We have your learning goals covered with Spring and Summer courses available. Enroll today!

G
Topic
First Poster
Last Poster
k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Sunday, Mar 2 - Jun 22
Friday, Mar 28 - Jul 18
Sunday, Apr 13 - Aug 10
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Tuesday, Mar 25 - Jul 8
Sunday, Apr 13 - Aug 10
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21


Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, Mar 23 - Jul 20
Monday, Apr 7 - Jul 28
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Sunday, Mar 16 - Jun 8
Wednesday, Apr 16 - Jul 2
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Monday, Mar 17 - Jun 9
Thursday, Apr 17 - Jul 3
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Sunday, Mar 2 - Jun 22
Wednesday, Apr 16 - Jul 30
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Tuesday, Mar 4 - Aug 12
Sunday, Mar 23 - Sep 21
Wednesday, Apr 23 - Oct 1
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Mar 16 - Sep 14
Tuesday, Mar 25 - Sep 2
Monday, Apr 21 - Oct 13
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Sunday, Mar 23 - Aug 3
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Friday, Apr 11 - Jun 27
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Sunday, Mar 16 - Aug 24
Wednesday, Apr 9 - Sep 3
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Wednesday, Mar 5 - May 21
Tuesday, Jun 10 - Aug 26

Calculus
Sunday, Mar 30 - Oct 5
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Sunday, Mar 23 - Jun 15
Wednesday, Apr 16 - Jul 2
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Friday, Apr 11 - Jun 27
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Tuesday, Mar 4 - May 20
Monday, Mar 31 - Jun 23
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Monday, Mar 24 - Jun 16
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Sunday, Mar 30 - Jun 22
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Tuesday, Mar 25 - Sep 2
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Sat & Sun, Apr 26 - Apr 27 (4:00 - 7:00 pm ET/1:00 - 4:00pm PT)
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
Mar 2, 2025
0 replies
H not needed
dchenmathcounts   44
N 16 minutes ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
16 minutes ago
IZHO 2017 Functional equations
user01   51
N 37 minutes ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
37 minutes ago
chat gpt
fuv870   2
N 39 minutes ago by fuv870
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
2 replies
fuv870
an hour ago
fuv870
39 minutes ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 41 minutes ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
41 minutes ago
No more topics!
Another NT FE
nukelauncher   59
N 4 hours ago by AshAuktober
Source: ISL 2019 N4
Find all functions $f:\mathbb Z_{>0}\to \mathbb Z_{>0}$ such that $a+f(b)$ divides $a^2+bf(a)$ for all positive integers $a$ and $b$ with $a+b>2019$.
59 replies
nukelauncher
Sep 22, 2020
AshAuktober
4 hours ago
Another NT FE
G H J
Source: ISL 2019 N4
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ATGY
2502 posts
#53
Y by
It seemed similar to 2019 APMO P1 initially but after working on it it is just an application of Dirichlet's..

Let $P(x,y)$ be the given assertion. $P(1, b)$ yields $1 + f(b)| 1 + bf(1)$. We know that $\text{gcd}(f(1), 1) = 1$, so by Dirichlet's, there exist infinitely many $b$ for which $1 + bf(1)$ is prime. Since $f(b) > 0$, we have $1 + f(b) = 1 + bf(1) \implies f(b) = bf(1)$. Now, we use such $b$ and substitute $P(a,b)$:
$$a + f(b)| a^2 + bf(a) \implies a + bf(1) | a^2 + bf(a) \implies a + bf(1) | f(1)(a^2 + bf(a)) - f(a)(a + bf(1)) \implies a + bf(1) | a^2f(1) - af(a)$$For sufficiently large $b$, $a + bf(1) > a^2f(1) - af(a) \implies a^2f(1) - af(a) = 0 \implies f(a) = ca$ where $c$ is some constant. Substituting this back into the assertion it clearly works so we are done.
This post has been edited 2 times. Last edited by ATGY, Jan 20, 2024, 7:25 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lelouchvigeo
172 posts
#54
Y by
Solved in 10 mins. Was an application of Dirichlet's. :D
Sketch
This post has been edited 1 time. Last edited by lelouchvigeo, Jan 20, 2024, 4:35 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shendrew7
787 posts
#56
Y by
We claim our only answer is $\boxed{f(x)=cx}$. Denote the assertion as $A(a,b)$. We first rewrite our condition as
\[a+f(b) \mid (a^2+bf(a)) - a(a+f(b)) = af(b) - bf(a).\]
Consider the infinite set of primes $p \equiv 1 \pmod{f(1)}$ greater than 2019, which exists by Dirichlet. Then
\[A\left(1, \frac{p-1}{f(1)}\right) \implies f\left(\frac{p-1}{f(1)}\right) = p-1.\]
Afterwards, we can choose a prime $q \neq f(1)$ and a sufficently large $p$ with $\gcd(q,p-1) = 1$. Then
\[A\left(q, \frac{p-1}{f(1)}\right) \implies f(q) = qf(1).\]
Thus we can finish by fixing an integer $k$, allowing $q$ to be sufficiently large, and using $A(k,q)$. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SHZhang
109 posts
#57
Y by
It is easy to see that all functions of the form $f(x) = cx$ for some $c \in \mathbb{Z}_{>0}$ works. We will show that these are the only possible functions.

Lemma: There are infinitely many $x \in \mathbb{Z}_{>0}$ such that $f(x) = xf(1)$.
Proof: Assume $x$ is sufficiently large. Setting $a = 1$ and $b = x$ in the condition, we have $f(x) + 1 \mid xf(1) + 1$. By Dirichlet's theorem, there are infinitely many $x$ such that $xf(1) + 1$ is prime. Since $f(x) + 1 > 1$, this forces $f(x) + 1 = xf(1) + 1$, so $f(x) = xf(1)$. $\square$

Now consider any integer $k > 1$; we claim that $f(k) = kf(1)$. To do that, consider setting $b = k$; then $a + f(k) \mid a^2 + kf(a)$ for sufficiently large $a$. By the lemma, $a + f(k) \mid a^2 + akf(1)$ for infinitely many $a$. Subtracting $a(a + f(k))$ from $a^2 + akf(1)$ gives $a + f(k) \mid a(kf(1) - f(k))$. Let $C = kf(1) - f(k)$; we want to show $C = 0$. The divisibility implies $\frac{Ca}{a + f(k)} = C - \frac{Cf(k)}{a + f(k)}$ is an integer for infinitely many $a$, so $y = \frac{Cf(k)}{a + f(k)}$ is an integer for infinitely many $a$. But if $C > 0$, then $y$ lies strictly between $0$ and $1$ for all large $a$; if $C < 0$ then $y$ lies strictly between $-1$ and $0$ for all large $a$. Therefore $C = 0$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathhotspot
70 posts
#58
Y by
@above As mentioned abpve , we can do use Dirchlet theorem,which basically is a corollary of Chebotarev density theorem (Galois Closure etc)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pikapika007
294 posts
#59
Y by
ooga booga force prime

The answer is $f(x) = Cx$ for any $x$. Let $C = f(1)$.

Claim: For any prime $p = 1 + kC$ (of which infinitely many $p$ and $k$ exist by Dirichlet), $f(k) = Ck$.

Proof. $P(1, k)$ gives $1 + f(k) \mid 1 + kC$, and since $1 + kC$ is prime, $1 + f(k) = 1 + kC$ so $f(k) = Ck$. $\square$

To finish, select a sufficiently large $k$ that satisfies the conditions in the claim and consider $P(a, k)$ for arbitrary $a \in \mathbb{N}$ - we have $a + Ck \mid a^2 + kf(a)$, or $a + Ck \mid C(a^2 + kf(a)) - f(a)(a + Ck) = Cn^2 - nf(n)$, and since $k$ is sufficiently large this forces $f(n) = Cn$.
This post has been edited 1 time. Last edited by pikapika007, Aug 3, 2024, 12:01 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bin_sherlo
661 posts
#60
Y by
Answer is $f(n)=cn \ \forall \ n\in Z^+$.
We have $a+f(b)|af(b)-bf(a)$ for $a+b>2019$. Let $f(1)=k$. Take a large enough prime $p$.
$P(1,p-k)\implies 1+f(p-k)|1+k(p-k)\implies f(p-k)\leq k(p-k)$
Plugging $p-k,1$ gives $p|(p-k)k-f(p-k)$. Note that $(p-k)k\geq f(p-k)$. Since $p|k^2+f(p-k),$ we get that $f(p-k)$ is unbounded.
If $k=1,$ then $p|p-1-f(p-1)$ where $0\leq p-1-f(p-1)<p-1$. Hence $f(p-1)=p-1$ for all large enough primes $p$. Plugging back yields
\[p-1+f(b)|(p-1+f(b))(f(b)-b)-(p-1)(f(b)-b)=f(b)(f(b)-b)\]If we let $p$ to go to positive infinity, we have $f(b)=b$.
If $k>1,$ then taking $a=p-k,b=1$ and $a=1,b=p-k$ give
\[p|(p-k)k-f(p-k) \ \text{and} \ f(p-k)+1|(p-k)k-f(p-k)\]If $p$ doesn't divide $f(p-k)+1,$ then
\[pf(p-k)+p|\underbrace{(p-k)k-f(p-k)}_{\geq 0}\]If $(p-k)k\neq f(p-k),$ then $f(p-k)<k$ since $pf(p-k)<LHS\leq |RHS|<(p-k)k<pk$. But this contradicts with $p|k^2+f(p-k)$ hence $(p-k)k=f(p-k)$ or $p|f(p-k)+1$.
Now assume that there exists infinitely many primes $p$ such that $f(p-k)=k(p-k)$. Plugging $p-k,b$ gives
\[p-k+f(b)|(p-k)(kb-f(b)) \ \text{and} \ p-k+f(b)|(p-k)(kb-f(b))+f(b)(kb-f(b))\]Thus, $p-k+f(b)|f(b)(kb-f(b))$ Since there are infinitely many $p$ where $f(p-k)=k(p-k),$ we can take $p$ sufficiently large which forces right hand side to be $0$ so $f(b)=kb$ for all $b$.
Suppose that there are finitely many primes which hold $f(p-k)=(p-k)k$. There exists some $N$ such that if $p\geq N$, then $p|f(p-k)+1$. But since $p|f(p-k)+k^2,$ we get that $p|k^2-1$ for infinitely many primes but $k^2-1>0$ is constant hence we get a contradiction.$\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cursed_tangent1434
548 posts
#61
Y by
We claim that the answer is
\[f(n)=nc\]for all $n \in \mathbb{N}$ and some $c\in \mathbb{Z}_{>0}$. It is clear that these solutions work. Now, we shall show that they are the only ones.

First, we consider $a\geq 2019$. Note that $1+a>2019$. Then, we have
\[1+f(a) \mid 1+af(1)\]which requires $f(a)\leq af(1)$ for all $a\geq 2019$. We further note that,
\begin{align*}
    a+f(1) &\mid a^2 + f(a)\\
    a+f(1) & \mid af(1)- f(a)
\end{align*}Thus, if $af(1)\neq f(a)$ we must have
\[f(a) \leq af(1) - f(a) - f(1)-a\]Further, say we have
\[a+f(1) \mid af(1)-f(a) - kf(1)-ka\]and $f(a) \leq af(1) - kf(1)-ka$ for some $k \in \mathbb{N}$. Then, note that we need from the divisibility,
\[a+f(1)\leq af(1)-f(a)-kf(1)-ka \implies f(a)\leq af(1) - f(a) - (k+1)f(1)-(k+1)a \]and we also have
\[a+f(1) \mid af(1)-f(a) - (k+1)f(1)-(k-1)a\]of which we know the RHS is positive. Now, this means that if $af(1)\neq f(a)$ we must have $f(a) \leq af(1) - kf(1)-ka$ for all $k$ which clearly cannot be true since $f(1),f(a)>0$.

Thus, we conclude that for all $a\geq 2019$, $f(a)=af(1)$. Now, to deal with smaller values.

Consider $1\leq b<2019$. Then, consider $a>|f(b)-2bf(1)|$ such that $\gcd(a,b)=1$ which is always possible. Now, we have
\begin{align*}
    b+f(a) &\mid b^2+af(b)\\
    af(1)+b &\mid af(b)- abf(1)\\
    af(1)+b &\mid a(f(b)-bf(1))\\
    a+b &\mid f(b)-bf(1)
\end{align*}since $\gcd(a+b,a)=\gcd(a,b)=1$. But, we know that
\[|(a+b)|=a+b >|f(b)-2bf(1)|+bf(1) \geq |f(b)-bf(1)|\]which is indeed impossible unless we have $f(b)=bf(1)$.

Thus, for all $n\leq 2018$ and all $n\geq 2019$ we have that $f(n)=nf(1)$ which is simply saying that the initially claimed solution is the only one.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Eka01
204 posts
#63
Y by
Put $a=1$ to get $1+f(b) | 1+bf(1)$. Call $f(1)=c$ for brevity. Then $f(b) \leq cb$. Also, by Dirichlet's, there exist infinitely many $b$ such that $cb+1$ is prime. This gives that $f(b)=cb$ for these $b$.

Take $b$ of the above form and very large in size to get
$$a+cb | a^2 +bf(a) \implies bf(a)-abc<0 \implies a+cb | abc-bf(a)>0$$Also set $b$ to be coprime to $a$.

Hence $a+cb | ac-f(a)$ but since $b$ is large, $f(a)=ac$ and we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ihatemath123
3426 posts
#64
Y by
Should've been a RELMO problem :wow: The answer is $f(x) = cx$ for any positive integer $c$.

Claim: Let $K$ be a fixed integer. There exist infinitely many integers $n$ such that $K \mid n$ and $f(n) = nf(1)$.
Proof: By Dirichlet's theorem, there exist infinitely many $n$ for which $1+ \tfrac{n}{K} \cdot (K f(1) ) = 1+nf(1)$ is prime. Taking $P(1,n)$ for such $n$ gives us
\[1+f(n) \mid 1+nf(1), \]and since $1+f(n) > 1$ it follows that $1+f(n) = 1+nf(1)$, implying the claim.

Claim: $f(x) = x$ for each $x$.
Proof: Fix $x$ and let $n$ be a sufficiently large multiple of $f(x)$ such that $f(n) = nf(1)$ – let $L = \tfrac{n}{f(x)}$. Taking $P(n,x)$ gives us
\begin{align*}
n + f(x) & \mid n^2 + nxf(1) \\
n + f(x) & \mid nxf(1) - nf(x) \\
f(x) \cdot (L+1) & \mid L f(x) (xf(1) - f(x)) \\
L+1 & \mid xf(1) - f(x) \\
f(x) & = xf(1).
\end{align*}
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
john0512
4170 posts
#65
Y by
We claim that the answer is $f(x)=$ @Catherine Xu, which clearly works. Let $f(1)=c$.

Claim: If $bc+1=p$ is a prime and $b$ is sufficently large, then $f(b)=p-1$. Plug in $a=1$ to get
$$1+f(b)\mid 1+bc.$$If $1+bc$ is prime, then $f(b)=p-1$, as desired (since $1+f(b)$ can't be $1$).

In particular, $ac+1$ is prime for infinitely many $a$ by Dirichlet's prime theorem. Thus, let $ac+1=p$ for sufficently large $a$, so
$$a+f(b)\mid a^2+abc.$$However,
$$\frac{a^2+abc}{a+f(b)}=a+bc-f(b)+\frac{f(b)^2-f(b)bc}{a+f(b)},$$so unless $f(b)^2-f(b)bc=0$, that is, $f(b)=bc$, this is a contradiction for sufficently large $a$. We are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
D4N13LCarpenter
13 posts
#66
Y by
We will prove that the only solution is $f(x)=xk$ for every $x$, which clearly works. Also, suppose that any $a+b>C$ satisfy the condition, instead of $a+b>2019$. Let $P(a, b)$ denote the given assertion. We begin by proving the following

Claim. We have $n\mid f(n)$ for every $n$
Proof. Notice that this is equivalent to proving that $\nu _{p}(n) \leq \nu_p(f(n))$ for every prime $p$. To do this, we will proceed by strong induction on $\nu_p(n)$.

Base Case.

If $\nu_p(n) = 1$, consider a sufficiently large $k$ such that $a=pk-f(n)$ satisfies $a+n>C$. $P(a,n)$ now yields $$pk\mid nf(pk-f(n))+f(n)^2,$$or equivalently, $p\mid f(n)^2$, which is what we wanted.

Inductive Step.

Assume $\nu_p(n)\leq \nu_p(f(n))$ if $\nu_p(n)<\alpha$ and, for the sake of contradiction, suppose that $\alpha= \nu_p(n)\geq \nu_p(f(n))=\beta$. Now consider sufficiently large $\alpha, t$ such that $a=p^{\alpha t} k-f(n)$ satisfies $a+n>C$ and let $\gamma = \nu_p(f(a))$. $P(a,n)$ then gives $$p^{\alpha t} k\mid nf(p^{\alpha t} k-f(n))+f(n)^2.$$Moreover, $\nu_p(nf(p^{\alpha t} k-f(n))+f(n)^2)\geq \min\{\nu_p(nf(p^{\alpha t} k-f(n))), \nu_p(f(n)^2)\}=\min\{\alpha+\gamma, 2\beta\}$, but note that $\alpha > \beta$ and $\gamma \geq \beta$ as $\gamma \geq \nu_p(f(p^{\alpha t} k-f(n)))=\beta$, so $\alpha + \gamma > 2\beta$. To be precise, this implies $$2\beta=\nu_p(nf(p^{\alpha t} k-f(n))+f(n)^2)\geq \alpha t$$which is a contradiction as $2\beta$ is fixed and $\alpha t$ can be arbitrarily large.

Now choose sufficiently large $a$ and let $f(a)=ak$. $P(1,k)$ yields $1+ak\mid 1+af(1),$ or equivalently $$1+ak\mid a(f(1)-k)\implies 1+ak\mid f(1)-k.$$However, this is clearly false if $a$ is large enough, unless $f(1)=k$.

We now have that $f(x)=xf(1)$ if $x>N$ for some $N$. We aim to prove that $f(x)=xf(1)$ even if $x\leq N$. Indeed, let $f(a)=at$ for a certain $a<N$ and choose a $b>N$ which is coprime with $a$. $P(a,b)$ then gives $$a+bf(1)\mid a^2+abt=a(a+bt).$$By letting $\gcd(a, k)=d$, this then rewrites as $$\frac{a+bk}{d}\mid a+bt,$$or equivalently $\frac{a+bk}{d}\mid b(t-k).$ However, $\gcd(a+bk, b)=\gcd(a,b)=1$ so $$\frac{a+bk}{d}\mid t-k.$$Nevertheless, $a+bk$ is unbounded, so this is only true if $t=k$, which ends the proof.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pie854
243 posts
#67
Y by
By Dirichlet, $1+kf(1)$ can be a prime infinitely often. Since $1+f(k)\mid 1+kf(1)$ it follows that $f(k)=kf(1)$ for infinitely many $k$. For any fixed $n>0$, take a large $k$ such that $f(k)=kf(1)$. We have $$n+kf(1) \mid n^2+kf(n) \implies n+kf(1)\mid n^2f(1)-nf(n).$$This forces $f(n)=nf(1)$.

Conversely, we can check that $f(n)=cn$ works.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
andrewthenerd
16 posts
#68
Y by
Solution: Click to reveal hidden text
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AshAuktober
917 posts
#69
Y by
The only sols are $f(b) = bf(1) \forall b$, which works.
Note that by plugging in $b$ such that $bf(1)+1$ is prime (there exist infinitely many by Dirichlet), we have $f(b) = bf(1)$ for all $b \in \mathcal{B}$, where $\mathcal{B}$ is defined as $b$ such that $bf(1)+1$ is prime.
Now for any $a$, note that of the values in $\gcd_{b \in \mathcal{B}}(a, b)$, some value $d$ shows up infinitely many times. Let $\mathcal{D} = \{b \in \mathcal{B} : \gcd(a, b) = d$.
Then we have \[x+yf(1) \mid y(af(1)-f(a))\]where $a = dx, b = dy$ for $b \in \mathcal{D}$, which implies $x+yf(1) \mid af(1)-f(a)$. Since $y$ is unbounded, $f(a) = af(1) \forall a$. $\square$
Z K Y
N Quick Reply
G
H
=
a