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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 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
How many ordered pairs of numbers can we find?
BR1F1SZ   2
N 5 minutes ago by BR1F1SZ
Source: 2024 Argentina TST P6
Given $2024$ non-negative real numbers $x_1, x_2, \dots, x_{2024}$ that satisfy $x_1 + x_2 + \cdots + x_{2024} = 1$, determine the maximum possible number of ordered pairs $(i, j)$ such that
\[
x_i^2 + x_j \geqslant \frac{1}{2023}.
\]
2 replies
BR1F1SZ
Jan 25, 2025
BR1F1SZ
5 minutes ago
IMO 2009, Problem 5
orl   87
N 10 minutes ago by asdf334
Source: IMO 2009, Problem 5
Determine all functions $ f$ from the set of positive integers to the set of positive integers such that, for all positive integers $ a$ and $ b$, there exists a non-degenerate triangle with sides of lengths
\[ a, f(b) \text{ and } f(b + f(a) - 1).\]
(A triangle is non-degenerate if its vertices are not collinear.)

Proposed by Bruno Le Floch, France
87 replies
orl
Jul 16, 2009
asdf334
10 minutes ago
1978 USAMO #1
Mrdavid445   54
N 23 minutes ago by Marcus_Zhang
Given that $a,b,c,d,e$ are real numbers such that

$a+b+c+d+e=8$,
$a^2+b^2+c^2+d^2+e^2=16$.

Determine the maximum value of $e$.
54 replies
Mrdavid445
Aug 16, 2011
Marcus_Zhang
23 minutes ago
The return of a legend inequality
giangtruong13   4
N an hour ago by Double07
Source: Legacy
Given that $0<a,b,c,d<1$.Prove that: $$ (1-a)(1-b)(1-c)(1-d) > 1-a-b-c-d $$
4 replies
giangtruong13
3 hours ago
Double07
an hour ago
degree of f=2^k
Sayan   15
N an hour ago by Gejabsk
Source: ISI 2012 #8
Let $S = \{1,2,3,\ldots,n\}$. Consider a function $f\colon S\to S$. A subset $D$ of $S$ is said to be invariant if for all $x\in D$ we have $f(x)\in D$. The empty set and $S$ are also considered as invariant subsets. By $\deg (f)$ we define the number of invariant subsets $D$ of $S$ for the function $f$.

i) Show that there exists a function $f\colon S\to S$ such that $\deg (f)=2$.

ii) Show that for every $1\leq k\leq n$ there exists a function $f\colon S\to S$ such that $\deg (f)=2^{k}$.
15 replies
1 viewing
Sayan
May 13, 2012
Gejabsk
an hour ago
Local-global with Fibonacci numbers
MarkBcc168   26
N an hour ago by pi271828
Source: ELMO 2020 P2
Define the Fibonacci numbers by $F_1 = F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n\geq 3$. Let $k$ be a positive integer. Suppose that for every positive integer $m$ there exists a positive integer $n$ such that $m \mid F_n-k$. Must $k$ be a Fibonacci number?

Proposed by Fedir Yudin.
26 replies
MarkBcc168
Jul 28, 2020
pi271828
an hour ago
Cauchy functional equations
syk0526   10
N an hour ago by imagien_bad
Source: FKMO 2013 #2
Find all functions $ f : \mathbb{R}\to\mathbb{R}$ satisfying following conditions.
(a) $ f(x) \ge 0 $ for all $ x \in \mathbb{R} $.
(b) For $ a, b, c, d \in \mathbb{R} $ with $ ab + bc + cd = 0 $, equality $ f(a-b) + f(c-d) = f(a) + f(b+c) + f(d) $ holds.
10 replies
syk0526
Mar 24, 2013
imagien_bad
an hour ago
Three circles are concurrent
Twoisaprime   21
N an hour ago by L13832
Source: RMM 2025 P5
Let triangle $ABC$ be an acute triangle with $AB<AC$ and let $H$ and $O$ be its orthocenter and circumcenter, respectively. Let $\Gamma$ be the circle $BOC$. The line $AO$ and the circle of radius $AO$ centered at $A$ cross $\Gamma$ at $A’$ and $F$, respectively. Prove that $\Gamma$ , the circle on diameter $AA’$ and circle $AFH$ are concurrent.
Proposed by Romania, Radu-Andrew Lecoiu
21 replies
Twoisaprime
Feb 13, 2025
L13832
an hour ago
IMO Shortlist 2011, Algebra 3
orl   45
N an hour ago by Ilikeminecraft
Source: IMO Shortlist 2011, Algebra 3
Determine all pairs $(f,g)$ of functions from the set of real numbers to itself that satisfy \[g(f(x+y)) = f(x) + (2x + y)g(y)\] for all real numbers $x$ and $y$.

Proposed by Japan
45 replies
orl
Jul 11, 2012
Ilikeminecraft
an hour ago
Hard FE with positive reals
egxa   8
N 2 hours ago by megarnie
Source: Turkey Olympic Revenge 2023 Shortlist A4
Find all functions $f:\mathbb{R^+}\to \mathbb{R^+}$ such that for all $x,y\in \mathbb{R^+}$
$f(xf(y)+y)=f(f(y))+yf(x)$
Proposed by Şevket Onur Yılmaz
8 replies
egxa
Jan 22, 2024
megarnie
2 hours ago
Like Father Like Son... (or Like Grandson?)
AlperenINAN   1
N 2 hours ago by hakN
Source: Turkey TST 2025 P4
Let $a,b,c$ be given pairwise coprime positive integers where $a>bc$. Let $m<n$ be positive integers. We call $m$ to be a grandson of $n$ if and only if, for all possible piles of stones whose total mass adds up to $n$ and consist of stones with masses $a,b,c$, it's possible to take some of the stones out from this pile in a way that in the end, we can obtain a new pile of stones with total mass of $m$. Find the greatest possible number that doesn't have any grandsons.
1 reply
AlperenINAN
Today at 6:09 AM
hakN
2 hours ago
Crazy number theory
MTA_2024   5
N 2 hours ago by bjump
Find all couple $(p;q)$ of primes (greater than 5) such that : $$pq \mid (5^q-3^q)(5^p-3^p)$$
5 replies
MTA_2024
4 hours ago
bjump
2 hours ago
hard number theory problem
Zavyk09   0
2 hours ago
Source: forgotten
Find all couple $(x, y)$ of positive integers such that:
$$2^n + 3^n \mid x^n + y^n, \forall n \in \mathbb{N}^*$$
0 replies
Zavyk09
2 hours ago
0 replies
Slightly weird points which are not so weird
Pranav1056   9
N 2 hours ago by Retemoeg
Source: India TST 2023 Day 4 P1
Suppose an acute scalene triangle $ABC$ has incentre $I$ and incircle touching $BC$ at $D$. Let $Z$ be the antipode of $A$ in the circumcircle of $ABC$. Point $L$ is chosen on the internal angle bisector of $\angle BZC$ such that $AL = LI$. Let $M$ be the midpoint of arc $BZC$, and let $V$ be the midpoint of $ID$. Prove that $\angle IML = \angle DVM$
9 replies
Pranav1056
Jul 9, 2023
Retemoeg
2 hours ago
Another NT FE
nukelauncher   60
N Yesterday at 12:50 AM by pi271828
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 $a+b>2019$.
60 replies
nukelauncher
Sep 22, 2020
pi271828
Yesterday at 12:50 AM
Another NT FE
G H J
Source: ISL 2019 N4
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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
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shendrew7
792 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 $A(k,q)$. $\blacksquare$
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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 $C = 0$.
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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)
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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
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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.$\blacksquare$
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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.
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Eka01
204 posts
#63
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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.
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ihatemath123
3430 posts
#64
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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*}
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john0512
4170 posts
#65
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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.
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D4N13LCarpenter
13 posts
#66
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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.
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pie854
243 posts
#67
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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.
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andrewthenerd
16 posts
#68
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Solution: Click to reveal hidden text
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AshAuktober
924 posts
#69
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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, $f(a) = af(1) \forall a$. $\square$
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pi271828
3363 posts
#70
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The answer is $f(x) = cx$, where $c$ is a positive integer. We fix $a$, and vary $b$ such that $\operatorname{gcd}(a, b) = 1$ and $1+bf(1)$ is a prime. Clearly from $P(1, b)$, this means $f(b) = bf(1)$. By Dirichlet's there are infinitely many such $b$. The divisibility condition clearly implies \begin{align*} a+f(b) \mid bf(a)-af(b) \\ \implies a+bf(1) \mid bf(a) - abf(1)\end{align*}Since $a$ and $b$ are relatively prime, we have $\operatorname{gcd}(b, a+bf(1)) = 1$, which means that \begin{align*} a+bf(1) \mid f(a) - af(1)\end{align*}Taking sufficiently large $b$ gives us that $f(a) - af(1) = 0$, done. $\square$
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