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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
Prove the inequality
Butterfly   0
7 minutes ago

Prove
$$x^2+y^2+7\ge 3(x+y)+\frac{9}{xy+2}~~(x,y>0).$$
0 replies
Butterfly
7 minutes ago
0 replies
Kaprekar Number
CSJL   5
N 8 minutes ago by Adywastaken
Source: 2025 Taiwan TST Round 1 Independent Study 2-N
Let $k$ be a positive integer. A positive integer $n$ is called a $k$-good number if it satisfies
the following two conditions:

1. $n$ has exactly $2k$ digits in decimal representation (it cannot have leading zeros).

2. If the first $k$ digits and the last $k$ digits of $n$ are considered as two separate $k$-digit
numbers (which may have leading zeros), the square of their sum is equal to $n$.

For example, $2025$ is a $2$-good number because
\[(20 + 25)^2 = 2025.\]Find all $3$-good numbers.
5 replies
CSJL
Mar 6, 2025
Adywastaken
8 minutes ago
Projective geometry
definite_denny   0
9 minutes ago
Source: IDK
Let ABC be a triangle and let DEF be the tangency point of incircirle with sides BC,CA,AB. Points P,Q are chosen on sides AB,AC such that PQ is parallel to BC and PQ is tangent to the incircle. Let M denote the midpoint of PQ. Let EF intersect BC at T. Prove that TM is tangent to the incircle
0 replies
definite_denny
9 minutes ago
0 replies
Problem 7 of RMO 2006 (Regional Mathematical Olympiad-India)
makar   11
N 15 minutes ago by SomeonecoolLovesMaths
Source: Functional Equation
Let $ X$ be the set of all positive integers greater than or equal to $ 8$ and let $ f: X\rightarrow X$ be a function such that $ f(x+y)=f(xy)$ for all $ x\ge 4, y\ge 4 .$ if $ f(8)=9$, determine $ f(9) .$
11 replies
makar
Sep 13, 2009
SomeonecoolLovesMaths
15 minutes ago
Hardest in ARO 2008
discredit   26
N 22 minutes ago by JARP091
Source: ARO 2008, Problem 11.8
In a chess tournament $ 2n+3$ players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least $ n$ next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
26 replies
discredit
Jun 11, 2008
JARP091
22 minutes ago
Inequality
Kei0923   2
N an hour ago by Kei0923
Source: Own.
Let $k\leq 1$ be a fixed positive real number. Find the minimum possible value $M$ such that for any positive reals $a$, $b$, $c$, $d$, we have
$$\sqrt{\frac{ab}{(a+b)(b+c)}}+\sqrt{\frac{cd}{(c+d)(d+ka)}}\leq M.$$
2 replies
Kei0923
Jul 25, 2023
Kei0923
an hour ago
PAMO 2023 Problem 2
kerryberry   6
N an hour ago by justaguy_69
Source: 2023 Pan African Mathematics Olympiad Problem 2
Find all positive integers $m$ and $n$ with no common divisor greater than 1 such that $m^3 + n^3$ divides $m^2 + 20mn + n^2$. (Professor Yongjin Song)
6 replies
kerryberry
May 17, 2023
justaguy_69
an hour ago
My Unsolved Problem
ZeltaQN2008   0
an hour ago
Source: IDK
Given a positive integer \( m \) and \( a > 1 \). Prove that there always exists a positive integer \( n \) such that \( m \mid (a^n + n) \).

P/s: I can prove the problem if $m$ is a power of a prime number, but for arbitrary $m$ then well.....
0 replies
ZeltaQN2008
an hour ago
0 replies
Computing functions
BBNoDollar   3
N an hour 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 $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )
3 replies
BBNoDollar
May 21, 2025
wh0nix
an hour ago
Computing functions
BBNoDollar   8
N an hour 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 $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )
8 replies
BBNoDollar
May 18, 2025
wh0nix
an hour ago
Find the remainder
Jackson0423   1
N 2 hours ago by wh0nix

Find the remainder when
\[
\frac{5^{2000} - 1}{4}
\]is divided by \(64\).
1 reply
Jackson0423
3 hours ago
wh0nix
2 hours ago
IMO 2018 Problem 1
juckter   170
N 2 hours ago by Adywastaken
Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.

Proposed by Silouanos Brazitikos, Evangelos Psychas and Michael Sarantis, Greece
170 replies
juckter
Jul 9, 2018
Adywastaken
2 hours ago
Nice "if and only if" function problem
ICE_CNME_4   2
N 2 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!
2 replies
ICE_CNME_4
Yesterday at 7:23 PM
wh0nix
2 hours ago
Minimum of this fuction
persamaankuadrat   1
N 2 hours ago by alexheinis
Source: KTOM January 2020
If $x$ is a positive real number, find the minimum of the following expression

$$\lfloor x \rfloor + \frac{500}{\lceil x\rceil^{2}}$$
1 reply
persamaankuadrat
4 hours ago
alexheinis
2 hours ago
integer functional equation
ABCDE   156
N May 21, 2025 by MathIQ.
Source: 2015 IMO Shortlist A2
Determine all functions $f:\mathbb{Z}\rightarrow\mathbb{Z}$ with the property that \[f(x-f(y))=f(f(x))-f(y)-1\]holds for all $x,y\in\mathbb{Z}$.
156 replies
ABCDE
Jul 7, 2016
MathIQ.
May 21, 2025
integer functional equation
G H J
G H BBookmark kLocked kLocked NReply
Source: 2015 IMO Shortlist A2
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Ilikeminecraft
658 posts
#164
Y by
The answer is $f\equiv-1, f\equiv x + 1,$ both of which clearly work.
let $P(x, y)$ be the assertion.
$P(x, f(x))\implies f(x - f(f(x))) = f(f(x)) - f(f(x)) -1 = -1.$
$P(x, x - f(f(x)))\implies f(x + 1) = f(x - f(x - f(f(x)))) = f(f(x)) - f(x - f(f(x))) - 1 = f(f(x)).$
$P(x, y), P(x - 1, y)\implies$
\begin{align*}
    f(x - f(y)) & = f(x + 1) - 1 - f(y) \\ 
    f(x - 1 - f(y)) & = f(x) - 1 - f(y) 
\end{align*}By subtracting, we get $f(x - f(y)) - f(x - 1 - f(y)) = f(x + 1) - f(x).$ Taking $x= f(y)$, we get $f(0) - f(-1) = f(f(y) + 1) - f(f(y)) = f(f(f(y))) - f(f(f(y - 1))).$ Hence, $f(f(f(x)))$ is a linear function since its finite difference is constant. We write $f(f(f(x))) = ax + b.$
$P(f(y), y)\implies f(0) = f(f(f(y))) - f(y) - 1\implies f(y) = ay + b + 1 - f(0),$ or $f$ is linear.

let $f(x) = ax + b.$ plug this in to get $ax - a^2y - ab + b = a^2x + ab + b - ay - b - 1.$ Thus, $a = 1, 0.$ If $a = 1, b = 1,$ and if $a = 0, b = -1.$
This post has been edited 1 time. Last edited by Ilikeminecraft, Jan 24, 2025, 11:13 PM
Reason: typo
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EpicBird08
1755 posts
#165
Y by
The only solutions are $f(x) = -1$ and $f(x) = x + 1$, which both work. Now we show that these are all the solutions. Let $P(x,y)$ denote the given assertion.

$P(0,f(0))$ gives $f(-f(f(0))) = -1.$ Letting $u = -f(f(0)),$ we see that $P(x,u)$ gives $$\boxed{f(f(x)) = f(x+1)}.$$Then $P(f(x),x)$ gives $$f(0) = f(f(f(x))) - f(x) - 1 = f(f(x+1)) - f(x) - 1 = f(x+2) - f(x) - 1.$$This implies that $$\boxed{f(x+2) = f(x) + f(0) + 1}.$$We now consider two cases.

Case 1: $f(0) = -1.$ The above boxed equation immediately gives $f(2k) = -1$ for all $k \in \mathbb{Z}.$ Let $f(1) = f(2k+1) = c$ for integers $k.$
Subcase 1: $c$ is odd. Then $P(2k,2k-1)$ gives $c = c - c - 1 \implies c = -1,$ yielding our first solution $f(x) = -1$ for all $x.$
Subcase 2: $c$ is even. Then $P(2k+1,2k+1)$ gives $c = -1 - c - 1 \implies c = -1,$ a contradiction.
Thus this case has been exhausted, and it yields one solution $\boxed{f(x) = -1}.$

Case 2: $f(0) \ne -1.$ Then the first boxed equation implies that if $f(x) \equiv x+1 \pmod{2},$ then $f(x) = x+1.$ By the second boxed equation, we know $f(2) = 2 f(0) + 1$ is odd, as is $2 + 1 = 3.$ Hence $f(2) = 3,$ immediately giving $f(0) = 1.$ Hence $f(x) = x+1$ for even $x.$ Let $f(x) = x + c$ for odd $x.$ Again, the first boxed equation implies $f(1+c) = f(f(1)) = f(2) = 3.$ Hence either $1+c+1 = 3$ or $1 + 2c = 3.$ Both cases give $c = 1,$ yielding our second solution $\boxed{f(x) = x + 1}$ for all $x.$

Hence the only solutions are those claimed at the beginning.
This post has been edited 2 times. Last edited by EpicBird08, Jan 29, 2025, 12:11 AM
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Ilikeminecraft
658 posts
#166
Y by
The solution is $f\equiv 1, x + 1.$

Take $y = f(x)$ to get that there exists $a = x - f(f(x))$ such that $f(a) = -1.$

Take $y = a$ to get that $f(x + 1) = f(f(x)).$

If $f$ is injective, we are done, as this implies $x + 1 = f(x),$ which indeed works.

Now, assume $f$ is not injective, and $f(a) = f(b)$ for some $a\neq b.$ Note that we can get $f(a + 1) = f(f(a)) = f(f(b)) = f(b + 1),$ and so thus, $f(x) = f(x + k(a - b))$ for some $k \in \mathbb Z.$

Take $x = f(y)$ to get $f(0) = f(f(f(y))) - f(y) - 1 = f(y + 2) - f(y) - 1$ to get that $f(2k)$ is linear and $f(2k + 1)$ is linear. By $f(x) = f(x + k(a - b)),$ we can also get that $f(2k) = c_2, f(2k + 1) = c_1$ for two constants(not necessarily the same). Finally, we do parity casework on $c_1, c_2:$
\begin{enumerate}
\item If $c_2\equiv 0\pmod 2,$ then taking $x\equiv y\equiv 0\pmod 2$ tells us $c_2 = -1,$ contradiction.
\item If $c_1 \equiv 0\pmod 2,$ then taking $x\equiv 0, y\equiv 1\pmod 2$ tells us $c_1 = -1,$ contradiction.
\item If $c_1 \equiv c_2 \equiv 1 \pmod 2,$ taking $x\equiv 0\pmod 2$ tells us $f(y) = -1,$ which finishes.
\end{enumerate}
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Bardia7003
22 posts
#167
Y by
Let $P(x,y)$ denote the given assertion.
$\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}}
P(f(x),x): f(0) = \underline{f(f(f(x))) - f(x) - 1 \quad (\RomanNumeral{1})} \\
P(x, f(x)):  \underline{f(x - f(f(x))) = -1 \quad (\RomanNumeral{2})} \\
P(x, x - f(f(x))): f(x - f(x - f(f(x)))) = f(f(x)) - f(x - f(f(x))) - 1 \xrightarrow{\RomanNumeral{2}}  \underline{f(x+1) = f(f(x)) \quad (\RomanNumeral{3})}$
So by $\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} (\RomanNumeral{3})$ we have that $f(f(f(x))) = f(f(x+1)) = f(x+2)$, and putting that in $\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} (\RomanNumeral{1})$ we have: $ \underline{f(0) + 1 = f(x+2) - f(x) \newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} \quad (\RomanNumeral{4})}$. Which means $f(2x)$ is linear. So $f(2x) = ax + b$, which $a = f(0) + 1 = b + 1$, So $f(2x) = (b+1)x + b$.
$f(2x+1)$ is also linear, so $f(2x + 1) = ax + c (a = f(0) + 1)$
We want to prove that $b$ is odd, so we assume otherwise. For an odd $x$, $f(f(2x)) = f((b+1)x + b)$, $(b+1)x$ is odd and $b$ is even so the value is odd: $f(f(2x)) = (b+1)(\frac{(b+1)x + b - 1}{2}) + c$. Also by $\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} (\RomanNumeral{3})$ we know $f(f(2x)) = f(2x + 1) = (b+1)x + c$
So $(b+1)x = (b+1)(\frac{(b+1)x + b - 1}{2})$. $b$ is even so $b+1\neq0$ and we can cross it out: $x = \frac{bx + x + b - 1}{2} \rightarrow x = bx + b -1 \rightarrow (1-b)x = (b - 1) \xrightarrow{b-1\neq0} x = -1$, but we had the equation for any odd $x$, hence contradiction. As a result, we proved $b$ is odd.
Now we know $f(2x+1) = ax+c \xrightarrow{\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} \RomanNumeral{3}} f(f(2x)) = ax + c \rightarrow f(ax + b) = ax + c \xrightarrow{x := 0} f(b) = c$
As we know $b$ is odd, so:
$c = f(b) = a(\frac{b-1}{2}) + c \rightarrow a(\frac{b-1}{2}) = 0$ so $b = 1$ or $a = 0$.
Case 1: $a = 0$. Then $b+1 = 0, f(2x) = b \rightarrow f(2x) = -1$ and $f(2x + 1) = c$. If c is odd, then by $\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} (\RomanNumeral{3})$ we know $f(c+1) = f(f(c)) = f(c) = c$ and $c + 1$ is even so $-1 = c = f(2x+1)$.
Hence $\boxed{f(x) = -1 \quad \forall x \in \mathbb{Z}}$ is the solution in this case, which indeed works.
Case 2: $b=1$. Then $f(2x) = 2x + 1, f(2x + 1) = 2x + c$. By $\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} (\RomanNumeral{2})$ we know $\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} f(2x - f(f(2x)) = -1 \xrightarrow{\RomanNumeral{3}} f(2x - f(2x + 1)) = -1 \rightarrow f(2x - 2x - c) = -1 \rightarrow f(-c) = -1$. If c is even then $-c +1 = -1 \rightarrow c = 2$, and we want to prove $c$ is even so we can conclude $c=2$, so we assume not. If c is odd then $f(f(c)) = f((c-1) + c) = ((2c-1)-1) + c$, and by $\newcommand{\RomanNumeral}[1]{\MakeUppercase{\romannumeral #1}} (\RomanNumeral{3})$ we know that $f(f(c)) = f(c+1) = (c+1) + 1$ so $3c - 2 = c + 2 \rightarrow c = 2$ but we supposed c is odd, contradiction.
So we proved $c$ is even and as a result $c=2$, as we proved. Now we have that $f(2x) = 2x+1, f(2x+1) = 2x+2$, so we can generally conclude $\boxed{f(x) = x+1 \quad \forall x \in \mathbb{Z}}$ in this case, which is also a solution.
So both cases are solved and we found all the solutions. $\blacksquare$
Please feel free to point out if I've made any mistakes through this proof, thanks. :)
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Ihatecombin
65 posts
#168
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A pretty unique problem, bit annoying to be honest lol. The only solutions are $f(x) \equiv -1$ and $f(x) \equiv x+1$.
You can use \(P(x,f(x))\) to get
\[f(x - f^2(x)) = -1\]Afterwards let \(f(r) = -1\), we can use \(P(x,r)\)
to get
\[f(x+1) = f^2(x)\]Notice that the equation can be transformed into
\[f(x-f(y)) = f(x+1) - f(y) - 1\]Let \(f(p+1) = k \neq -1\) (otherwise \(f(x) \equiv -1\), which works).
Notice that if we use \(P(k+p+1,p+1)\), then we must have
\[f(p+1) = f(k+p+2) -k-1 \Longrightarrow 2k+1 =  f(k+p+2)\]Since \(k \neq -1\), we must have that \(2k+1 \neq k\) and \(2k+1\) is also in the image of \(f(x)\),
thus \(2(2k+1) + 1\) is also in the image and so on. Therefore the set containing all the integers which are in the image of \(f(x)\) is infinite.
Now let \(n\) be a number such that \(f(n) = k \neq -1\) for some \(k\). Substituting \(P(x,n)\) we have
\[f(x-k) = f(x+1)-k-1\]we already know that if \(k\) is in the image of \(f(x)\), then \(2k+1\) is in the image. Let \(f(m) = 2k+1\), we can substitute \(P(x,m)\) to get
\[f(x-2k-1) = f(x+1) - 2k - 2\]Thus we must have
\[f(x-2k-1) + k + 1= f(x-k) \Longrightarrow f(x) = f(x-k-1) +k+1\]for some \(k \neq -1\).

Notice that since \(f(x+1) = f^2(x)\), if \(f(x)\) is injective then we are done \(f(x) \equiv x+1\).
Therefore assume \(f(x) \not\equiv -1\) and \(f(x) \not\equiv x+1\) and \(f(x)\) is not injective. Let \(f(a) = f(b)\), notice that since \(f(x+1) = f^2(x)\), we must have
\[f(a+1) = f^2(a) = f^2(b) = f(b+1)\]similarly
\[f(a+2) = f^2(a+1) = f^2(b+1) = f(b+2) \Longrightarrow f(a) = f(b) = f(2b-a)\]By induction we have
\[f(a) = f(xb-(x-1)a)\]However substituting \(k+1\), we have
\[f(a) = f([k+1](b-a) + a)\]But since \(f(x) = f(x-k-1) + k +1\), we easily get
\[f(a) = f([k+1](b-a) + a) = f(a) + (k+1)(b-a)\]This is a clear contradiction.
This post has been edited 1 time. Last edited by Ihatecombin, Mar 13, 2025, 1:28 PM
Reason: Apparently \textbf{} doesn't cause bold text in AOPS
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Adywastaken
60 posts
#169
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$y=f(x)\implies f(x-f(f(x)))=-1$
$y=x-f(f(x))\implies f(x+1)=f(f(x))$
$x=f(y)-1\implies f(-1)+1=f(f(f(x)-1))-f(x)=f(f(x))-f(x)=f(x+1)-f(x)$
So, $f=mx+c$, and matching coefficients,
$m^2=m, 2mc=c+1$
So, $(m,c)=(0,-1),(1,0)$
$f(x) \equiv -1$ or $f(x)=x+1$
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Jakjjdm
4 posts
#170
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The only solutions are $f(x) \equiv -1$ and $f(x) = x + 1$. Let $P(m,n)$ be the main equation plugging $x = m$ and $y = n$. $P(x,f(x)) : f(x - f(f(x))) = - 1$, so let k be an integer such that $f(k) = -1$, so $P(x,k) : f(x +1) = f(f(x))$. $P(f(y) - 1, y) : f(-1) + 1 = f(f(f(y) - 1)) - f(y) = f(y + 1) - f(y)$, so the function is linear. Now, Just check the cases when $f(x) \equiv c$, giving $f(x) \equiv -1$, and check when $f(x) = x + c$, that gives $f(x) = x + 1$, and we're done.
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ezpotd
1287 posts
#171
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We claim the only such functions that work are $f(x) = x + 1$ and $f(x) = -1 $. It is obvious to see that both of these work.

Set $y = f(x)$, then $f$ is $-1$ at some input, then set $y$ such that $f(y) = -1$, so $f(x + 1) = f(f(x))$. If $f$ is injective, we are done. Otherwise, take two inputs $a,b$ such that $f(a) = f(b), a \neq b$. Then we also have $f(a +1) = f(f(a)) = f(f(b)) =  f(b + 1)$, applying this inductively gives $f$ periodic with period $ b -a$. Assume $f$ is periodic, thus it has finite range. Now consider the range $R$ of $f$, take the largest element $l$ in $R$ and the smallest element $s$ in $R$, then since $f(f(x)) = f(x + 1)$, $f(f(x))$ has the same range $R$, so set $x$ with $f(f(x)) = l, y$ with $f(y) = s$, then we have $l - s - 1 \in R$. We must then have $s \ge -1$ otherwise a number at least $l + 1$ would be in $R$, contradiction, but since $-1 \in R$ we have $s = -1$. Likewise $l = -1$, so $f$ must be constantly $-1$.
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youochange
180 posts
#172
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ABCDE wrote:
Determine all functions $f:\mathbb{Z}\rightarrow\mathbb{Z}$ with the property that \[f(x-f(y))=f(f(x))-f(y)-1\]holds for all $x,y\in\mathbb{Z}$.
can someone pls veify my sol :maybe:
f(x)=x+1
This post has been edited 2 times. Last edited by youochange, May 13, 2025, 12:31 PM
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youochange
180 posts
#173
Y by
Bummmmmp
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pco
23515 posts
#174 • 1 Y
Y by youochange
youochange wrote:
ABCDE wrote:
Determine all functions $f:\mathbb{Z}\rightarrow\mathbb{Z}$ with the property that \[f(x-f(y))=f(f(x))-f(y)-1\]holds for all $x,y\in\mathbb{Z}$.
can someone pls veify my sol :maybe:
Let $P(x,y)$ be the assertion.

Let $f(a)=0$ for a special $a \in \mathbb Z$
...
And what if such $a$ does not exist ?
(which is the case in the forgotten - in your post - solution $f\equiv -1$)
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heheman
1004 posts
#175
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BRO MY COMPUTER JUST RESTARTED AND DELETED ALL MY PROGRESS
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heheman
1004 posts
#176
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Answers are $f(x) = -1$ for all $x$ or $f(x) = x+1$ for all $x$.

Put $y=f(x)$ to see some $f(u) = -1$.
Put $(x, u)$ to see that $f(f(x)) = f(x+1)$.
If $f$ is injective, then $f(x) = x+1$. Otherwise, some $f(a)=f(b)$ with $a \ne b$.
Then $f(f(a)) = f(f(b))$ implies that $f(a+1) = f(b+1)$ etc. so $f$ is periodic.
Suppose $g$ and $l$ are the greatest and least values in the period. Let $x_g$ and $x_l$ be very large values of $x$ that achieves $g, l$ when plugging them into $f$ (we want $x_g, x_l$ to be large so $x-f(y)$ stays in the periodic part). Note that $(x_l-1, x_g)$ and $(x_g-1, x_l$ gives $l-g-1$ and $g-l-1$ are both in the period. If $l < -1$ the second expression is larger than $g$, contradiction. If $g > -1$, it would make the first expression less than $l$, contradiction. Then we have $-1 \le l \le g \le -1$, so $l = g = -1$. So $f(x) = -1$.
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heheman
1004 posts
#177
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comment
This post has been edited 1 time. Last edited by heheman, May 21, 2025, 9:02 PM
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MathIQ.
44 posts
#178
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Sol:Click to reveal hidden text
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