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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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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
x^2 + 3y^2 = 8n + 4
Ink68   0
5 minutes ago
Let $n$ be a positive integer. Let $A$ be the number of pairs of $(x,y)$ satisfying $x^2 + 3y^2 = 8n + 4$ for odd values of $x$. Let $B$ be the number of pairs of $(x,y)$ satisfying $x^2 + 3y^2 = 8n + 4$. Prove that $A = \frac {2}{3} B$.
0 replies
Ink68
5 minutes ago
0 replies
At least k points of S equidistant from P
orl   9
N 26 minutes ago by Twan
Source: IMO 1989/3 , ISL 20, ILL 66
Let $ n$ and $ k$ be positive integers and let $ S$ be a set of $ n$ points in the plane such that

i.) no three points of $ S$ are collinear, and

ii.) for every point $ P$ of $ S$ there are at least $ k$ points of $ S$ equidistant from $ P.$

Prove that:
\[ k < \frac {1}{2} + \sqrt {2 \cdot n}
\]
9 replies
orl
Nov 19, 2005
Twan
26 minutes ago
Gergonne point Harmonic quadrilateral
niwobin   1
N 35 minutes ago by on_gale
Triangle ABC has incircle touching the sides at D, E, F as shown.
AD, BE, CF concurrent at Gergonne point G.
BG and CG cuts the incircle at X and Y, respectively.
AG cuts the incircle at K.
Prove: K, X, D, Y form a harmonic quadrilateral. (KX/KY = DX/DY)
1 reply
niwobin
Yesterday at 8:17 PM
on_gale
35 minutes ago
Find the minimum
sqing   8
N an hour ago by sqing
Source: China Shandong High School Mathematics Competition 2025 Q4
Let $ a,b,c>0,abc>1$. Find the minimum value of $ \frac {abc(a+b+c+8)}{abc-1}. $
8 replies
sqing
Yesterday at 9:12 AM
sqing
an hour ago
Interesting inequalities
sqing   3
N an hour ago by sqing
Source: Own
Let $ a,b >0 $ and $ a^2-ab+b^2\leq 1 $ . Prove that
$$a^4 +b^4+\frac{a }{b +1}+ \frac{b }{a +1} \leq 3$$$$a^3 +b^3+\frac{a^2}{b^2+1}+ \frac{b^2}{a^2+1} \leq 3$$$$a^4 +b^4-\frac{a}{b+1}-\frac{b}{a+1} \leq 1$$$$a^4+b^4 -\frac{a^2}{b^2+1}- \frac{b^2}{a^2+1}\leq 1$$$$a^3+b^3 -\frac{a^3}{b^3+1}- \frac{b^3}{a^3+1}\leq 1$$
3 replies
sqing
May 9, 2025
sqing
an hour ago
Marking vertices in splitted triangle
mathisreal   2
N an hour ago by sopaconk
Source: Mexico
Let $n$ be a positive integer. Consider a figure of a equilateral triangle of side $n$ and splitted in $n^2$ small equilateral triangles of side $1$. One will mark some of the $1+2+\dots+(n+1)$ vertices of the small triangles, such that for every integer $k\geq 1$, there is not any trapezoid(trapezium), whose the sides are $(1,k,1,k+1)$, with all the vertices marked. Furthermore, there are no small triangle(side $1$) have your three vertices marked. Determine the greatest quantity of marked vertices.
2 replies
mathisreal
Feb 7, 2022
sopaconk
an hour ago
distance of a point from incircle equals to a diameter of incircle
parmenides51   5
N an hour ago by Captainscrubz
Source: 2019 Oral Moscow Geometry Olympiad grades 8-9 p1
In the triangle $ABC, I$ is the center of the inscribed circle, point $M$ lies on the side of $BC$, with $\angle BIM = 90^o$. Prove that the distance from point $M$ to line $AB$ is equal to the diameter of the circle inscribed in triangle $ABC$
5 replies
parmenides51
May 21, 2019
Captainscrubz
an hour ago
f(a + b) = f(a) + f(b) + f(c) + f(d) in N-{O}, with 2ab = c^2 + d^2
parmenides51   8
N 5 hours ago by TiagoCavalcante
Source: RMM Shortlist 2016 A1
Determine all functions $f$ from the set of non-negative integers to itself such that $f(a + b) = f(a) + f(b) + f(c) + f(d)$, whenever $a, b, c, d$, are non-negative integers satisfying $2ab = c^2 + d^2$.
8 replies
parmenides51
Jul 4, 2019
TiagoCavalcante
5 hours ago
Functional Inequality Implies Uniform Sign
peace09   33
N 5 hours ago by ezpotd
Source: 2023 ISL A2
Let $\mathbb{R}$ be the set of real numbers. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that \[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]for every $x,y\in\mathbb{R}$. Assume that the inequality is strict for some $x_0,y_0\in\mathbb{R}$.

Prove that either $f(x)\geqslant 0$ for every $x\in\mathbb{R}$ or $f(x)\leqslant 0$ for every $x\in\mathbb{R}$.
33 replies
peace09
Jul 17, 2024
ezpotd
5 hours ago
Labelling edges of Kn
oVlad   1
N 6 hours ago by TopGbulliedU
Source: Romania Junior TST 2025 Day 2 P3
Let $n\geqslant 3$ be an integer. Ion draws a regular $n$-gon and all its diagonals. On every diagonal and edge, Ion writes a positive integer, such that for any triangle formed with the vertices of the $n$-gon, one of the numbers on its edges is the sum of the two other numbers on its edges. Determine the smallest possible number of distinct values that Ion can write.
1 reply
oVlad
May 6, 2025
TopGbulliedU
6 hours ago
c^a + a = 2^b
Havu   8
N Yesterday at 8:29 PM by MathematicalArceus
Find $a, b, c\in\mathbb{Z}^+$ such that $a,b,c$ coprime, $a + b = 2c$ and $c^a + a = 2^b$.
8 replies
Havu
May 10, 2025
MathematicalArceus
Yesterday at 8:29 PM
Concurrence of lines defined by intersections of circles
Lukaluce   1
N Yesterday at 8:28 PM by sarjinius
Source: 2025 Macedonian Balkan Math Olympiad TST Problem 2
Let $\triangle ABC$ be an acute-angled triangle and $A_1, B_1$, and $C_1$ be the feet of the altitudes from $A, B$, and $C$, respectively. On the rays $AA_1, BB_1$, and $CC_1$, we have points $A_2, B_2$, and $C_2$ respectively, lying outside of $\triangle ABC$, such that
\[\frac{A_1A_2}{AA_1} = \frac{B_1B_2}{BB_1} = \frac{C_1C_2}{CC_1}.\]If the intersections of $B_1C_2$ and $B_2C_1$, $C_1A_2$ and $C_2A_1$, and $A_1B_2$ and $A_2B_1$ are $A', B'$, and $C'$ respectively, prove that $AA', BB'$, and $CC'$ have a common point.
1 reply
Lukaluce
Apr 14, 2025
sarjinius
Yesterday at 8:28 PM
Factorial Divisibility
Aryan-23   45
N Yesterday at 8:28 PM by MathematicalArceus
Source: IMO SL 2022 N2
Find all positive integers $n>2$ such that
$$ n! \mid \prod_{ p<q\le n, p,q \, \text{primes}} (p+q)$$
45 replies
Aryan-23
Jul 9, 2023
MathematicalArceus
Yesterday at 8:28 PM
Multiple of multinomial coefficient is an integer
orl   14
N Yesterday at 8:23 PM by mickeymouse7133
Source: Romanian Master in Mathematics 2009, Problem 1
For $ a_i \in \mathbb{Z}^ +$, $ i = 1, \ldots, k$, and $ n = \sum^k_{i = 1} a_i$, let $ d = \gcd(a_1, \ldots, a_k)$ denote the greatest common divisor of $ a_1, \ldots, a_k$.
Prove that $ \frac {d} {n} \cdot \frac {n!}{\prod\limits^k_{i = 1} (a_i!)}$ is an integer.

Dan Schwarz, Romania
14 replies
orl
Mar 7, 2009
mickeymouse7133
Yesterday at 8:23 PM
IMO Shortlist 2011, Number Theory 3
orl   48
N May 11, 2025 by Markas
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
48 replies
orl
Jul 11, 2012
Markas
May 11, 2025
IMO Shortlist 2011, Number Theory 3
G H J
Source: IMO Shortlist 2011, Number Theory 3
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orl
3647 posts
#1 • 8 Y
Y by Davi-8191, anantmudgal09, jhu08, Adventure10, Mango247, BorivojeGuzic123, cubres, and 1 other user
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
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MellowMelon
5850 posts
#2 • 10 Y
Y by theproductof6by9, Anar24, jhu08, Adventure10, BorivojeGuzic123, Newmaths, and 4 other users
We claim there exists a positive integer $d$ dividing $n$, an $a \in \mathbb{Z}$, and an $e \in \{-1,1\}$ such that $f(x) = e \cdot x^d + a$ for all $x$. It can be easily checked that all of these solutions work.

First, notice that if $f(x)$ is a solution, then $f(x) + a$ is a solution for any $a$. So WLOG $f(0) = 0$. Now take $x = 1$, $y = 0$ in the given equation. We get $f(1) \mid 1$, so $f(1) = \pm 1$. If $f(x)$ is a solution, then $-f(x)$ is a solution, so WLOG $f(1) = 1$. We also get $f(-1) \mid 1$ by taking $x = -1$ and $y = 0$. Now take $x = -1$ and $y = 1$. We get $f(-1) - 1 \mid 2$, so $f(-1)$ must be $-1$.

Consider any odd prime $p$. We have $f(p) - f(0) \mid p^n$, so therefore $f(p)$ is either $p^d$ or $-p^d$ for some $d$. First consider the case that $f(p) = -p^d$. We have $f(p) - f(-1) \mid p^n + 1$, so $p^d - 1 \mid p^n + 1$. If we write $n = qd + r$ using the division algorithm, we get $p^d - 1 \mid p^r + 1$. We know $0 \leq r < d$, so $0 < p^r + 1 < p^d - 1$ since $p > 2$, a contradiction. So this case is impossible

Then we must have $f(p) = p^d$. Again write $n = qd + r$ using the division algorithm. We get $p^d - 1 \mid (-1)^q p^r - 1$. Since $0 \leq r < d$, this is only possible if $r = 0$. Therefore, $d$ divides $n$.

Let $b$ be an arbitrary integer, and let $q$ be a prime number greater than $b^n + 2|f(b)|^n$.
We know $f(q) = q^d$ for some $d$ dividing $n$. Consider $x = b$ and $y = q$. We get $f(b) - q^d \mid b^n - q^n$. Writing $q^n$ as $(q^d)^{n/d}$, we get $f(b) - q^d \mid b^n - (f(b))^{n/d}$. Suppose that $b^n - (f(b))^{n/d}$ is nonzero. Then we know that
\[|b^n - (f(b))^{n/d}| \leq b^n + |f(b)|^n < q - |f(b)|^n \leq q^d - f(b).\]
This is a contradiction, so therefore $b^n - (f(b))^{n/d} = 0$, and $f(b) = b^d$. If we apply this process to two arbitrary integers $b,c$, taking the same large $q$ for both of them, we obtain that $f(b) = b^d$ and $f(c) = c^d$ for the same $d$. Therefore there exists a single $d$ dividing $n$ such that $f(x) = x^d$ for all integers $x$. Reversing our transformations in the second paragraph, we get the set of solutions described in the first paragraph, so we are done.
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leshik
433 posts
#3 • 3 Y
Y by jhu08, Adventure10, and 1 other user
This problem has been posted before: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=453800
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hal9v4ik
368 posts
#4 • 4 Y
Y by jhu08, Adventure10, Mango247, and 1 other user
Can someone post IMO Shortlist 2011 on contest page?
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KittyOK
349 posts
#5 • 3 Y
Y by jhu08, Adventure10, Mango247
The formulation of the problem is not quite clear. If we plug $x=y$ in the condition we get the nonsense $0$ divides $0$.
Evidently, the official solution tacitly uses that $f$ is injective and that the condition applies only with distinct $x$ and $y$.
A more interesting interpretation is as follows:
Quote:
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 $y$ such that $f(x)$ is not equal to $f(y)$, the difference $f(x)-f(y)$ divides $x^n-y^n.$
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ACCCGS8
326 posts
#6 • 2 Y
Y by jhu08, Adventure10
$0$ divides $0$ is actually true, because there exists an integer $k$ such that $(0)(k)=0$. So I don't think there is any confusion in the problem.
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mavropnevma
15142 posts
#7 • 4 Y
Y by jhu08, Adventure10, Mango247, and 1 other user
And injectivity of $f$ results from the given relation(s), since assuming $f(x)=f(y)$ for some $x\neq y$ leads to $0 \mid x^n-y^n \neq 0$ (don't forget, $n$ is odd), so it does not need to be tacitly implied.
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harazi
5526 posts
#8 • 3 Y
Y by jhu08, Adventure10, Mango247
Maybe a more interesting question: find all functions $f$ defined on $\textbf{positive integers}$ with integer values and with the same property as in the problem.
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eaglet
6 posts
#9 • 3 Y
Y by jhu08, Adventure10, Mango247
mavropnevma wrote:
And injectivity of $f$ results from the given relation(s), since assuming $f(x)=f(y)$ for some $x\neq y$ leads to $0 \mid x^n-y^n \neq 0$ (don't forget, $n$ is odd), so it does not need to be tacitly implied.

I just wanted to express my mild surprise at this explanation. If my understanding is correct, the statement $ 0\mid x^{n}-y^{n}\neq 0 $ is nonsense rather than false, and the responsibility of eliminating such nonsense statements lies not with the problem solver, but with the problem proposer. In this particular case, the proposer could have defined $f$ to be injective, a definition that would have left no ambiguity in the problem. Now, I could be totally inaccurate on this explanation attempt, and if so, please kindly enlighten me.

On another note, I would like to confirm that KittyOK's interpretation is solvable, and indeed a much more interesting problem than the one originally intended.

And to KittyOK, my warmest congratulations once again for your gold medal at IMO 2012. :)
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mavropnevma
15142 posts
#10 • 5 Y
Y by eaglet, jhu08, Adventure10, Mango247, and 1 other user
My pleasure. Of course, the statement should have contained the domain of the functional relation excluding its diagonal, so

Let $n \geq 1$ be an odd integer. Determine all functions $f$ from the set of integers to itself, such that for all distinct integers $x$ and $y$, the difference $f(x)-f(y)$ divides $x^n-y^n.$

Now, one will ask oneself: what if $f(x) = f(y)$ for some $x\neq y$. Then $v = x^n-y^n \neq 0$, since $n$ is given odd, so the given relation would be $0\mid v$.
I never said this is false; if you prefer to call it nonsense it's ok with me. What matters is that it's not TRUE, so by contradicting the given relation forbids $f(x) = f(y)$ when $x\neq y$. Now, why should the proposer have ruled out this consequence of the relation, by redundantly stating "$f$ is injective" ?

Another example. Say I write: Let $A$ be a set of positive integers and let $M$ be a number such that $a\leq M$ for all $a\in A$. Why should I "warn" the solver by stating "$A$ is finite", when it's a trivial consequence of the givens?
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Isogonics
185 posts
#11 • 2 Y
Y by jhu08, Adventure10
I have a question about MellowMelon's Solution, in the $f(p)$ part :
MellowMelon wrote:
If we write $n = qd + r$ using the division algorithm, we get $p^d - 1 \mid p^r + 1$. We know $0 \leq r < d$, so $0 < p^r + 1 < p^d - 1$ since $p > 2$, a contradiction. So this case is impossible.

Well, I think this is not true when $p=3, r=0, d=1$ , so setting $p$ any odd prime is wrong.
(But in this solution "infinitely many prime" works, so only "any odd prime" needs to be corrected.)
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junioragd
314 posts
#12 • 3 Y
Y by IgorM, jhu08, Adventure10
Here is another solution(More precise,a way to finish via Dirichlet's theoerem instead of bounding):
All solutions are if the form $a*x^b$+$c$,where $a$ is $1$ or $-1$,$d$ divides $n$ and $c$ is an arbirtary integer.Note that,if $f(x)$ is a solution,then $f(x)+a$ is also a solution and if $f(x)$ is a solution,then $-f(x)$ is a solution.Now,WLOG $f(0)=0$ and P$(1,1)$ gives $f(1)=1$.P$(2,0)$ gives $f(2)$ divides $2^n$,so $f(2)=2^k$($k<n$) and P$(2,1)$ gives $2^k-1$ divides $2^n$-$1$,so we get $k$ divides $n$.Now,pluging any prime bigger than 2^(n^2) we get $f(p)=p^k$.Now,pick an arbirtary integer $m$ and suppose $f(m)=m^k$ doesn't hold.Now,let $n=ak$.
Lemma: The equation $r^a$ congruent $l$ $modp$ always has a solution if $GCD(a,p-1)=1$($p$ is a prime).
Proof:Pick two distinct integers $m,n$ such that $p$ doesn't divide $m-n$.Suppose $p$ divides $m^a-n^a$.Now,from Ferma's little theorem we have that $p$ divides m^(p-1)-n^(p-1),now by LTE we have GCD((m^a-n^a),(m^p-1-n^p-1))=$m-n$,which is a contradiction.

Pick a prime $r$ such that $r$ doesn't divide $f(m)^a$-$m^n$ and such that $k$ doesn't divide $r-1$.By Dirichlet,such a prime exists
Now,by Dirichlet there will exist a prime $q$ such that $r$ divides $q^k-f(m)$,now P$(q,m)$ gives that $r$ divides $f(m^a)$-$m^n$,which is a contradiction,so we get $f(m)=m^k$.
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JuanOrtiz
366 posts
#13 • 2 Y
Y by jhu08, Adventure10
First notice that if $n<0$ then $f(x)-f(0)|x^n$ but the left side is an integer and $x^n$ isn't an integer for all $x$, and therefore there exists no such function $f$. If $n=0$, then if $x=0$, $0^0$ is undefined and so the problem's statement is undefined. Now assuming $n>0$, I claim that the functions that work are $f(x)=\Gamma+x^{\Omega}$ and $f(x)=\Gamma-x^{\Omega}$, for any positive divisor $\Omega | n$ and any integer $\Gamma$.

Firstly notice that we can multiply $f$ by $-1$ and shift it by a constant and it still works, and therefore WLOG $f(1) \ge 0$ and $f(0)=0$. Notice $f(1)-0|1-0$ and so $f(1)=1$. Now take $p$ be any odd prime. Notice that $f(p)|p^n$ and therefore $f(p)=p^{\Omega}$ or $-p^{\Omega}$ for some $0\le {\Omega} \le n$. In the second case, we get that $p^{\Omega}+1 | p^n-1$ and so $p+1 | p^n-1$ (or $\Omega=0 \Rightarrow f(p)=-1$) but notice that since $n$ is odd, $p+1 | p^n+1$ and so $p+1 | 2$, impossible. Therefore, $f(p)=p^{\Omega}$ for some $1 \le \Omega \le n$, or $f(p)=-1$. So $p^{\Omega}-1 | p^n-1$. It is a well known fact that $\text{mcd}(a^b-1,a^c-1) = a^{\text{mcd}(b,c)}-1$ for $a,b,c$ positive integers and $a>1$, which can be proven by Euclid's Algorithm easily. Therefore $\Omega | n$.

Now let $S$ be the set of positive integers that divide $n$. Notice that if $f(a)=f(b)=-1 \Rightarrow 0 | a-b \Rightarrow -1$ and so $f(p)=-1$ for at most one value of $p$. For all other odd primes $p$, $\Omega \in S$ and since $S$ is finite, there exists an infinite set $S_{\Omega}$ of prime numbers such that $f(p)=p^{\Omega}$ for all $p \in S_{\Omega}$. Now take any integer $x \neq -1,0-1$ and notice that for any such $p$, $p^{\Omega}-f(x) | p^n-x^n$, but also $p^{\Omega}-f(x) | p^n-f(x)^{\frac{n}{\Omega}}$ and therefore $p^{\Omega}-f(x) | x^n-f(x)^{\frac{n}{\Omega}}$. This implies that

$x^n-f(x)^{\displaystyle\frac{n}{\Omega}}$

has an infinite number of divisors, which clearly implies that said number is $0$. Therefore $f(x)={\Omega}$ for all $x \neq 0,1,-1$. But this equality is also true for $x=0,1$ as we had previously established. Now all that is left is to prove $f(-1)=-1$. But notice $f(-1)|-1$ and so $f(-1)=-1,1$. We also have $1-f(-1)|2$ and so $f(-1)=-1$. Therefore $f(x)=x^{\Omega}$ for all $x$, where $\Omega$ is a fixed positive divisor of $n$.

Taking into account that we shifted and multiplied $f$ by $-1$ at the beginning, this gives us the final answers $\boxed{f(x)=\Gamma+x^{\Omega}}$ and $\boxed{f(x)=\Gamma-x^{\Omega}}$, for any positive divisor $\Omega | n$ and any integer $\Gamma$.
This post has been edited 6 times. Last edited by JuanOrtiz, Apr 2, 2015, 10:03 PM
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anantmudgal09
1980 posts
#14 • 4 Y
Y by jhu08, SatisfiedMagma, Adventure10, Mango247
Answer. The only solutions are $f(x)=cx^d+e$ where $c \in \{-1, 1\}$, $d \ge 0$ is an odd divisor of $n$ and $e \in \mathbb{Z}$ for all integers $x$.

Solution. It is clear that these satisfy the conditions. We will show that no other function works. Indeed, suppose a function $f$ other than those specified works.

Injectivity of $f$ follows since $f(a)=f(b) \Longrightarrow a^n=b^n$ and as $n$ is odd, $a=b$. Shifting by a constant; let $f(0)=0$, and for all integers $a$ we get $f(a) \mid a^n$. As $f(1) \mid 1$, scaling by a $\pm 1$ factor, we let $f(1)=1$. For all primes $p$, $f(p) \mid p^n \Longrightarrow f(p)=\pm \, p^k$ for some $k >0$.

Firstly, let $f(p)=-p^k$ and observe that $p^k+1 \mid p^n-1$. Write $n=kl+r$ for $0 \le r<k$ and note that $$p^n \equiv p^r\cdot (-1)^l \pmod {p^k+1} \Longrightarrow p^k+1 \le \left|p^r\cdot (-1)^l-1\right| \le p^r+1,$$which is clearly false. It follows that $f(p)=p^{k_p}$ for all primes $p$ with $1 \le k_p \le n$. Evidently, there exists $1 \le d \le n$ such that $k_p=d$ holds for infinitely many primes $p$.

Fix a prime $p_1$ with $k_{p_1}=d$ and vary $p$ with $k_p=d$. Write $n=md+r$ for $0 \le r<d$ and note that $$p^d-p_1^d \mid p^n-p_1^n \Longrightarrow p_1^{md}\cdot \left(p^r-p_1^r\right) \equiv 0 \pmod {p^d-p_1^d}.$$This is fails to occur unless $d \mid n$.

Finally, fix an integer $x$ with $f(x) \ne x^d$ and vary prime $p$ with $k_p=d$. It follows that $$p^d-f(x) \mid p^n-x^n \Longrightarrow p^d-f(x) \mid f(x)^{\frac{n}{d}}-x^n,$$which fails to hold by taking $p$ sufficiently large. The initial claim holds. $\, \square$
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H.R.
172 posts
#15 • 2 Y
Y by jhu08, Adventure10
Answer. All functions f of the form f(x) = εxd + c, where ε is in {1, −1}, the integer d is a
positive divisor of n, and c is an integer.
Solution. Obviously, all functions in the answer satisfy the condition of the problem. We will
show that there are no other functions satisfying that condition.
Let f be a function satisfying the given condition. For each integer n, the function g defined
by g(x) = f(x) + n also satisfies the same condition. Therefore, by subtracting f(0) from f(x)
we may assume that f(0) = 0.
For any prime p, the condition on f with (x, y) = (p, 0) states that f(p) divides p
n
. Since the
set of primes is infinite, there exist integers d and ε with 0 ≤ d ≤ n and ε ∈ {1, −1} such that
for infinitely many primes p we have f(p) = εpd
. Denote the set of these primes by P. Since a
function g satisfies the given condition if and only if −g satisfies the same condition, we may
suppose ε = 1.
The case d = 0 is easily ruled out, because 0 does not divide any nonzero integer. Suppose
d ≥ 1 and write n as md + r, where m and r are integers such that m ≥ 1 and 0 ≤ r ≤ d − 1.
Let x be an arbitrary integer. For each prime p in P, the difference f(p)−f(x) divides p
n −x
n
.
Using the equality f(p) = p
d
, we get
p
n − x
n = p
r
(p
d
)
m − x
n ≡ p
r
f(x)
m − x
n ≡ 0 (mod p
d − f(x))
Since we have r < d, for large enough primes p ∈ P we obtain
|p
r
f(x)
m − x
n
| < pd − f(x).
Hence p
rf(x)
m − x
n has to be zero. This implies r = 0 and x
n = (x
d
)
m = f(x)
m. Since m is
odd, we obtain f(x) = x
d
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G
H
=
a