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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
Six variables
Nguyenhuyen_AG   1
N 21 minutes 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
21 minutes ago
Anything real in this system must be integer
Assassino9931   3
N 26 minutes 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
26 minutes ago
Interesting inequalities
sqing   3
N 35 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0 , (a+k )(b+c)=k+1.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2k-3+2\sqrt{k+1}}{3k-1}$$Where $ k\geq \frac{2}{3}.$
Let $ a,b,c\geq 0 , (a+1)(b+c)=2.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 2\sqrt{2}-1$$Let $ a,b,c\geq 0 , (a+3)(b+c)=4.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{7}{4}$$Let $ a,b,c\geq 0 , (3a+2)(b+c)= 5.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2(2\sqrt{15}-5)}{3}$$
3 replies
sqing
Yesterday at 1:29 PM
sqing
35 minutes ago
Interesting inequalities
sqing   1
N 40 minutes ago by sqing
Source: Own
Let $ a,b>0 $ and $ a+b\leq 1  $ . Prove that
$$\left(\frac{1}{a^3}-1\right)\left(\frac{1}{b^3}-1\right)-k\left(\frac{a}{b}+\frac{b}{a}\right) \geq 49-2k$$Where $24\geq k\in N^+.$
$$\left(\frac{1}{a^3}-1\right)\left(\frac{1}{b^3}-1\right) \geq 49$$$$\left(\frac{1}{a^3}-1\right)\left(\frac{1}{b^3}-1\right)-25\left(\frac{a}{b}+\frac{b}{a}\right) \geq -\frac{13}{12}$$$$\left(\frac{1}{a^3}-1\right)\left(\frac{1}{b^3}-1\right)-26\left(\frac{a}{b}+\frac{b}{a}\right) \geq -\frac{10}{3}$$$$\left(\frac{1}{a^3}-1\right)\left(\frac{1}{b^3}-1\right)-27\left(\frac{a}{b}+\frac{b}{a}\right) \geq -\frac{23}{4}$$
1 reply
sqing
44 minutes ago
sqing
40 minutes ago
Tetrahedron
4everwise   3
N Yesterday at 10:43 PM by aidan0626
Four balls of radius 1 are mutually tangent, three resting on the floor and the fourth resting on the others. A tetrahedron, each of whose edges have length $s$, is circumscribed around the balls. Then $s$ equals

$\text{(A)} \ 4\sqrt 2 \qquad \text{(B)} \ 4\sqrt 3 \qquad \text{(C)} \ 2\sqrt 6 \qquad \text{(D)} \ 1+2\sqrt 6 \qquad \text{(E)} \ 2+2\sqrt 6$
3 replies
4everwise
Jan 1, 2006
aidan0626
Yesterday at 10:43 PM
Concurrent in a pyramid
vanstraelen   0
Yesterday at 7:13 AM

Given a pyramid $(T,ABCD)$ where $ABCD$ is a parallelogram.
The intersection of the diagonals of the base is point $S$.
Point $A$ is connected to the midpoint of $[CT]$, point $B$ to the midpoint of $[DT]$,
point $C$ to the midpoint of $[AT]$ and point $D$ to the midpoint of $[BT]$.
a) Prove: the four lines are concurrent in a point $P$.
b) Calulate $\frac{TS}{TP}$.
0 replies
vanstraelen
Yesterday at 7:13 AM
0 replies
Triangle on a tetrahedron
vanstraelen   2
N Friday at 7:51 PM by ReticulatedPython

Given a regular tetrahedron $(A,BCD)$ with edges $l$.
Construct at the apex $A$ three perpendiculars to the three lateral faces.
Take a point on each perpendicular at a distance $l$ from the apex such that these three points lie above the apex.
Calculate the lenghts of the sides of the triangle.
2 replies
vanstraelen
Friday at 2:43 PM
ReticulatedPython
Friday at 7:51 PM
shadow of a cylinder, shadow of a cone
vanstraelen   2
N Friday at 6:33 PM by vanstraelen

a) Given is a right cylinder of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the botom base?

b) Given is a right cone of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the base?
2 replies
vanstraelen
Friday at 3:08 PM
vanstraelen
Friday at 6:33 PM
Cube Sphere
vanstraelen   4
N Friday at 2:37 PM by pieMax2713

Given the cube $\left(\begin{array}{ll} EFGH \\ ABCD \end{array}\right)$ with edge $6$ cm.
Find the volume of the sphere passing through $A,B,C,D$ and tangent to the plane $(EFGH)$.
4 replies
vanstraelen
Friday at 1:10 PM
pieMax2713
Friday at 2:37 PM
parallelogram in a tetrahedron
vanstraelen   1
N Friday at 12:19 PM by vanstraelen
Given a tetrahedron $ABCD$ and a plane $\mu$, parallel with the edges $AC$ and $BD$.
$AB \cap \mu=P$.
a) Prove: the intersection of the tetrahedron with the plane is a parallelogram.
b) If $\left|AC\right|=14,\left|BD\right|=7$ and $\frac{\left|PA\right|}{\left|PB\right|}=\frac{3}{4}$,
calculates the lenghts of the sides of this parallelogram.
1 reply
vanstraelen
May 5, 2025
vanstraelen
Friday at 12:19 PM
Regular tetrahedron
vanstraelen   7
N May 6, 2025 by ReticulatedPython
Given the points $O(0,0,0),A(1,0,0),B(\frac{1}{2},\frac{\sqrt{3}}{2},0)$
a) Determine the point $C$, above the xy-plane, such that the pyramid $OABC$ is a regular tetrahedron.
b) Calculate the volume.
c) Calculate the radius of the inscribed sphere and the radius of the circumscribed sphere.
7 replies
vanstraelen
May 4, 2025
ReticulatedPython
May 6, 2025
volume 9f a pentagonal base pyramid circumscribed around a right circular cone
FOL   1
N May 6, 2025 by Mathzeus1024
A pentagonal base pyramid is circumscribed around a right circular cone, whose height is equal to the radius of the base. The total surface area of the pyramid is d times greater than that of the cone. Find the volume of the pyramid if the lateral surface area of the cone is equal to $\pi\sqrt{2}$.
1 reply
FOL
Jul 22, 2023
Mathzeus1024
May 6, 2025
Geometry books
T.Mousavidin   4
N Apr 30, 2025 by compoly2010
Hello, I wanted to ask if anybody knows some good books for geometry that has these topics in:
Desargues's Theorem, Projective geometry, 3D geometry,
4 replies
T.Mousavidin
Apr 29, 2025
compoly2010
Apr 30, 2025
Tetrahedrons and spheres
ReticulatedPython   4
N Apr 28, 2025 by soryn
Let $OABC$ be a tetrahedron such that $\angle{AOB}=\angle{AOC}=\angle{BOC}=90^\circ.$ A sphere of radius $r$ is circumscribed about tetrahedron $OABC.$ Given that $OA=a$, $OB=b$, and $OC=c$, prove that $$r^2+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9\sqrt[3]{4}}{4}$$with equality at $a=b=c=\sqrt[3]{2}.$
4 replies
ReticulatedPython
Apr 21, 2025
soryn
Apr 28, 2025
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
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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
8863 posts
#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
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#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
Reason: LaTeX fixing
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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.
This post has been edited 1 time. Last edited by OronSH, Feb 21, 2025, 8:59 PM
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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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