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

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

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
A geometry problem
Lttgeometry   1
N 12 minutes ago by Funcshun840
Triangle $ABC$ has two isogonal conjugate points $P$ and $Q$. The circle $(BPC)$ intersects circle $(AP)$ at $R \neq P$, and the circle $(BQC)$ intersects circle $(AQ)$ at $S\neq Q$. Prove that $R$ and $S$ are isogonal conjugates in triangle $ABC$.
Note: Circle $(AP)$ is the circle with diameter $AP$, Circle $(AQ)$ is the circle with diameter $AQ$.
1 reply
Lttgeometry
Today at 4:03 AM
Funcshun840
12 minutes ago
interesting functional equation
tabel   0
13 minutes ago
Source: random romanian contest
Determine all functions \( f : (0, \infty) \to (0, \infty) \) that satisfy the functional equation:
\[
f(f(x)(1 + y)) = f(x) + f(xy), \quad \forall x, y > 0.
\]
0 replies
tabel
13 minutes ago
0 replies
Inequality em981
oldbeginner   15
N an hour ago by sqing
Source: Own
Let $a, b, c>0, a+b+c=3$. Prove that
\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]
15 replies
1 viewing
oldbeginner
Sep 22, 2016
sqing
an hour ago
Functional equation
shobber   19
N 2 hours ago by Unique_solver
Source: Canada 2002
Let $\mathbb N = \{0,1,2,\ldots\}$. Determine all functions $f: \mathbb N \to \mathbb N$ such that
\[ xf(y) + yf(x) = (x+y) f(x^2+y^2)  \]
for all $x$ and $y$ in $\mathbb N$.
19 replies
shobber
Mar 5, 2006
Unique_solver
2 hours ago
Prove the inequality
Butterfly   0
2 hours ago
Let $a,b,c$ be real numbers such that $a+b+c=3$. Prove $$a^3b+b^3c+c^3a\le \frac{9}{32}(63+5\sqrt{105}).$$
0 replies
Butterfly
2 hours ago
0 replies
Functional equation
shactal   1
N 2 hours ago by ariopro1387
Let $f:\mathbb R\to \mathbb R$ a function satifying $$f(x+2xy) = f(x) + 2f(xy)$$for all $x,y\in \mathbb R$.
If $f(1991)=a$, then what is $f(1992)$, the answer is in terms of $a$.
1 reply
shactal
4 hours ago
ariopro1387
2 hours ago
interesting diophantiic fe in natural numbers
skellyrah   5
N 2 hours ago by skellyrah
Find all functions \( f : \mathbb{N} \to \mathbb{N} \) such that for all \( m, n \in \mathbb{N} \),
\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]
5 replies
skellyrah
Yesterday at 8:01 AM
skellyrah
2 hours ago
Non-linear Recursive Sequence
amogususususus   3
N 2 hours ago by SunnyEvan
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
3 replies
amogususususus
Jan 24, 2025
SunnyEvan
2 hours ago
Inspired by 2025 Beijing
sqing   6
N 3 hours ago by sqing
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
6 replies
sqing
Yesterday at 4:56 PM
sqing
3 hours ago
Serbian selection contest for the IMO 2025 - P4
OgnjenTesic   2
N 3 hours ago by sqing-inequality-BUST
Source: Serbian selection contest for the IMO 2025
For a permutation $\pi$ of the set $A = \{1, 2, \ldots, 2025\}$, define its colorfulness as the greatest natural number $k$ such that:
- For all $1 \le i, j \le 2025$, $i \ne j$, if $|i - j| < k$, then $|\pi(i) - \pi(j)| \ge k$.
What is the maximum possible colorfulness of a permutation of the set $A$? Determine how many such permutations have maximal colorfulness.

Proposed by Pavle Martinović
2 replies
OgnjenTesic
May 22, 2025
sqing-inequality-BUST
3 hours ago
Nice "if and only if" function problem
ICE_CNME_4   14
N 3 hours ago by wh0nix
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )

Please do it at 9th grade level. Thank you!
14 replies
ICE_CNME_4
Friday at 7:23 PM
wh0nix
3 hours ago
2-var inequality
sqing   1
N 3 hours ago by sqing
Source: Own
Let $ a,b> 0 , ab(a+b+1) =3.$ Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{24}{(a+b)^2} \geq 8$$$$ \frac{a}{b^2}+\frac{b}{a^2}+\frac{49}{(a+  b)^2} \geq \frac{57}{4}$$Let $ a,b> 0 ,  (a+b)(ab+1) =4.$ Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{40}{(a+b)^2} \geq 12$$$$\frac{a}{b^2}+\frac{b}{a^2}+\frac{76}{(a+ b)^2}  \geq 21$$
1 reply
sqing
4 hours ago
sqing
3 hours ago
Balkan Mathematical Olympiad
ABCD1728   1
N 3 hours ago by ABCD1728
Can anyone provide the PDF version of the book "Balkan Mathematical Olympiads" by Mircea Becheanu and Bogdan Enescu (published by XYZ press in 2014), thanks!
1 reply
ABCD1728
Yesterday at 11:27 PM
ABCD1728
3 hours ago
area of quadrilateral
AlanLG   1
N 4 hours ago by Altronrren
Source: 3rd National Women´s Contest of Mexican Mathematics Olympiad 2024 , level 1+2 p5
Consider the acute-angled triangle \(ABC\). The segment \(BC\) measures 40 units. Let \(H\) be the orthocenter of triangle \(ABC\) and \(O\) its circumcenter. Let \(D\) be the foot of the altitude from \(A\) and \(E\) the foot of the altitude from \(B\). Additionally, point \(D\) divides the segment \(BC\) such that \(\frac{BD}{DC} = \frac{3}{5}\). If the perpendicular bisector of segment \(AC\) passes through point \(D\), calculate the area of quadrilateral \(DHEO\).
1 reply
AlanLG
Jun 14, 2024
Altronrren
4 hours ago
f(2) = 7, find all integer functions [Taiwan 2014 Quizzes]
v_Enhance   59
N May 18, 2025 by MathIQ.
Find all increasing functions $f$ from the nonnegative integers to the integers satisfying $f(2)=7$ and \[ f(mn) = f(m) + f(n) + f(m)f(n) \] for all nonnegative integers $m$ and $n$.
59 replies
v_Enhance
Jul 18, 2014
MathIQ.
May 18, 2025
f(2) = 7, find all integer functions [Taiwan 2014 Quizzes]
G H J
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v_Enhance
6877 posts
#1 • 12 Y
Y by son7, kirillnaval, mathematicsy, megarnie, Anshul_singh, HamstPan38825, Jufri, itslumi, Adventure10, Sedro, MS_asdfgzxcvb, and 1 other user
Find all increasing functions $f$ from the nonnegative integers to the integers satisfying $f(2)=7$ and \[ f(mn) = f(m) + f(n) + f(m)f(n) \] for all nonnegative integers $m$ and $n$.
This post has been edited 1 time. Last edited by v_Enhance, Jul 19, 2014, 7:37 PM
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cobbler
2180 posts
#2 • 9 Y
Y by son7, Adventure10, Mango247, MS_asdfgzxcvb, and 5 other users
Adding 1 to both sides gives $f(mn)+1=f(m)+f(n)+f(m)f(n)+1=(f(m)+1)(f(n)+1)$.
Define $g(x)=f(x)+1$, thus $g(mn)=g(m)g(n)$. Note that $g(x)=x^3$ for $x\in \{0,1,2\}$.
Taking logs of both sides yields $\ln g(mn)=\ln g(m)g(n)=\ln g(m)+\ln g(n)$. Define $h(x)=\log g(x)$,
$\therefore h(mn)=h(m)+h(n)$. Trivial solution is $h(x)=0\ \forall x\in \mathbb{N}$, but this doesn't satisfy the condition.

I know the rest is wrong, but I g2g so I might as well post it since it seems to give the right answer.

Assume that $h(x)$ is differentiable $x>0$, thus differentiating wrt $x$ we obtain $nh'(mn)=h'(m)$,
which for $m=1$ becomes $h'(n)=\tfrac{h'(1)}{n}$ or $h(n)=\int \tfrac{h'(1)}{n}\ dn = h'(1)\ln |n|+C$
$\implies g(n)=e^{h'(1)\ln |n|+C}=|n|^{h'(1)}e^C=n^{h'(1)}e^C$ since $n\ge 0$ by definition.
Putting $n=1$ yields $g(1)=e^C$ or $1=e^C\iff C=0$. Putting $n=2$ yields $g(2)=2^{h'(1)}$ or
$8=2^{h'(1)}\iff h'(1)=3$. It follows that $g(n)=n^3\ \forall n\in \mathbb{N}\implies \boxed{f(n)=n^3-1\ \forall n\in \mathbb{N}}$.
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v_Enhance
6877 posts
#3 • 4 Y
Y by HamstPan38825, mathematicsy, Adventure10, Assassino9931
Oops I seem to have left out the condition that $f$ is increasing. Sorry about that... Can a mod please edit that in? [EDIT: never mind, computer obtained]

Without the increasing condition I think the solution family is just to pick a value for each $f(p)$.
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fclvbfm934
759 posts
#4 • 8 Y
Y by kprepaf, Kezer, JasperL, sa2001, missgirl, EulersTurban, Adventure10, Mango247
Let $g(x) = f(x)+1$, so we have $g(mn) = g(m)g(n)$ for all non-negative $m,n$, and $g(2) = 8$. Substitute $m = 0$ and we get $g(0) = g(0)g(n)$ which tells us that $g(0) = 0$. Letting $m = 1$ we get $g(n) = g(1)g(n)$ so $g(1) = 1$. It is also easy to see that $g(2^n) = 2^{3n}$.

Since $g$ is completely multiplicative, we only have to define $g$ over the set of prime numbers. I will now show that $g(p) = p^3$ for all primes $p$. Here, we have to use that increasing condition. Let $a=g(p)$ and given a certain integer $s$, let $r$ be an integer such that suppose $2^r < p^s < 2^{r+1}$, which means $r = \lfloor s\log_2{p} \rfloor$. Because $g$ is increasing, we know that $g(2^r) = 2^{3r} < g(p)^s$ which means that
\[g(p) > 2^{\frac{3r}{s}} = 2^{\frac{3(s\log_2{p} - \{ s\log_2{p} \})}{s}} = \frac{p^3}{\sqrt[s]{2^{3 \{ s\log_2{p} \}}}}\]
Clearly, if we make $s$ arbitrarily large, we can get as close to $p^3$ as we want; specifically, we can get it so that $\frac{p^3}{\sqrt[s]{2^{3\{ s\log_2{p} \}}}} > p^3 - 1$ for sure, which means that $g(p) \ge p^3$. Through a similar argument, we can get an upper bound on $g(p)$ as tight as we want to $p^3$, getting $g(p) < p^3 + 1$, telling us that $g(p) \le p^3 \Rightarrow g(p) = p^3$.

So now that we have defined $g(p) = p^3$ for all primes $p$, it becomes evident that $g(n) = n^3$ for all $n \in \mathbb{Z}_0^+$, which implies that $f(n) = n^3 - 1$ is the only possible solution. It is quite easy to test this:
\[m^3n^3 - 1 = m^3 - 1 + n^3 - 1 + (m^3-1)(n^3-1)\]
which is clearly true if you just expand it.
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droid347
2679 posts
#5 • 5 Y
Y by Generic_Username, Kezer, Korgsberg, sa2001, Adventure10
First, we define $g(x)=f(x)+1$. After adding 1 to both sides, the statement becomes $g(mn)=g(m)g(n)$, and $m=0\implies g(0)=0$ or $g(n)=1$ for all $x$, the latter of which is a contradiction to $f$ increasing (because $g$ is increasing as well). Then, $m=1\implies g(n)=g(n)g(1)$ so take some $n$ such that $g(n)\neq 0$ (which must exist because $g$ is increasing) and we get $g(1)=1$. This, together with the condition, implies $g$ is totally multiplicative and is thus defined by its values on the primes.

We will now show that $g(p)=p^3$ for all primes $p$. Assume not; first we will assume $g(p)>p^3$ and produce a contradiction. If there exist $k, p$ such that $2^k>p^n$ but yet $g(2^k)<g(p^n)$, then we contradict $f$ increasing. Because $f$ is totally multiplicative, $f(a^b)=f(a)^b$ so the second inequality becomes $g(2)^k=8^k<g(p)^n$. Taking the log base 2 of both inequalities, we want to find $n,k$ where $k>n\log_2 p, 3k<n\log_ 2 g(p)$ so it suffices to find $n,k$ where \[\log_2 p^3 <\frac{3k}{n}<\log_2 g(p).\]However, a rational $\frac{a}{b}$ must exist between these two values because the rationals are dense in the reals, and we can take $k=a, n=3b$ to finish. We can repeat the same argument with the reverse inequalities to show that $g(p)<p^3$ yields a contradiction, so we must have $g(p)=p^3$ for all primes and thus $g(x)=x^3$ for all integer $x$. Therefore, the only solution is $f(x)=x^3-1$ and we can check that this works to finish:
\[\begin{aligned} f(m) + f(n) + f(m)f(n)&=m^3-1+n^3-1+(m^3-1)(n^3-1)\\ &=(mn)^3+1-m^3-n^3+m^3+n^3-2\\ &=(mn)^3-1\\ &=f(mn).\end{aligned}\]
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anantmudgal09
1980 posts
#6 • 9 Y
Y by Wizard_32, Kalpa, sa2001, kirillnaval, hakN, Beginner2004, Adventure10, Mango247, parola
Posting as I like this idea a lot! :)
v_Enhance wrote:
Find all increasing functions $f$ from the nonnegative integers to the integers satisfying $f(2)=7$ and \[ f(mn) = f(m) + f(n) + f(m)f(n) \]for all nonnegative integers $m$ and $n$.

Set $g(x)=f(x)+1$ and obtain $g(2)=8$ and $g$ is multiplicative, monotone increasing. Let $p$ be a prime and pick $x, y$ as positive integers such that $$p^x>2^y \iff \frac{y}{x} <\log_2 p \Longrightarrow g(p)^x=g(p^x) \ge g(2^y)=g(2)^y=2^{3y} \Longrightarrow g(p)>2^{3\frac{y}{x}}.$$Since rationals are dense in $\mathbb{R}$ we can take the LHS as close to $p^3$ as we want, so in fact, $g(p) \ge p^3$. A similar argument with upper bounds yields $g(p)=p^3$ thus $g(n)=n^3$ for all $n>1$. It's clear to see $g(1)=1$ and $g(0)=0$ so $g(n)=n^3$ for all $n$, consequently $f(n)=n^3-1$ for all $n$.
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zhcosin
11 posts
#7 • 2 Y
Y by Adventure10, Mango247
make $g(x)=1+f(x)$, so we have $g(mn)=g(m)g(n)$
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winnertakeover
1179 posts
#8 • 2 Y
Y by Adventure10, Mango247
Slightly different...

Write $g(m) = f(m)+1$. Then we have $g(2)=8$ and $g(mn)=g(m)g(n)$. Note $g(0)=g(0)^2$ and $g(1)=g(1)^2$, and since $g(0)<g(1)$, we have $g(0)=0$ and $g(1)=1$. Now fix and $m$ and assume WLOG that $m$ is not a power of 2. Note that $$g\left(2^{\lfloor \log_2 m^k \rfloor}\right) < g(m^k) < g\left(2^{\lfloor \log_2 m^k \rfloor + 1}\right)$$Invoking multiplicity, $$g(2)^{\lfloor k\log_2 m \rfloor} < g(m)^k < g(2)^{\lfloor k\log_2 m \rfloor + 1}.$$Hence, $$g(2)^{(\lfloor k\log_2 m \rfloor)/k} < g(m) < g(2)^{(\lfloor k\log_2 m \rfloor + 1)/k}.$$It's obvious that $g(m)=g(2)^{\log_2 m}=m^3$ satisfies the inequality. Now we claim there is only one integer between the two bounds and that will finish the problem. It suffices to show that we may pick some $k$ such that $8^{(\lfloor k\log_2 m \rfloor + 1)/k} - 8^{(\lfloor k\log_2 m \rfloor)/k} < 1$. Note that $$8^{(\lfloor k\log_2 m \rfloor)/k}\left( 8^{1/k}-1\right) < m^3(8^{1/k}-1)$$Clearly, we may choose some $k$ such that $8^{1/k} \le \frac{1}{m^3}+1$, and this $k$ satisfies the required bounds. Hence, $g(m)=m^3$, and we have $f(m)=m^3-1$, and it is easy to check this satisfies the original constraints.
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Grizzy
920 posts
#9 • 3 Y
Y by DrYouKnowWho, centslordm, Mango247
Solution
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brianzjk
1201 posts
#10 • 1 Y
Y by centslordm
incorrect solution
This post has been edited 1 time. Last edited by brianzjk, Nov 6, 2020, 1:58 PM
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justin1228
301 posts
#11 • 2 Y
Y by kirillnaval, endless_abyss
This was a pretty hard problem. I did not solve it entirely myself and got quite a few hints.
We claim that the only solutions are $f(x)=2^{3k}-1$
Using Simon's favorite Factoring Trick, we have:
$$[f(m)+1][f(n)+1]=f(mn)+1$$Let $g=f+1$. Then $g$ is completely multiplicative.

Claim:$g(2^k)=8^k$

Proof:
$g(2^k)=\underbrace {g(2) \cdot g(2) \ldots \cdot g(2) }_\text {k times}=g(2)^k=2^{3k=8^k}$
Now, we bound $g$ on prime powers. Let $ g(p)=q$ for a prime $p$, and $g(p^k)=q^k$.
Then notice that we have for some $a$ and $b$, $g(2^a) \le g(p^k) \le g(2^b)$ since $g$ is strictly increasing. Let us take $a= \lfloor {klog_2p} \rfloor$ and $b= \lceil {klog_2p} \rceil$.


This yields $$8^{\lfloor {klog_2p} \rfloor} \le q^k \le 8^{\lceil {klog_2p} \rceil}$$Taking log base 2,
$$ 3{\lfloor {klog_2p} \rfloor} \le log_2 q^k \le 3{\lceil {klog_2p} \rceil}$$and thus:
$${\lfloor {klog_2p} \rfloor} \le k log_2 \sqrt[3]{q} \le {\lceil {klog_2p} \rceil}$$Using trivial but useful inequalities (basic properties of floor and ceiling function), we have that:

$$ klog_2-1 \le  k log_2 \sqrt[3]{q} \le klog_2p+1$$and thus,
$$ log_2-1/k \le log_2 \sqrt[3]{q} \le log_2p+1/k$$Now taking $lim_{k \rightarrow \infty}$, we can conclude that $\sqrt[3]{q}=p \implies q=p^3$, as desired.
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IAmTheHazard
5003 posts
#12 • 1 Y
Y by centslordm
The answer is $f(x)=x^3-1$, which clearly works. Now we show nothing else works.
Add $1$ to both sides to get
$$f(mn)+1=f(m)f(n)+f(m)+f(n)+1=(f(m)+1)(f(n)+1).$$Then define $g: \mathbb{Z}_{\geq 0} \to \mathbb{Z}$ such that $g(x)=f(x)+1$. The equation then becomes $g(mn)=g(m)g(n)$. We also have $g(2)=8$ and $g$ is strictly increasing. Now, we have to show that $g(x)=x^3$ is the only solution to this functional equation.
First, putting $m=0,n=2$ gives $g(0)=8g(0)$, hence $g(0)=0$. Also, putting in $m=n=1$ gives $g(1)=1$ since $g$ is strictly increasing. Hence, for positive integers $x$, $g(x)$ is positive. Further, note that from $g(2)=2^3$ and $g(mn)=g(m)g(n)$, we find that $g(2^k)=2^{3k}$, where $k$ is a positive integer.
Now suppose $g(x)>x^3$, where $x>2$, and let $g(x)=(ax)^3$ where $a>1$ is real. We can easily see that $g(x^k)=(ax)^{3k}$. Now pick some $k$ such that $a^{3k}>8$, and choose $y$ such that $2^{y-1}<x^k<2^y$. Since $g$ is strictly increasing, it follows that
$$g(x^k)<g(2^y) \iff (ax)^{3k}<2^{3y},$$but we also have $(ax)^{3k}>8(x^k)^3>8(2^{y-1})^3=2^{3y}$, which is a contradiction. Similarly, if $g(x)<x^3$, letting $g(x)=(ax)^3$ for $0<x<1$ and picking $y$ such that $2^y<x^k<2^{y+1}$ and $a^k<\tfrac{1}{8}$ yields a contradiction. It follows that $g(x)=x^3$ for all nonnegative integers $x$, which is the desired result. $\blacksquare$
This post has been edited 1 time. Last edited by IAmTheHazard, Aug 24, 2021, 3:10 PM
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circlethm
98 posts
#13
Y by
Solution. Let $g(x) = f(x) + 1$. Then $g$ has to be increasing and satisfy
$$
g(mn) = g(m)g(n).
$$
Lemma (Erdős). If $g: \mathbb{N} \rightarrow \mathbb{R}$ is totally multiplicative and increasing then $g(n) = n^{\alpha}$, for some $\alpha$.
Proof. Suppose that $f(2) = 2^{\alpha}$, and take $n > 2$ such that $f(n) = n^{\beta}$. Then for any $\ell \in \mathbb{N}$, we can write
$$
2^a \leq n^{\ell} \leq 2^{a + 1},
$$for $a = \lfloor \log_2(n) \ell \rfloor$. Since $f$ is increasing, evaluating the function at each of the above must still satisfy the inequality
$$
\begin{aligned}
2^{\alpha a} \leq n^{\beta \ell} &\leq 2^{\alpha(a + 1)} \\
\implies \lfloor \log_2(n) \ell\rfloor \leq \frac{\beta}{\alpha} \log_2(n) \ell &\leq \lceil \log_2(n) \ell\rceil.
\end{aligned}
$$This implies the inequality
$$
\log_2(n) \ell \left|\frac{\beta}{\alpha} - 1\right| \leq 1.
$$But this has to hold for all $\ell \in \mathbb{N}$, and thus we must have $\beta = \alpha$, and we're done.

Since $g(n) = n^{\alpha}$ for some $\alpha$, and $g(2) = 8 = 2^3$, we must have $g(n) = n^{3}$, and thus $f(n) = n^{3} - 1$, which clearly works.
This post has been edited 1 time. Last edited by circlethm, Sep 2, 2021, 3:12 PM
Reason: Equality is possible
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DrYouKnowWho
184 posts
#14
Y by
\walkthrough Answer: $f(n) \equiv n^3-1$
\begin{walk}
\ii Add $1$ to both sides and factorize
\ii Introduce $g(m) = f(m)+1$, notice that $g$ is thus completely multiplicative.
\ii Notice that $g(m^k) = g(m)^k$ for all $m,k\in \mathbb{N}$
\ii Use the strcitly increasing condition on the fact that $g(2^{\lfloor \log_2p^k\rfloor})<g(p^k)$
\ii Show that the previous step implies $\frac{\lfloor k\log_2p\rfloor\cdot 3}{k}<\log_2g(p)$
\ii Repeat similarly for $g(p^l)<g(2^{\lceil \log_2p^l\rceil})$
\ii Notice that we would arrive with the following, which holds for all positive integers $k,l$: \[3(\log_2p-\frac{1}{k})<\frac{3\lfloor k\log_2p\rfloor}{k}<\log_2g(p)<\frac{3\lceil l\log_2p\rceil}{l}<3(\log_2p+\frac{1}{l})\]\ii Notice that by varying $k,l$, we can make $\log_2g(p)$ arbitrarily close to $\log_2p^3$, which is what we wanted.
\end{walk}
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DottedCaculator
7357 posts
#15 • 1 Y
Y by centslordm
Let $g(x)=f(x)+1$. Then, $g(mn)=g(m)g(n)$ and $g(2)=8$. We claim that $g(x)=x^3$. Since $g$ is multiplicative, it suffices to show $g(p)=p^3$ for all primes. Suppose that $g(p)\neq p^3$. Then, $g(2^k)=8^k$ and $g(p^k)=g(p)^k$. Therefore, $2^i<p^j$ if and only if $8^i<g(p)^j$, so $i\log2<j\log p$ if and only if $i\log8<j\log(g(p))$. This is equivalent to $\frac{3i\log2}j<\log(p^3)$ if and only if $\frac{3i\log2}j<\log(g(p))$. Since $g(p)\neq p^3$, this means that $\log(g(p))\neq3\log p$. However, there exists integer $i$ and $j$ such that $\frac{3i\log2}j$ is between $\log(p^3)$ and $\log(g(p))$, which is a contradiction. Therefore, we must have $g(p)=p^3$ for all primes $p$, so $g(x)=x^3$, which satisfies the conditions. Therefore, the only function $f$ that satisfies the original equation is $\boxed{f(x)=x^3-1}$.
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