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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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1 viewing
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
Inspired by old results
sqing   1
N a minute ago by sqing
Source: Own
Let $ a,b,c\geq 0,ab+bc+ca-4abc=\frac{1}{8}$. Prove that
$$ a^2+b^2+c^2+ab+bc+ca   \geq \frac{3}{8}$$Let $ a,b,c\geq 0,a^2+b^2+c^2+4abc=\frac{1}{4}$. Prove that
$$ a+b+c+ab+bc+ca- 4(13-8\sqrt{2})abc\leq \frac{1}{8}+\frac{1}{\sqrt 2}$$
1 reply
1 viewing
sqing
6 minutes ago
sqing
a minute ago
3-var inequality
sqing   3
N 6 minutes ago by pooh123
Source: Own
Let $ a,b\geq  0 ,a^3-ab+b^3=1  $. Prove that
$$  \frac{1}{2}\geq     \frac{a}{a^2+3 }+ \frac{b}{b^2+3}   \geq  \frac{1}{4}$$$$  \frac{1}{2}\geq     \frac{a}{a^3+3 }+ \frac{b}{b^3+3}   \geq  \frac{1}{4}$$$$  \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2}  \geq  \frac{1}{3}$$$$  \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2}  \geq  \frac{1}{3}$$Let $ a,b\geq  0 ,a^3+ab+b^3=3  $. Prove that
$$  \frac{1}{2}\geq     \frac{a}{a^2+3 }+ \frac{b}{b^2+3}   \geq  \frac{1}{4}(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{3}-1)$$$$  \frac{1}{2}\geq     \frac{a}{a^3+3 }+ \frac{b}{b^3+3}   \geq  \frac{1}{2\sqrt[3]{9}}$$$$  \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2}  \geq  \frac{4\sqrt[3]{3}+3\sqrt[3]{9}-6}{17}$$$$  \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2}  \geq  \frac{\sqrt[3]{3}}{5}$$
3 replies
sqing
Today at 4:32 AM
pooh123
6 minutes ago
Geometry hard
Lukariman   0
6 minutes ago
Given triangle ABC inscribed in circle (O). The bisector of angle A intersects (O) at D. Let M, N be the midpoints of AB, AC respectively. OD intersects BC at P and AD intersects MN at S. The circle circumscribed around triangle MPS intersects BC at Q different from P. Prove that QA is tangent to (O).
0 replies
1 viewing
Lukariman
6 minutes ago
0 replies
Great similarity
steven_zhang123   0
7 minutes ago
Source: a friend
As shown in the figure, there are two points $D$ and $E$ outside triangle $ABC$ such that $\angle DAB = \angle CAE$ and $\angle ABD + \angle ACE = 180^{\circ}$. Connect $BE$ and $DC$, which intersect at point $O$. Let $AO$ intersect $BC$ at point $F$. Prove that $\angle ACE = \angle AFC$.
0 replies
steven_zhang123
7 minutes ago
0 replies
Integer polynomial w factorials
Solilin   0
7 minutes ago
Source: 9th Thailand MO
Let $a_1, a_2, ..., a_{2012}$ be pairwise distinct integers. Show that the equation $(x -a_1)(x - a_2)...(x - a_{2012}) = (1006!)^2$ has at most one integral solution.
0 replies
Solilin
7 minutes ago
0 replies
Coloring plane in black
Ryan-asadi   2
N 17 minutes ago by Primeniyazidayi
Source: Iran Team Selection Test - P3
..........
2 replies
Ryan-asadi
3 hours ago
Primeniyazidayi
17 minutes ago
B.Stat & B.Math 2022 - Q8
integrated_JRC   6
N an hour ago by Titeer_Bhar
Source: Indian Statistical Institute (ISI) - B.Stat & B.Math Entrance 2022
Find the minimum value of $$\big|\sin x+\cos x+\tan x+\cot x+\sec x+\operatorname{cosec}x\big|$$for real numbers $x$ not multiple of $\frac{\pi}{2}$.
6 replies
integrated_JRC
May 8, 2022
Titeer_Bhar
an hour ago
AD=BE implies ABC right
v_Enhance   117
N an hour ago by cj13609517288
Source: European Girl's MO 2013, Problem 1
The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$. The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$. Prove that, if $AD=BE$, then the triangle $ABC$ is right-angled.
117 replies
v_Enhance
Apr 10, 2013
cj13609517288
an hour ago
IMO Genre Predictions
ohiorizzler1434   64
N an hour ago by ariopro1387
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
64 replies
ohiorizzler1434
May 3, 2025
ariopro1387
an hour ago
3-var inequality
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b>0 $ and $\frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \leq \frac{1}{2} . $ Prove that
$$a^2+ab+b^2\geq 3$$$$a^2-ab+b^2 \geq 1 $$Let $ a,b>0 $ and $\frac{1}{a^3+3}+ \frac{1}{b^3+ 3}\leq \frac{1}{2} . $ Prove that
$$a^3+ab+b^3 \geq 3$$$$ a^3-ab+b^3\geq 1 $$
1 reply
sqing
2 hours ago
sqing
an hour ago
Iranians playing with cards module a prime number.
Ryan-asadi   2
N 2 hours ago by AshAuktober
Source: Iranian Team Selection Test - P2
.........
2 replies
Ryan-asadi
4 hours ago
AshAuktober
2 hours ago
An analytic sequence
Ryan-asadi   1
N 2 hours ago by AshAuktober
Source: Iran Team Selection Test - P1
..........
1 reply
Ryan-asadi
4 hours ago
AshAuktober
2 hours ago
Geometry
gggzul   6
N 2 hours ago by Captainscrubz
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
6 replies
gggzul
Yesterday at 8:22 AM
Captainscrubz
2 hours ago
Need help on this simple looking problem
TheGreatEuler   0
2 hours ago
Show that 1+2+3+4....n divides 1^k+2^k+3^k....n^k when k is odd. Is this possible to prove without using congruence modulo or binomial coefficients?
0 replies
TheGreatEuler
2 hours ago
0 replies
f(2) = 7, find all integer functions [Taiwan 2014 Quizzes]
v_Enhance   58
N Apr 25, 2025 by Ilikeminecraft
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 $n$.
58 replies
v_Enhance
Jul 18, 2014
Ilikeminecraft
Apr 25, 2025
f(2) = 7, find all integer functions [Taiwan 2014 Quizzes]
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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
5001 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
7348 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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