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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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[*]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 circle tangent to the circumcircle, excircles related
kosmonauten3114   0
5 minutes ago
Source: My own, maybe well-known
Let $ABC$ be a scalene triangle with excircles $\odot(I_A)$, $\odot(I_B)$, $\odot(I_C)$. Let $\odot(A')$ be the circle which touches $\odot(I_B)$ and $\odot(I_C)$ and passes through $A$, and whose center $A'$ lies outside of the excentral triangle of $\triangle{ABC}$. Define $\odot(B')$ and $\odot(C')$ cyclically. Let $\odot(O')$ be the circle externally tangent to $\odot(A')$, $\odot(B')$, $\odot(C')$.

Prove that $\odot(O')$ is tangent to the circumcircle of $\triangle{ABC}$ at the anticomplement of the Feuerbach point of $\triangle{ABC}$.
0 replies
kosmonauten3114
5 minutes ago
0 replies
3 var inequality
SunnyEvan   0
13 minutes ago
Let $ a,b,c \in R $ ,such that $ a^2+b^2+c^2=4(ab+bc+ca)$Prove that :$$ \frac{53}{2}-9\sqrt{14} \leq \frac{8(a^3b+b^3c+c^3a)}{27(a^2+b^2+c^2)^2} \leq \frac{53}{2}+9\sqrt{14} $$
0 replies
SunnyEvan
13 minutes ago
0 replies
Bounds on degree of polynomials
Phorphyrion   4
N 25 minutes ago by Kingsbane2139
Source: 2020 Israel Olympic Revenge P3
For each positive integer $n$, define $f(n)$ to be the least positive integer for which the following holds:

For any partition of $\{1,2,\dots, n\}$ into $k>1$ disjoint subsets $A_1, \dots, A_k$, all of the same size, let $P_i(x)=\prod_{a\in A_i}(x-a)$. Then there exist $i\neq j$ for which
\[\deg(P_i(x)-P_j(x))\geq \frac{n}{k}-f(n)\]
a) Prove that there is a constant $c$ so that $f(n)\le c\cdot \sqrt{n}$ for all $n$.

b) Prove that for infinitely many $n$, one has $f(n)\ge \ln(n)$.
4 replies
Phorphyrion
Jun 11, 2022
Kingsbane2139
25 minutes ago
A point on BC
jayme   7
N 42 minutes ago by jayme
Source: Own ?
Dear Mathlinkers,

1. ABC a triangle
2. 0 the circumcircle
3. D the pole of BC wrt 0
4. B', C' the symmetrics of B, C wrt AC, AB
5. 1b, 1c the circumcircles of the triangles BB'D, CC'D
6. T the second point of intersection of the tangent to 1c at D with 1b.

Prove : B, C and T are collinear.

Sincerely
Jean-Louis
7 replies
jayme
Today at 6:08 AM
jayme
42 minutes ago
Zack likes Moving Points
pinetree1   73
N an hour ago by NumberzAndStuff
Source: USA TSTST 2019 Problem 5
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$. A line through $H$ intersects segments $AB$ and $AC$ at $E$ and $F$, respectively. Let $K$ be the circumcenter of $\triangle AEF$, and suppose line $AK$ intersects $\Gamma$ again at a point $D$. Prove that line $HK$ and the line through $D$ perpendicular to $\overline{BC}$ meet on $\Gamma$.

Gunmay Handa
73 replies
pinetree1
Jun 25, 2019
NumberzAndStuff
an hour ago
Domain and Inequality
Kunihiko_Chikaya   1
N an hour ago by Mathzeus1024
Source: 2018 The University of Tokyo entrance exam / Humanities, Problem 1
Define on a coordinate plane, the parabola $C:y=x^2-3x+4$ and the domain $D:y\geq x^2-3x+4.$
Suppose that two lines $l,\ m$ passing through the origin touch $C$.

(1) Let $A$ be a mobile point on the parabola $C$. Let denote $L,\ M$ the distances between the point $A$ and the lines $l,\ m$ respectively. Find the coordinate of the point $A$ giving the minimum value of $\sqrt{L}+\sqrt{M}.$

(2) Draw the domain of the set of the points $P(p,\ q)$ on a coordinate plane such that for all points $(x,\ y)$ over the domain $D$, the inequality $px+qy\leq 0$ holds.
1 reply
Kunihiko_Chikaya
Feb 25, 2018
Mathzeus1024
an hour ago
JBMO TST Bosnia and Herzegovina 2020 P1
Steve12345   3
N an hour ago by AylyGayypow009
Determine all four-digit numbers $\overline{abcd}$ which are perfect squares and for which the equality holds:
$\overline{ab}=3 \cdot \overline{cd} + 1$.
3 replies
Steve12345
Aug 10, 2020
AylyGayypow009
an hour ago
Problem3
samithayohan   116
N an hour ago by fearsum_fyz
Source: IMO 2015 problem 3
Let $ABC$ be an acute triangle with $AB > AC$. Let $\Gamma $ be its circumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$. Let $M$ be the midpoint of $BC$. Let $Q$ be the point on $\Gamma$ such that $\angle HQA = 90^{\circ}$ and let $K$ be the point on $\Gamma$ such that $\angle HKQ = 90^{\circ}$. Assume that the points $A$, $B$, $C$, $K$ and $Q$ are all different and lie on $\Gamma$ in this order.

Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.

Proposed by Ukraine
116 replies
samithayohan
Jul 10, 2015
fearsum_fyz
an hour ago
geometry problem
invt   0
an hour ago
In a triangle $ABC$ with $\angle B<\angle C$, denote its incenter and midpoint of $BC$ by $I$, $M$, respectively. Let $C'$ be the reflected point of $C$ wrt $AI$. Let the lines $MC'$ and $CI$ meet at $X$. Suppose that $\angle XAI=\angle XBI=90^{\circ}$. Prove that $\angle C=2\angle B$.
0 replies
invt
an hour ago
0 replies
the locus of $P$
littletush   10
N an hour ago by SuperBarsh
Source: Italy TST 2009 p2
$ABC$ is a triangle in the plane. Find the locus of point $P$ for which $PA,PB,PC$ form a triangle whose area is equal to one third of the area of triangle $ABC$.
10 replies
littletush
Mar 10, 2012
SuperBarsh
an hour ago
Abelkonkurransen 2025 3b
Lil_flip38   3
N 2 hours ago by Adywastaken
Source: abelkonkurransen
An acute angled triangle \(ABC\) has circumcenter \(O\). The lines \(AO\) and \(BC\) intersect at \(D\), while \(BO\) and \(AC\) intersect at \(E\) and \(CO\) and \(AB\) intersect at \(F\). Show that if the triangles \(ABC\) and \(DEF\) are similar(with vertices in that order), than \(ABC\) is equilateral.
3 replies
Lil_flip38
Mar 20, 2025
Adywastaken
2 hours ago
I got stuck in this combinatorics
artjustinhere237   3
N 2 hours ago by GreekIdiot
Let $S = \{1, 2, 3, \ldots, k\}$, where $k \geq 4$ is a positive integer.
Prove that there exists a subset of $S$ with exactly $k - 2$ elements such that the sum of its elements is a prime number.
3 replies
artjustinhere237
May 13, 2025
GreekIdiot
2 hours ago
Geo metry
TUAN2k8   5
N 2 hours ago by Melid
Help me plss!
Given an acute triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively. Lines $BD$ and $CE$ intersect at point $F$. The circumcircles of triangles $BDF$ and $CEF$ intersect at a second point $P$. The circumcircles of triangles $ABC$ and $ADE$ intersect at a second point $Q$. Point $K$ lies on segment $AP$ such that $KQ \perp AQ$. Prove that triangles $\triangle BKD$ and $\triangle CKE$ are similar.
5 replies
TUAN2k8
May 6, 2025
Melid
2 hours ago
power of a point
BekzodMarupov   2
N 2 hours ago by BekzodMarupov
Source: lemmas in olympiad geometry
Epsilon 1.3. Let ABC be a triangle and let D, E, F be the feet of the altitudes, with D on BC, E on CA, and F on AB. Let the parallel through D to EF meet AB at X and AC at Y. Let T be the intersection of EF with BC and let M be the midpoint of side BC. Prove that the points T, M, X, Y are concyclic.
2 replies
BekzodMarupov
Yesterday at 5:41 AM
BekzodMarupov
2 hours ago
Bijection on the set of integers
talkon   19
N Apr 18, 2025 by AN1729
Source: InfinityDots MO 2 Problem 2
Determine all bijections $f:\mathbb Z\to\mathbb Z$ satisfying
$$f^{f(m+n)}(mn) = f(m)f(n)$$for all integers $m,n$.

Note: $f^0(n)=n$, and for any positive integer $k$, $f^k(n)$ means $f$ applied $k$ times to $n$, and $f^{-k}(n)$ means $f^{-1}$ applied $k$ times to $n$.

Proposed by talkon
19 replies
talkon
Apr 9, 2018
AN1729
Apr 18, 2025
Bijection on the set of integers
G H J
G H BBookmark kLocked kLocked NReply
Source: InfinityDots MO 2 Problem 2
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talkon
276 posts
#1 • 10 Y
Y by SAUDITYA, ThE-dArK-lOrD, SHREYAS333, anantmudgal09, Smita, Ankoganit, Mathuzb, Omeredip, GeoMetrix, Adventure10
Determine all bijections $f:\mathbb Z\to\mathbb Z$ satisfying
$$f^{f(m+n)}(mn) = f(m)f(n)$$for all integers $m,n$.

Note: $f^0(n)=n$, and for any positive integer $k$, $f^k(n)$ means $f$ applied $k$ times to $n$, and $f^{-k}(n)$ means $f^{-1}$ applied $k$ times to $n$.

Proposed by talkon
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SAUDITYA
250 posts
#2 • 3 Y
Y by zephyr7723, Adventure10, Mango247
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rmtf1111
698 posts
#4 • 2 Y
Y by Adventure10, Mango247
Let $f^{-1}=g$. Let $c=f(0)$. Suppose that $c\neq 0$. Let $P(m,n)$ be the assertion $f^{f(m+n)}(mn) = f(m)f(n)$.
$$P(0,g(n))\implies f^n(0)=nf(0)=cn \implies c(n+1)=f^{n+1}(0)=f(cn)\implies f^k(nc)=f^{n+k}(0)=c(n+k)$$$$P(m+c,-c)\implies f^{f(m)}(-c(m+c))=0\implies c(f(m)-m-c)=0 \implies f(m)=m+c$$Let $c=0$. Letting $n=-m$ we have that $f(m)f(-m)=-m^2 \implies f(1)f(-1)=-1$. Suppose that $f(1)=1$. Let $p$ be a prime.
$$P(p,-p) \implies f(p)f(-p)=-p^2 \stackrel{\text{injectivity}}{\implies} f(-p)=-f(p)$$$$f(-m)=-f(m) \stackrel{\text{P(m,-1)}}{\Longleftrightarrow} f^{f(m-1)-1}(-m)=-m \Longleftrightarrow f^{f(-m-1)-1}(m)=m \stackrel{\text{P(m,1)}}{\Longleftarrow} f(-m-1)=-m-1$$Now apply Cauchy induction with base cases large primes, thus we have that $f(n)=-f(-n)\implies f(n)=\pm n$. If there exists $u\in \mathbb{Z}/\{0\}$ such that $f(u)=-u$, then by letting $m=u$ and $n\equiv_2 u$ we see that $f(n)=-n$, and now by looking at $P(2k+1-u,u)$ we see that $u$ must be even and we have the solution $f(m)=(-1)^{m+1}m$. If there doesn't exist such $u$, then $f(m)=m$. The case $f(1)=-1$ can be treated in the same way, but using simple induction instead of Cauchy, the latter case producing any solutions.
To sum up: the solutions are $f(m)=m+c$ , $f(m)=(-1)^{m+1}m$ , where $c\in \mathbb{Z}$ is a constant.
This post has been edited 1 time. Last edited by rmtf1111, Apr 9, 2018, 6:20 PM
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anantmudgal09
1980 posts
#5 • 1 Y
Y by Adventure10
I wonder what was the proposer's motivation :)
#2 wrote:
Determine all bijections $f:\mathbb Z\to\mathbb Z$ satisfying
$$f^{f(m+n)}(mn) = f(m)f(n)$$for all integers $m,n$.

Note: $f^0(n)=n$, and for any positive integer $k$, $f^k(n)$ means $f$ applied $k$ times to $n$, and $f^{-k}(n)$ means $f^{-1}$ applied $k$ times to $n$.

#2
This post has been edited 2 times. Last edited by anantmudgal09, Apr 9, 2018, 9:20 PM
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talkon
276 posts
#6 • 2 Y
Y by anantmudgal09, Adventure10
Answer: the infinite family of functions $n\mapsto n+c$ for any integer $c$, and the function $n\mapsto (-1)^{n+1}n$.

Official Solution. We consider two cases, depending on the value of $f(0)$.

Case 1: $f(0)=0$.

By plugging in $(m,n)=(k,-k)$, we have $-k^2 = f(k)f(-k)$
for all integers $k$. Since $f$ is bijective, by induction on $k$, $\{f(k),f(-k)\} = \{k,-k\}$
for all positive integers $k$. Hence $f(f(n))=n$ for all integers $n$.

Now suppose that $m,n$ are integers with the same parity, so $f(m+n)$ is even. Hence,
$$mn=f^{f(m+n)}(mn) = f(m)f(n),$$so either both $f(m)=m$ and $f(n)=n$ or $f(m)=-m$ and $f(n)=-n$. Therefore there are four solutions left to check: $n\mapsto n, n\mapsto -n, n\mapsto (-1)^nn$, and $n\mapsto (-1)^{n+1}n$, and by considering $m$ even and $n$ odd, we can see that only two work: $n\mapsto n$ and $n\mapsto (-1)^{n+1}n$.

Case 2: $f(0)\neq 0$.

Plug in $(m,n)=(f^{-1}(k),0)$ to get $f^k(0)=kf(0)$ for all integers $k$. In particular, when $k=-1$ we have $f^{-1}(0) = -f(0)$. Now substitute in $(m,n)=(m,-f(0))$ to get, for all integers $m$,
$$f^{f(m-f(0))}(-mf(0)) = 0. \qquad \text{\_\_\_ (1)}$$
Now note that the orbit $\ldots\to f^{-2}(0)\to f^{-1}(0)\to 0\to f(0)\to f(f(0))\to\ldots $ contains all multiples of $f(0)$, so it is unbounded and not periodic. Hence from

$$f^{f(n)-n-f(0)}(0) = f^{f(n)}\big(f^{-n-f(0)}(0)\big)= f^{f((n+f(0))-f(0))}\big((-n-f(0))\cdot f(0)\big) = 0 $$where the second equation follows from $f^k(0)=kf(0)$ and the third equation follows from (1), we have $f(n)-n-f(0)=0$ for all integers $n$. Hence the function $f$ must be of the form $n\mapsto n+c$ for some constant $c$, and it's easy to see that all such functions work. $\blacksquare$

Comments. There are several possible ways to proceed in Case 2. For example, another way is to plug in $m=0, f(0)$ and $2f(0)$.

anantmudgal09 wrote:
I wonder what was the proposer's motivation :)

Let's say I've always liked FEs with expressions like $f^{f(x)}$, so from the equation
$$mn + k(m+n+k) = (m+k)(n+k)$$there is the natural equation
$$f^{f(m+n)}(mn) = f(m)f(n).$$As for the bijection condition, first I actually thought about the above equation in $\mathbb Z^+$ without getting anything, and when I revisited it later I just somehow thought that making it a bijection will allow changing the equation to $\mathbb Z$ and that worked out perfectly. The solution $n\mapsto (-1)^{n+1}n$ wasn't planned but it just popped out, which in my opinion makes it even nicer.
This post has been edited 1 time. Last edited by talkon, Apr 12, 2018, 7:42 PM
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yayups
1614 posts
#7 • 2 Y
Y by Adventure10, Mango247
We claim the solutions are $f(x)=x+c$ and $f(x)=(-1)^{x+1}x$, both of which can easily be checked to work.

Now suppose that $f$ is some solution to the FE $P(m,n)$. Let's start by solving the problem assuming $f(0)\ne 0$. Note that
\[P(m,0)\implies f^{f(m)}(0)=f(m)f(0)\implies \boxed{f^x(0)=xf(0)}\]for all $x$, since $f$ is bijective so $f(m)$ is any integer $x$.

Let $a=f^{-1}(0)=-f(0)$. Then, we have
\[f^{f(m+a)}(ma)=0\implies ma=f^{-f(m+a)}(0)\implies -mf(0)=-f(0)f(m+a),\]so $f(m+a)=m$, or $f(m)=m-a=m+f(0)$, so $f(x)=x+c$ for some $c\ne 0$.

We will now assume that $f(0)=0$. We have that
\[P(n,-n)\implies f^0(-n^2)=f(n)f(-n)\implies\boxed{f(n)f(-n)=-n^2}.\]In particular, $f(1)f(-1)=-1$, so $\{f(1),f(-1)\}=\{1,-1\}$.

We claim that $\{f(n),f(-n)\}=\{n,-n\}$, where we prove this by induction. The base case of $n=1$ is clear. Now suppose $\{f(m),f(-m)\}=\{m,-m\}$ for all $m<n$. Then, if $f(n)\ne\pm n$, we must have one of $f(n)$ or $f(-n)$ have absolute value $0<m<n$. But if $f(\pm n)=\pm m$, then $\pm n=\pm m$ for some choice of signs since $\pm m=f(\pm m)$, which is a contradiction. Thus, $\{f(n),f(-n)\}=\{n,-n\}$. This also implies that $f^2(x)=x$.

Thus, if $m$ and $n$ are of the same parity, then $P(m,n)$ implies $mn=f(m)f(n)$, so $f(m)/m$ and $f(n)/n$ are the same. Therefore, by fixing the values of $f(1)$ and $f(2)$, we uniquely determine $f$. The four choices give us
\[f(x)\equiv x,-x,(-1)^xx,(-1)^{x+1}(x).\]One can check that only the first and the last work, which solves the problem.
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william122
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#8 • 2 Y
Y by Adventure10, Mango247
Denote the assertion as $P(m,n)$. Suppose that $f(0)=0$. Then, $P(m,-m)$ yields $-m^2=f(m)f(-m)$. Plugging in $m=1$, we get $f(1)f(-1)=-1$, so $\{f(1),f(-1)\}=\{-1,1\}$. Similarly, $f(2)f(-2)=-4$. However, since neither can have magnitude $1$, due to injectivity, we must have $\{f(2),f(-2)\}=\{2,-2\}$. We can induct upwards to get $f(x)=\pm x\forall x$, and $f(f(x))=x$. Now, if we consider $P(m,n)$ for $2|m+n$, note that LHS is just $mn$, so $f(m),f(n)$ must be $m,n$ or $-m,-n$, implying that we take the same sign for all numbers of the same sign. Checking the 4 cases yields the solutions $f(x)=x$ and $f(x)=\begin{cases}x\text{ if }x\equiv 1\pmod 2\\ -x\text{ otherwise}\end{cases}$.

If $f(a)=0$ for $a\neq 0$, consider $P(a,n)$. We have $f^{f(a+n)}(an)=0$. Thus, there exists some $k$ for all multiples of $a$, $an$, such that $f^k(an)=0$. Furthermore, since $f$ is a bijection, $f(a+n)$ cycles through all the integers, implying that $f^k(0)$ for any $k$ is always a multiple of $a$. Now, consider $P(am,a(1-m))$. Similar to before, we get the relation $a^2m(1-m)=f(am)f(a(1-m))$, and using the fact that $a|f(am),f(a(1-m))$, we get a similar induction as before, yielding $f(ax)=-ax$ or $(a-1)x$. Now, if we take $2|x+y$, and consider $P(ax,ay)$, we get $a^2xy=f(ax)f(ay)$. Checking all 4 cases (i.e. $f(ax)=-ax,(a-1)x$, $f(ay)=-ay,(a-1)y$), we find that the only solution is if $f(ax)=(a-1)x$, $f(ay)=(a-1)y$. So, we get that $f(ax)=(a-1)x\forall x$. Finally, note that we had $f^{f(a+n)}(an)=0$. The previous statement tells us, however, that $f^n(an)=0$. As $0$ does not occur in a cycle (i.e. there does not exist $k$ such that $f^k(0)=0$), we must have that $n=f(a+n)$, or $f(x)=x-a$. All $a$ produce valid solutions.

Thus, our solution set is $f(x)=x+c$ for $c\in\mathbb{Z}$, and $f(x)=\begin{cases}x\text{ if }x\equiv 1\pmod 2\\ -x\text{ otherwise}\end{cases}$.
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Idio-logy
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#9 • 1 Y
Y by Adventure10
We divide into two cases: $f(0)=k\neq 0$ or $f(0)=0$.

If $f(0)=k\neq 0$, plug in $m=0 \Longrightarrow f^n(0)=kn$ $\forall n\in\mathbb{Z}$, which implies $f(nk)=(n+1)k$. Specially, $f(-k)=0$. Plug in $n=-k$, then we have $f^{f(m-k)}(-mk)=0$, and since $f$ is bijective we have $f(m-k)=m$. This gives us one solution: $f(m)=m+k$ for all $m$.

If $f(0)=0$, then plug in $n=-m$ we get $f(m)f(-m)=-m^2$. By induction we see that $|f(m)|=m$ and $f(m)=-f(-m)$. If $m+n$ is even, then by the original equation $mn=f(m)f(n)$, meaning that if $m\equiv n\mod 2$, then $f(m)$ and $f(n)$ have the same signs. If $m+n$ is odd (let's say $m$ is even and $n$ is odd), then $f(mn)=f(m)f(n)$. This gives us two solutions $f(x)=x$ and $f(x)=(-1)^{x+1}x$.
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IndoMathXdZ
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#10 • 1 Y
Y by Adventure10
The solutions are $f(n) = n + c$ and $f(n) = (-1)^{n + 1} \cdot n$. It is easy to check that both of them satisfy. Now we'll prove that those are the only ones.
Let $P(m,n)$ denote the assertion of $m$ and $n$ to the given functional equation.
\[ P(m,n) : f^{f(m+n)} (mn) = f(m) f(n) \]We'll consider two cases:
$\textbf{Case 01.}$ When $f(0) = 0$.
\[ P(m,-m) : -m^2 = f^{f(0)}(-m^2) = f(m)f(-m) \]Simple induction gives us $\{ f(m),  f(-m) \} = \{ -m, m \}$ for all $m \in \mathbb{N}$. In both case, we get $f(f(m)) = m$.

Take $m,n$ such that $m+n$ even, then
\[ mn = f^{f(m+n)} (mn) = f(m) f(n) \]
Suppose that $f(1) = 1$, then we have
\[ f^{f(m+1) - 1} (m) = m \]If $m$ is odd, then $f(m+1) - 1$ is odd, this gives us $f(m) = m$.
Hence, if $f(2) = 2$, then $f(n) = n$ for all even $n$. This gives us the solution $\boxed{f(x) = x}$ which is indeed true.

If $f(2) = -2$, then $f(n) = -n$ for all even $n$. This gives us the solution $\boxed{f(x) = (-1)^{x+1} \cdot x}$ which is indeed true.
Suppose that $f(1) = -1$, then $f(x) = -x$ for all odd $x$. Similar reasoning as before, we gain two possible functions: $f(x) = -x$ or $f(x) = (-1)^x \cdot x$.
Take $m$ and $n$ having different parity, this gives us
\[ mn = f^{f(m+n)} (mn) = f(m) f(n) = -mn \]for the latter case and
\[ -mn = f^{f(m+n)}(mn) = f(m) f(n) = mn \]for the former case.
$\textbf{Case 02.}$ When $f(0) \not= 0$, then suppose that $f(c) = 0$ for $c \not= 0$.
$P(k,0)$ gives us
\[ f^{f(k)}(0) = f(k)f(0) \]By surjectivity, we have $f^m (0) = mf(0)$ for all $m \in \mathbb{Z}$.
\[ f^{f(m+n)}(mn) = f(m)f(n) \]$P(m, -f(0))$ gives us
\[ f^{f(m - f(0))} (-mf(0)) = f(m) f(-f(0)) = f(m)f(f^{-1} (0)) = 0 \]We then have
\[ f^{f(m - f(0))} ( f^{-m} (0) ) = 0 \]Hence, $0 = f^{f(m - f(0)) - m} (0) = ( f(m - f(0)) - m)f(0) $, which gives us $f(m - f(0)) = m$ for all $m \in \mathbb{Z}$. This gives the solution $\boxed{ f(n) = n + c}$, which satisfies the original equation.
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jj_ca888
2726 posts
#11 • 2 Y
Y by cosmicgenius, Mango247
Solution w/ cosmicgenius


Let $P(n, m)$ be the assertion. We repeatedly take advantage of the fact that $f$ is bijective. $P(f^{-1} (a), 0)$ gives $f^a (0) = af(0)$ for all integers $a$. Thus $f^{-1} (0) = -f(0)$ and $f(-f(0))= 0$. Furthermore, $P(-f(0), x)$ gives
\[ f^{f(-f(0)+x)} (-f(0)x) = 0.\]But the LHS can be simplified as follows:\begin{align*}
f^{f(-f(0)+x)} (-f(0)x) &= f^{f(-f(0)+x)} (-xf(0))\\&= f^{f(-f(0)+x)} (f^{-x} (0))\\&= f^{-x + f(-f(0)+x)} (0)\\&= f(0) (-x + f(-f(0)+x)) = 0\end{align*}where we repeatedly took advantage of $P(f^{-1}(a), 0)$. Hence, for each individual $x$, we must have $f(x - f(0)) = x$ or $f(0)=0$. If $f(0) \neq 0$, then we must have $f(x - f(0)) = x$ for all $x \in \mathbb{Z}$, which yields the solutions$$f(x) = x + c, c \in \{\mathbb{Z}^+, \mathbb{Z}^-\}$$for nonzero integers $c$.

If $f(0) = 0$, then $P(x, -x)$ gives $-x^2 = f(x)f(-x)$. From $x = 1$, note that $f(1)f(-1) = -1$, hence $f(-1), f(1) \in \{-1, 1\}$ since $f$ brings integers to integers. In fact, we can show that, for all integers $n$, $f(-n), f(n) \in \{-n, n\}$. We only have to consider positive $n$, since then negative $n$ must follow by symmetry and $f(0) = 0$ is already assumed.

We will use strong induction. Our base cases of $n = 1, 2$ are obvious. For the inductive step, assume that for all positive integers $n < k$, $f(-n), f(n) \in \{-n, n\}$. $P(k, -k)$ yields $-k^2 = f(k)f(-k)$. If $|f(k)| > k$, then $|f(-k)| < k$, a contradiction since all integers in the range $(-k, k)$ are already attained by $f$ at some value $v \in (-k, k)$. For the same reason, $|f(k)| < k$ cannot hold. Hence, we must have $|f(k)| = |f(-k)| = k$ as desired. $\square$

Thus, for all $n \in \mathbb{Z}$, $f(n) = -n$ or $n$, so $f(f(n)) = f^2(n) = n$. Now, back to the original assertion $P$, consider all pairs $(m, n)$ with same parity $\implies m+n$ even. $f^{f(m+n)}(mn) = mn = f(m)f(n)$. Either $f(m) = m$ and $f(n) = n$, or $f(m) = -m$ and $f(n) = -n$. Now, we just have to consider all four possibilities regarding $f(1) \in \{-1, 1\}$ and $f(2) \in \{-2, 2\}$. Checking all such possibilities, we see that it is only possible for $f(1) = 1$ and $f(2) = 2$, or $f(1) = 1$ and $f(2) = -2$, hence the solutions $f(x) = x$ and $f(x) = x$ when $x$ odd and $f(x) = -x$ when $x$ even.

in summary, our solutions are $\boxed{f(x) = x + c, c \in \mathbb{Z}}$, and $$\boxed{f(x)=\begin{cases} x &\text{ if } x \text{ is odd} \\ -x &\text{ if } x \text{ is even} \end{cases}}$$, as desired. $\blacksquare$
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pad
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#13
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Let $P(m,n)$ denote the FE. Note $P(m,0)$ gives $f^{f(m)}(0)=f(m)f(0)$, and since $f$ is surjective, $f^x(0)=xf(0)$ for any $x$. In particular,
  • $f^{-1}(0)=-f(0)$, i.e. $f(-f(0))=0$, and
  • $0=f^{-x}(xf(0))$, so $f^y(-yf(0))=0$ for any $y$.
Now, $P(m+f(0),-f(0))$ gives \[ f^{f(m)}\big( -f(0)(m+f(0)) \big) =0. \]But $f^{f(m)}(-f(m)f(0))=0$ by using the second bullet above. Since $f$ is bijective, this implies $-f(m)f(0)=-f(0)[m+f(0)]$. If $f(0)\not = 0$, then $f(m)=m+f(0)$, i.e. $f(m)=m+c$ for some constant $c$. It is easy to check that this works in general.
Now assume $f(0)=0$. $P(n,-n)$ gives \[ -n^2=f(n)f(-n).\]Let $g(n)=|f(n)|$. Then $g(n)\ge 0$ and $n^2=g(n)g(-n)$ for all $n$. We claim $g(n)=|n|$ for each $n$. Induct on $|n|$. For $n=1$, $1=g(1)g(-1)$, so $g(1)=g(-1)=1$. WLOG $n$ is positive. We have $g(n)=k$ and $g(-n)=n^2/k$ for some positive $k\mid n^2$. If $k\not = n$, then WLOG $k<n$. But $g(k)=|k|=k$, contradicting injectivity. Hence $g(n)=|n|$, i.e. $f(n)=\pm n$ for each $n$.
  • So $\{f(n),f(-n)\}=\{n,-n\}$, which implies $f(-n)=-f(n)$.
  • If $f(n)=n$, then $f(f(n))=f(n)=n$, and if $f(n)=-n$, then $f(f(n))=f(-n)=-f(n)=n$. In all cases, $f(f(n))=n$.
  • Now, if $m+n$ even, then $mn=f(m)f(n)$.
So if $s(n)=f(n)/n$, then $s(m)=s(n)$ for $m\equiv n \pmod2$. If $s(1)=1$ and $s(2)=1$, then $f(n)=n$. If $s(1)=1$ and $s(2)=-1$, then $s(n)=(-1)^{n+1}$, i.e. $f(n)=(-1)^{n+1}n$. The other two cases are symmetric.
In conclusion, the solutions are \[ f(n)=n+c, \qquad f(n)=(-1)^{n+1}n,\]for any $c$, both of which work.

Remarks
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Eyed
1065 posts
#14 • 2 Y
Y by kevinmathz, RedFlame2112
The two solutions are $f(x) = x + c$, or $f(x) = x$ for odd $x$ and $f(x) = -x$ for even $x$. It's not hard to see that these both work.

If $f(x) = 1, f(y) = 0$, we have $f(m(x-m)) = f(m)f(x-m)$, $m(y-m) = f(m)f(y-m)$
$f(y(x-y)) = 0$, so by bijectivity, $y(x-y) = y$. Either $y = 0, x = y+1$.

If $y = 0$, we have $-m^{2} = f(m)f(-m)$, so $f(1) = \pm1$ and $x = \pm1$. If $x = -1$, then $f(m(-1-m)) = f(m)f(-1-m)$, plug in $m = 1$ for contradict.
Thus, $y = 0, x = 1$, $f(0) = 0, f(1) = 1, f(-1) = -1$. inducktively, prove that $f(x) = \pm x$ (true through bijectivity), (induckt on $x$). Repeat proof that if $x$ odd, then $f(x) = x$, otherwise $f(x)$ can be either $x$ or $-x$.

If $x = y+1$, we have $x(y-x) = f(x)f(y-x)$ implies $f(-1) = -x$, also $1(y-1) = f(1)f(y-1)$ so $x | y-1, y+1 | y-1$. Take some $n$ such that $f(n) = -1$, then $f^{-1}(0) = f(0)(-1) \Rightarrow f(0) = -y$. Observe that
\[f^{f(n)}(0) = f(n)f(0) = f(n)\cdot -y, f^{c}(0) = c\cdot -y\]We can prove $f(ky) = (k-1)y$ for integer $k$, by inducktion. Our base case is $k = 1$, which is true since $f(y) = 0$. Now, assume it's true for $k$, we will prove that it is true for $k + 1$. Using the observation, since $f^{-k}(0) = ky$, this means
\[f^{-1}(f^{-k}(0)) = f^{-1}(ky) = f^{-k-1}(0) = (k+1)y\]This implies $f((k+1)y) = ky$.

Now we can prove $f(ky + 1) = (k-1)y + 1$. This is because, consider $m = y+1, n = (k-1)y$. Now,
\[f^{f(ky + 1)}(y(k-1)(y+1)) = y((k-1)(y+1) - f(ky+1))\]\[= f(y(k-1))f(y+1) = y(k-2)\]This means $ky - y + k - 1 - (k-2) = f(ky+1) = (k-1)y + 1$. A useful corollary is $f^{r}(ky + 1) = (k-r)y + 1$.

Now, we will use inducktion on $r$. to show that $f(ky + r) = y(k-1) + r$. Our base case is already proven, for $r = 0, 1$. Assume that it is true for all $i < r$, such that $f^{c}(ky + i) = (k-c)y + i$. We prove the same for $r$. We have:
\[f^{f(ky + r)}(((k-1)y + (r-1))(y + 1)) = f((k-1)y+(r-1))f(y+1)\]\[= f((k-1)y+(r-1)) = (k-2)y + r-1\]\[ = f^{f(ky + r)}(y((k-1)y + (r-1) +(k-1)) + (r-1)) \]\[= y((k-1)y + (r-1) + (k-1) - f(ky + r)) + r-1\]\[\Longrightarrow f(ky + r) = (k-1)y + (r-1) + (k-1) - (k-2) = (k-1)y + r\]Thus, our inducktive step is proven.

We conclude that $f(x) = x - y$. Thus, our only two solutions are the ones listed here.
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DottedCaculator
7356 posts
#15 • 2 Y
Y by centslordm, RedFlame2112
Solution
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VicKmath7
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#16
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Solution
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CircleInvert
654 posts
#17
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Observe that $f(x)=x+k$ works. Furthermore, if $f(0)=k\ne 0$, then we get $f^{f(m)}(0)=kf(m)$. Now, as $f$ is a bijection, we can replace $f(m)$ with $m$, and we get $f^{m}(0)=km$. Then $f(x)=x+k$ holds for $x=km$. Then plugging in $n=-k$, we get $f^{f(m-k)}(-km)=0=f^{m}(-km)$. so $f(m-k)=m$ for all $m$, thus implying that we have $f(x)=x+k$. Hence this is the only solution unless $f(0)=0$.

We now take the $f(0)=0$ case: then $m=-n$ gives $-m^2=f(m)f(-m)$. Now, we claim that we have $|f(\pm k)|=k$ where $k\ge 0$. Suppose not. First observe that for $k=0$ this holds and for $k=1$ this must also hold from $m=1$ implying $-1=f(m)f(-m)$ and $1$ and $-1$ are the only divisors of $-1$. Thus, taking the smallest counterexample for $k$, we have $k\ge 2$. Then $f(k)f(-k)=-k^2$. Now, as $k$ is the smallest counterexample, we know that $|f(k)|\ge k$ as all values with lower absolute value are mapped to values with those absolute values and are thus in orbits with values of lower absolute value. Similarly, $|f(-k)|\ge k$. Now, at least one of these inequalities must be strict in order for $k$ to be a counterexample, but this contradicts $f(k)f(-k)=-k^2$. Thus, we have $|f(\pm k)|=k$. Now, we just need to determine when $f(x)=x$ and when $f(x)=-x$. Observe that if $m$ and $n$ have the same parity, the original FE just says that $mn=f(m)f(n)$ indicating that either there is a sign change for both $m$ and $n$ or for neither. Thus, sign change or not is dependent only on the parity of the input. For $m+n$ odd, the original FE just says $f(mn)=f(m)f(n)$. Now say $m$ is the even one and $n$ is the odd one; then $f(mn)$ and $f(m)$ are either both sign changed or both left alone so $f(n)=n$ (assuming we choose $m$ to be not $0$ which we may do). Thus, $f$ of any odd number is equal to that number and then we can either have the same be true for evens or we can have it such that $f$ of any even is negative that even. This is two possible solutions for $f$; we see that both work.
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cj13609517288
1922 posts
#18
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The answer is $\boxed{f(x)=x+c}$ for all integers $c$, along with
\[\boxed{f(x)=\begin{cases}x&x\text{ is odd}\\-x&x\text{ is even}\end{cases}}\,.\]These "clearly" work(the last function is easier to check after the reductions in case 2). For the proof, we split into cases.

Case 1. $f(0)\ne 0$. Let $f(0)=c$. Then $P(m,0)$ gives
\[f^{f(m)}(0)=cf(m)\Longrightarrow f^m (0)=cm.\]Thus the chain through $0$ goes $\cdots\rightarrow -2c\rightarrow -c\rightarrow 0\rightarrow c\rightarrow 2c\rightarrow\cdots$.

Now $P(m,c)$ gives
\[f^{f(m+c)}(mc)=f(m)f(c)\Longrightarrow mc+cf(m+c)=2cf(m)\Longrightarrow m+f(m+c)=2f(m).\]Let $g(x):=f(x)-x-c$. Then
\[m+g(m+c)+m+2c=2g(m)+2m+2c\Longrightarrow g(m+c)=2g(m).\]Thus by $\nu_2$ we have $g\equiv 0$, so $f(x)=x+c$ for all $x$, which works.

Case 2. $f(0)=0$. Then $P(m,-m)$ gives $-m^2=f(m)f(-m)$. Therefore, $\{f(1),f(-1)\}=\{1,-1\}$, so $\{f(2),f(-2)\}=\{2,-2\}$, so $\{f(3),f(-3)\}=\{3,-3\}$, etc, so $f(m)=\pm m$ for all $m$. Therefore, $f^2(m)=m$ for all $m$, so the two possible values of $f(m+n)$ do the same thing. Therefore,
\[f^{m+n}(mn)=f(m)f(n).\]Let $g(n):=\frac{f(n)}{n}$ be defined for all nonzero integers. Then
\[g(mn)^{m+n}=g(m)g(n).\]If $m$ and $n$ have the same parity, then $g(m)g(n)=1$. Therefore, $g$ is identical for each individual parity. Now let $m=1$ and $n=2$, then
\[g(2)=g(1)g(2)\Longrightarrow g(1)=1,\]as desired. $\blacksquare$
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Ilikeminecraft
656 posts
#19
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We do casework on if $f(0) = 0$ or not.

If $f(0) = 0,$ take $m = 1, n = -1$ to get $(f(1), f(-1)) = (1, -1), (-1, 1).$ Take $(m, n) = (2, - 1)$ to get $f(-2) = f(2)f(-1).$ By induction, we can force $f(k)\in\{k, -k\}$ and $f$ is an involution. We do casework on $f(1), f(2)$:

If $f(1) = 1, f(2) = 2:$ take $(2, 2^{k})$ to get that $2^{k + 1} = f(2)f(2^k),$ and so $f(2^k) = 2^k.$ take $(1, 2\ell - 1)$ to get $f(2\ell - 1) = 2\ell - 1.$ Then, take $(2^k, 2\ell - 1)$ to get $f\equiv x$
If $f(1) = -1, f(2) = 2:$ take $(2, 4)$ to get that $f(4) = 4$. take $(4, 1)$ to get $f(4) = f(4)f(1),$ contradiction.
if $f(1) = 1, f(2) = -2:$ take $(2, 2^k)$ to get $2^{k + 1} = f(2)f(2^k),$ so $f(2^k) = -2^k.$ $(1, 2\ell - 1)$ to get $f(2\ell - 1) = 2\ell - 1.$ take $(2^k, 2\ell - 1), f(2^k(2\ell - 1)) = -2^k(2\ell - 1).$ thus, $f\equiv (-1)^xx$.
if $f(1) = -1, f(2) = -2:$ do a ``merging'' of case 1 and case 3 to get $f\equiv -x.$

Now, assume $f(0) = k \neq 0.$ Take $m = 0$ and we get $f^{f(n)}(0) = kf(n),$ or $f^n(0) = nk.$ Furthermore, $f(-k) = 0$(from $n = -1$). Take:
\begin{align*}
    f^{f(n) - n - k}(0) & = f^{f(n)}(f^{-n-k}(0)) \\
    & = f^{f((n + k) - k)}(-(n + k)k) \\
    & = 0
\end{align*}so $f(n) = n + k.$
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MathLuis
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#20
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Denote $P(m,n)$ to be the assertion of the given F.E.
From $P(0,n)$ and bijectivity we get $f^t(0)=f(0) \cdot t$ for all $t \in \mathbb Z$ and from here we can get that $f(t \cdot f(0))=(t+1)f(0)$ but also that $f^{u}(t \cdot f(0))=(t+u)f(0)$ for all $t,u, \in \mathbb Z$. Setting $u=1, t=-1$ gives $f(-f(0))=0$.
Now from $P(m+f(0), -f(0))$ we get that $f^{f(m)}(-(m+f(0))f(0))=f(m+f(0))f(-f(0))=0$ and thus $(f(m)-m-f(0))f(0)=0$, now if $f(0) \ne 0$ then trivially we have $f(x)=x+c$ for all $x \in \mathbb Z$ and some integer $c \ne 0$ which works.
Now if $f(0)=0$ were to hold then $P(m,-m)$ gives $f(m)f(-m)=-m^2$, now let $m=p$ for $p$ being a prime number then this means for all primes except on the case of whether $f(-p)=1$ or $f(p)=1$ for exctly one prime $p$, it holds that $\{ f(p), f(-p) \}= \{ p, -p \}$.
This should hold for all primes $p$ because $P(1,-1)$ gives $f(1)f(-1)=-1$ and therefore one of these has to be $1$ and in either case they are not a prime number, now notice that from induction and these finding we can now easly conclude that $\{f(n), f(-n) \}= \{ n, -n \}$ for all $n \in \mathbb Z$, and thus this means $f(f(x))=x$ for all integers $x$ and now from $P(m,n)$ for $m \equiv n \pmod 2$ gives that $f(n)f(m)=mn$ and this can only mean either $f(m)=m, f(n)=n$ or $f(m)=-m, f(n)=-n$ and for $m-n$ odd we have that $f(n)f(m)=f(mn)$ so setting $m$ to be a non-zero even number and $n=1$ gives that $f(1)=1$ and therefore we either have $f(x)=x$ for all $x \in \mathbb Z$ or $f(x)=(-1)^{x+1} x$ for all $x \in \mathbb Z$ which can be seen to work as well, thus we are done :cool:.
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HamstPan38825
8866 posts
#21
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The answers are $f(n) = n + c$ for integers $c$ and the intriguing $f(n) = (-1)^{n+1}n$.

Based on the solution set, we will split into cases based on whether $f(0) = 0$.

First Case: Suppose that $f(0) = 0$. Then setting $m=-n$ yields $f(n) f(-n) = -n^2$. Because $f$ is bijective, it is easy to see that $\{f(1), f(-1)\} = \{-1, 1\}$ and inductively $\{f(n), f(-n)\} = \{n, -n\}$ for all positive integers $n$, otherwise the smaller of the two yields a contradiction by injectivity.

In particular, $f(f(n)) = n$ for all positive integers $n$, so if $m + n$ is even, it follows that $f(m)f(n) = mn$. It follows that $\tfrac{f(n)}n$ is constant for $n$ of the same parity, and it is not hard to show by setting $m+n$ odd that either $f(n) = n$ for all $n$ or $f(n) = -n$ only when $n$ is even.

Second Case: Suppose now that $f(0) = c \neq 0$. By setting $m = 0$, we get $f^{f(n)}(0) = cf(n)$, implying that $f^n(0) = cn$ for all integers $n$ as $f$ is bijective. In particular, $f(kc) = (k+1)c$ for all integers $k$, and thus $f(-c) = 0$.

Now, for an integer $a$, letting $(m, n) = (-c, a)$ now yields \[0 = f^{f(a-c)}(-ac) = -ac + f(a-c)c\]which implies $f(a-c) = a$, as needed.
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AN1729
17 posts
#22
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Solved with rjp08

The solutions are $f(x)=x+a$ and $f(x)=(-1)^{x-1}x$

Denote the problem condition as $P(m,n)$

Case 1: $f(0)=a\neq 0$
$P(0,f^{-1}(n)) \implies f^n(0) = nf(0)$
Thus we get the chain $\dots \rightarrow -2a \rightarrow -a \rightarrow 0 \rightarrow a \rightarrow 2a \rightarrow \dots$
Now, $P(-a,n) \implies f^{f(n-a)}(-an) = 0 \implies -an + af(n-a)=0$
But $a \neq 0 \implies f(n-a)=n    \forall n \in \mathbb{Z}$

Case 2: $f(0)=0$
$P(m,-m) \implies -m^2 = f(m)f(-m)$
By induction on $|{m}|$, we get that $\forall m$ we have $\{f(m),f(-m)\} = \{m,-m\}$
Define $g : \mathbb{Z} \rightarrow \{-1,1\} $such that $f(n)=ng(n)$
Thus, we have that $f(f(x))=x$
Now, if $m \equiv n \pmod {2}$, $m+n$ and hence $f(m+n)$ is even. $\implies mn= f(m)f(n) \implies g(m)=g(n)$
Verifying cases for $g(1)= \pm 1$ and $g(2) = \pm 1$, we get the only valid solutions as $f(x) = x$ and $f(x) = (-1)^{x-1}x$
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