\[
p^k-q^l \mid p^x - q^y  \quad \text{where} \quad x<k
\]But now we can just say that we pick $p$ which is large enough so and $q$ small enough, in that way we can make it so that $p^k-q^l > p^x - q^y $
Thus we conclude that $k=l$, now we are looking at (also assuming $p\neq q$)
\[
p^k-q^k \mid p^n-q^n
\]We can now proceed by the Euclidean algorithm or we can simply just scream out cyclotomic polynomials and conclude $k\mid n$

Claim:
$f(p^a)=p^{ak}+f(0)$ for any integer $a$ and for $k\mid n$

Proof:
This is basically as above, only exception is a little uglier exponents

Claim:
$f(pq)=(pq)^k+f(0)$ for some $k\mid n$

Proof:
Okay so we have some structure for primes, but let's extend this to integers which are made up of $2$ primes. By simillar methods as above, we have that $f(pq)=p^k\cdot q^l+f(0)$, and once again, consider the assertion $P(pq,q)$ we get
\[
p^kq^l-q^r\mid p^nq^n-q^n \iff q^r(p^kq^{l-r}-1)\mid q^n(p^n-1) 
\]Now we have $p^kq^{l-r}-1\mid (p^n-1)$, since we can take $q$ to be sufficiently large, we either have $p=1$ (which is a no) or $l=r$

Remark:
We immediately saw that we must have $l\geq r$, in the other case we get \[
p^k-q^{r-l} \mid p^n-1
\]But this is absurd since we can make $p^n-1$ have infinitely many divisors by moving $q$ around

Now assume we consider the assertion $P(pq,p)$, we get something which looks like
\[
p^kq^l-p^r\mid p^nq^n-p^n 
\]By the same reasoning as above we can conclude that $r=l$, hence proving that the degree of $p$ and $q$ is the same.

Claim:
$f(p^aq^b)=(p^aq^b)^k+f(0)$ for $k\mid n$

Proof:
By considering $P(p^aq^b,0)$ we get that $f(p^aq^b)=p^xq ^y$ such that $x\leq an$ and $y \leq bn$. Now we consider the assertion $P(p^aq^b, p^a)$:
\[
p^x(p^{ak-x}-q^y)\mid p^{an}(q^{an}-1) 
\]Since $p^{ak-x}-q^y \nmid p^{an}$ we have
\[
p^{ak-x}-q^y \mid q^{an}-1
\]But since we can select a huge $p$ we have that either $q=1$ (a no), or $ak=x$ (which is a yes)


Claim:
$f(n)=(n)^k+f(0)$ for $k\mid n$

Proof:
Let $n=p_1^{\alpha_1}\cdot p_2^{\alpha_2} \cdots p_m^{\alpha_m}$

This is a generalization of the above. We consider the following assertions
\[
P(n, p_1^{\alpha_1}) \quad P(n,p_2^{\alpha_2}) \quad \dots \quad P(p_m^{\alpha_m})
\]We basically prove the claim for every prime $p$, as seen when we have only $2$ primes, this simply works because we can always selects a huge prime that is not present in the denominator.


Hence finally we can conclude: $f(x)=\pm x^k+c$ for $k\mid n$ and any integer $c$
This post has been edited 1 time. Last edited by Zsnim, Jan 7, 2025, 8:09 PM
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math004
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#47
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Note that if $f$ is a solution then so is $-f+a$ for some constant $a.$ Thus, we can suppose WLOG that $f(0)=0$ and $f(1)=1.$

\[P(p,0) : f(p) \mid p^n\]which implies that $f(p)=\pm p^k$ for some non negative integer $k.$ Now, $P(1,p)$ gives that $1\pm p^k \mid p^n -1.$ If $f(p)=-p^k,\quad O_{1-p^k}(p)\mid (2k,n)$ but does not divide $k$ which is impossible. Hence $f(p)$ is a power of $p$ for all primes $p.$ Moreover, \[p^k-1\mid p^n-1 \implies O_{p^k-1}(p)\mid n  \iff k \mid n.\]By piegonhole principle, there is a infinity of prime numbers and a fixed divisor of $n$ named $c,$ such that $f(p)=p^c.$ Now, fix $x$ and take a large enough such a prime and observe that
\[0\equiv  x^n-p^n =x^n -{p^{c}}^{\frac{n}{c}} \equiv x^n-f(x)^{\frac{n}{c}}\pmod{f(x)-p^c}\]For large enough $p^c,$ we have $x^n=f(x)^{n/c} \iff f(x)=x^c.$ Whence the answer is $f\equiv \pm x^c+a  $ for some constant $a$ and $c$ divisor of $n.$
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OronSH
1745 posts
#49 • 3 Y
Y by megarnie, megahertz13, gvh300
We claim $f(x)=(-1)^ix^d+c$ for $d\mid n$ are the solutions, which clearly work.

Shift so that $f(0)=0$. Then $P(p,0)$ implies $f(p)=\pm p^d\mid p^n$. Additionally $P(\pm 1,0)$ gives $P(\pm 1)=\pm 1$ so the product of $P(p,\pm 1)$ gives $p^{2d}-1\mid p^{2n}-1$. By a euclidean algorithm argument, $2d\mid 2n$ so $d\mid n$. Thus there exists some $i\in\{0,1\}$ and $d\mid n$ for which $f(p)=(-1)^ip^d$ for infinitely many primes $p$.

Then $P(x,p)$ gives $f(x)-(-1)^ip^d\mid x^n-p^n$, but $f(x)-(-1)^ip^d\mid f(x)^{\frac nd}-(-1)^ip^n$ so combining these we have $f(x)-(-1)^ip^d\mid f(x)^{\frac nd}-(-1)^ix^n$ for infinitely many $p$, implying $f(x)^{\frac nd}-(-1)^ix^n=0$, or $f(x)=(-1)^ix^d$ for all $x$. Shifting back gives the desired.
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InterLoop
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#50
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solution
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Ilikeminecraft
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#51
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Observe that if $f(x)$ works, then so do $f(x) + c, -f(x) + c$. Hence, assume that $f(0) = 0, f(1) = 1. $ Let $p$ be a prime. We have that plugging in $x = p, y = 0,$ we have that $f(p)\mid p^n.$ Hence, assume that $f(p) = \pm p^d.$ Plugging in $x = p, y = 1,$ we have that $1\pm p^d \mid p^n - 1.$ Clearly, $f(p) = p^d$ since $n$ is odd. Furthermore, we have that $p^d -1\mid p^n - 1\implies d \mid n.$

Now, pick $x = p, y = k.$ It follows that $p^d - f(k)\mid p^n - k^n.$ However, $k^n - p^n \equiv k^n - f(k)^{\frac nd}\pmod{p^d - f(k)}.$ By taking some really large $p,$ it follows that $k^n = f(k)^{\frac nd},$ and thus, $f(k) = k^d.$
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