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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
Basic ideas in junior diophantine equations
Maths_VC   5
N 9 minutes ago by Royal_mhyasd
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
5 replies
Maths_VC
May 27, 2025
Royal_mhyasd
9 minutes ago
Find values for a, b ,c
Ferum_2710   2
N 11 minutes ago by Jupiterballs
Source: Romania JBMO tst 2023 day2 p4
Let $M \geq 1$ be a real number. Determine all natural numbers $n$ for which there exist distinct natural numbers $a$, $b$, $c > M$, such that
$n = (a,b) \cdot (b,c) + (b,c) \cdot (c,a) + (c,a) \cdot (a,b)$
(where $(x,y)$ denotes the greatest common divisor of natural numbers $x$ and $y$).
2 replies
Ferum_2710
Apr 30, 2023
Jupiterballs
11 minutes ago
Center lies on circumcircle of other
Philomath_314   41
N 23 minutes ago by Adywastaken
Source: INMO P1
In triangle $ABC$ with $CA=CB$, point $E$ lies on the circumcircle of $ABC$ such that $\angle ECB=90^{\circ}$. The line through $E$ parallel to $CB$ intersects $CA$ in $F$ and $AB$ in $G$. Prove that the center of the circumcircle of triangle $EGB$ lies on the circumcircle of triangle $ECF$.

Proposed by Prithwijit De
41 replies
Philomath_314
Jan 21, 2024
Adywastaken
23 minutes ago
find question
mathematical-forest   7
N 36 minutes ago by whwlqkd
Are there any contest questions that seem simple but are actually difficult? :-D
7 replies
mathematical-forest
Thursday at 10:19 AM
whwlqkd
36 minutes ago
Inspired by a cool result
DoThinh2001   1
N 36 minutes ago by arqady
Source: Old?
Let three real numbers $a,b,c\geq 0$, no two of which are $0$. Prove that:
$$\sqrt{\frac{a^2+bc}{b^2+c^2}}+\sqrt{\frac{b^2+ca}{c^2+a^2}}+\sqrt{\frac{c^2+ab}{a^2+b^2}}\geq 2+\sqrt{\frac{ab+bc+ca}{a^2+b^2+c^2}}.$$
Inspiration
1 reply
DoThinh2001
Today at 12:08 AM
arqady
36 minutes ago
Crossing ٍٍChords
matinyousefi   1
N an hour ago by Trenod
Source: Iranian Combinatorics Olympiad 2020 P3
$1399$ points and some chords between them is given.
$a)$ In every step we can take two chords $RS,PQ$ with a common point other than $P,Q,R,S$ and erase exactly one of $RS,PQ$ and draw $PS,PR,QS,QR$ let $s$ be the minimum of chords after some steps. Find the maximum of $s$ over all initial positions.
$b)$ In every step we can take two chords $RS,PQ$ with a common point other than $P,Q,R,S$ and erase both of $RS,PQ$ and draw $PS,PR,QS,QR$ let $s$ be the minimum of chords after some steps. Find the maximum of $s$ over all initial positions.

Proposed by Afrouz Jabalameli, Abolfazl Asadi
1 reply
matinyousefi
Apr 24, 2020
Trenod
an hour ago
Nice NT with powers of two
oVlad   7
N an hour ago by SimplisticFormulas
Source: Romania TST 2024 Day 1 P3
Let $n{}$ be a positive integer and let $a{}$ and $b{}$ be positive integers congruent to 1 modulo 4. Prove that there exists a positive integer $k{}$ such that at least one of the numbers $a^k-b$ and $b^k-a$ is divisible by $2^n.$

Cătălin Liviu Gherghe
7 replies
oVlad
Jul 31, 2024
SimplisticFormulas
an hour ago
Inequality in triangle
Nguyenhuyen_AG   0
2 hours ago
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]
0 replies
Nguyenhuyen_AG
2 hours ago
0 replies
D,E,F are collinear.
TUAN2k8   2
N 2 hours ago by TUAN2k8
Source: Own
Help me with this:
2 replies
TUAN2k8
May 28, 2025
TUAN2k8
2 hours ago
Combinatorial identity
MehdiGolafshan   4
N 2 hours ago by watery
Let $n$ is a positive integer. Prove that
$$\sum_{k=0}^{n-1}\frac{1}{k+1}\binom{n-1}{k} = \frac{2^n-1}{n}.$$
4 replies
MehdiGolafshan
Jan 16, 2023
watery
2 hours ago
JBMO Shortlist 2023 G7
Orestis_Lignos   7
N 3 hours ago by tilya_TASh
Source: JBMO Shortlist 2023, G7
Let $D$ and $E$ be arbitrary points on the sides $BC$ and $AC$ of triangle $ABC$, respectively. The circumcircle of $\triangle ADC$ meets for the second time the circumcircle of $\triangle BCE$ at point $F$. Line $FE$ meets line $AD$ at point $G$, while line $FD$ meets line $BE$ at point $H$. Prove that lines $CF, AH$ and $BG$ pass through the same point.
7 replies
Orestis_Lignos
Jun 28, 2024
tilya_TASh
3 hours ago
Reflected point lies on radical axis
Mahdi_Mashayekhi   5
N 3 hours ago by Mahdi_Mashayekhi
Source: Iran 2025 second round P4
Given is an acute and scalene triangle $ABC$ with circumcenter $O$. $BO$ and $CO$ intersect the altitude from $A$ to $BC$ at points $P$ and $Q$ respectively. $X$ is the circumcenter of triangle $OPQ$ and $O'$ is the reflection of $O$ over $BC$. $Y$ is the second intersection of circumcircles of triangles $BXP$ and $CXQ$. Show that $X,Y,O'$ are collinear.
5 replies
Mahdi_Mashayekhi
Apr 19, 2025
Mahdi_Mashayekhi
3 hours ago
Find the value
sqing   18
N 3 hours ago by Yiyj
Source: 2024 China Fujian High School Mathematics Competition
Let $f(x)=a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$ $a_i\in\{-1,1\} ,i=0,1,2,\cdots,6 $ and $f(2)=-53 .$ Find the value of $f(1).$
18 replies
sqing
Jun 22, 2024
Yiyj
3 hours ago
Number Theory
fasttrust_12-mn   14
N 3 hours ago by Namisgood
Source: Pan African Mathematics Olympiad P1
Find all positive intgers $a,b$ and $c$ such that $\frac{a+b}{a+c}=\frac{b+c}{b+a}$ and $ab+bc+ca$ is a prime number
14 replies
fasttrust_12-mn
Aug 15, 2024
Namisgood
3 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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ABCDE
1963 posts
#1 • 30 Y
Y by mathmaths, doxuanlong15052000, Davi-8191, tenplusten, mathleticguyyy, Amir Hossein, A-Thought-Of-God, MathLuis, mathematicsy, chessgocube, HWenslawski, centslordm, donotoven, megarnie, jhu08, RedFlame2112, ImSh95, vic_52math, OronSH, Adventure10, Mango247, kimyager, lian_the_noob12, ItsBesi, WiseTigerJ1, Sedro, aidan0626, cubres, ray66, Funcshun840
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 $x,y\in\mathbb{Z}$.
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Kezer
986 posts
#2 • 10 Y
Y by Vietjung, Plops, chessgocube, HWenslawski, centslordm, donotoven, megarnie, lego_man, Adventure10, WiseTigerJ1
That was also Problem 5 in this year's German TSTST (VAIMO).
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Aiscrim
409 posts
#6 • 9 Y
Y by baladin, A-Thought-Of-God, chessgocube, centslordm, donotoven, megarnie, Adventure10, gatnghiep, WiseTigerJ1
The only functions that work are $f\equiv -1$ and $f(x)=x+1,\ \forall x\in \mathbb{Z}$.

Let $P(x,y)$ be the assertion that $f(x-f(y))=f(f(x))-f(y)-1$. From $P(x,f(x))$, we get that there is an $a\in \mathbb{Z}$ s.t. $f(a)=-1$. From $P(x,a)$ we get $f(x+1)=f(f(x))$, so $f(f(f(x)))=f(f(x+1))=f(x+2)$.

$P(f(x),x)$ yields $f(0)=f(f(f(x)))-f(x)-1\Leftrightarrow f(x+2)=f(x)+f(0)+1$. It's easy to show by induction that $f(2k)=f(0)+k(f(0)+1)\ (*)$

Looking at $P(f(8),f(4))$ and using $(*)$, we get $f(2f(0)+2)=3f(0)+2$. As $2f(0)+2$ is even we can use again $(*)$ to get $f(0)\in \{-1,1\}$.

If $f(0)=-1$, by $(*)$ and induction we get $f(2k)=-1,\ \forall k\in \mathbb{Z}$. Also note $f(2k+1)=f(f(2k))=f(-1)$. From $P(1,0)$ we get $f(-1)=-1$, so $f(2k+1)=-1,\ \forall k\in \mathbb{Z}$, whence $f\equiv -1$.

If $f(0)=1$, by $(*)$ and induction we get $f(2k)=2k+1,\ \forall k\in \mathbb{Z}$. Also note $f(2k+1)=f(2k-1)+f(0)+1=f(2k-1)+2$, so by induction $f(2k+1)=f(1)+2k\ (**)$
If $f(1)$ is odd, by $(**)$, $f(f(1))=f(1)+2\cdot \dfrac{f(1)-1}{2}=2f(1)-1$; $f(f(1))=f(2)=3$, so $f(1)=2$, contradiction. If $f(1)$ is even, by $(*)$, $f(f(1))=\dfrac{f(1)}{2}(f(0)+1)+f(0)=f(1)+1$; but $f(f(1))=f(2)=3$, so $f(1)=2$, whence $f(2k+1)=2k+2,\ \forall k\in \mathbb{Z}$, so $f(x)=x+1,\ \forall x\in \mathbb{Z}$.
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ABCDE
1963 posts
#8 • 4 Y
Y by chessgocube, centslordm, Adventure10, WiseTigerJ1
Let $A$ be the range of $f$, and let $A+1=\{a+1\mid a\in A\}$. The condition implies that for all $a,b\in A$, $a-b-1,a+b+1\in A$, so for all $m,n\in A+1$, $m+n,m-n\in A+1$. Hence, $A+1$ is the set of all integer multiples of some nonnegative integer $k$.

If $k=0$, then $A=\{-1\}$, so $f(x)=-1$. We can check that this works.

If $k>0$, let $g(x)=\frac{f(x)+1}{k}$. Since $A$ is the set of integers that are $-1\pmod{k}$, $g$ is a surjective map from integers to integers. Substituting, we have that $g(x-kg(y)+1)=g(kg(x)-1)-g(y)$. Since $g$ is surjective, we in fact have $g(x-kz+1)=g(kg(x)-1)-z$, where $z=f(y)$ can be any integer. Setting $z=0$, we have that $g(x+1)=g(kg(x)-1)$.

Now, if $g(a)=g(b)$ for some $a\neq b$, then $g(a+1)=g(b+1)=g(kg(a)-1)$, so if $g$ is not injective then it is periodic for large inputs. But fixing $x$, for some constants $c$ and $d$ we have that $g(c-kz)=d-z$, so for sufficiently negative values of $z$, and thus sufficiently large values of $c-kz$, $g$ is unbounded, a contradiction. Hence, $g$ is injective, so we can conclude that $g(x+1)=g(kg(x)-1)\implies g(x)=\frac{x+2}{k}$. Since $g$ maps integers to integers, $k=1$, and we obtain a solution of $g(x)=x+2$ or $f(x)=x+1$, which works.

Thus, our solutions are $f(x)=-1$ and $f(x)=x+1$.
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Darn
996 posts
#9 • 43 Y
Y by SCP, doxuanlong15052000, PRO2000, Tommy2000, Kayak, Dukejukem, Fermat_Theorem, Myriam2003, ayan_mathematics_king, Bassiskicking, Lifefunction, Siddharth03, Imayormaynotknowcalculus, mathleticguyyy, Luchitha2, A-Thought-Of-God, mijail, microsoft_office_word, lneis1, Smkh, chessgocube, centslordm, Timmy456, Elnuramrv, myh2910, Bapun, Mathlover_1, megarnie, Chokechoke, hqxaev, Ibrahim_K, Adventure10, sabkx, D_S, jimmy_li, bin_sherlo, DrYouKnowWho, wizixez, EpicBird08, aidan0626, DroneChaudhary, Ismayil_Orucov, ZZzzyy
Here's an alternate solution.
The only solutions are $f(x)=\boxed{x+1}$ and $f(x)=\boxed{-1}$.

Let $P(x,y)$ be the assertion that \[f(x-f(y))=f(f(x))-f(y)-1.\]$P(x,f(x))$ implies that \[ f(x-f(f(x)))=-1, \]so $-1$ must be in the range of $f$.

Let $f(a)=-1$ for some integral $a$. Then from $P(x,a)$ we get \[ f(x+1)=f(f(x)). \quad (\star).  \]Now from $P(f(x)-1,x)$ we find that \begin{align*} f(-1)+1 &= f(f(f(x)-1)))-f(x) \\ &= f(f(x))-f(x) \\ &= f(x+1)-f(x), \end{align*}implying that $f(x+1)-f(x)$ is constant for all $x\in\mathbb{Z}$. Since the domain of $f$ is $\mathbb{Z}$, this means that $f(x)$ must be of the form $kx+c$ for some constants $k,c$.
It remains to solve for $f$. From $(\star)$, we obtain \[ k(x+1)+c = k(kx+c)+c, \]so \[ k(x+1)=k(kx+c). \]If $k\not=0$, then \[ kx+c=x+1, \]meaning that $f(x)=x+1$. Otherwise, $f$ is constant, so from the hypothesis we get \[ c=c-c-1=-1,\]so $f(x)=-1$.
Both of these functions satisfy the original condition, and so we have shown that $\boxed{-1\text{\;and\;}x+1}$ are indeed the only solutions to $f$.
$\square$
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navi_09220114
487 posts
#10 • 4 Y
Y by JoelBinu, centslordm, Adventure10, Mango247
The only solutions are $f(n)=-1$ for all $n\in \mathbb{Z}$, and $f(n)=n+1$ for all $n\in \mathbb{Z}$.

To verify that these are solutions, the first solution gives $-1=-1-(-1)-1$, which is true. The second solution gives $x-(y+1)+1=(x+1)+1-(y+1)-1$, which is also true. Now we will prove that these are the only soluitons.

To start, take $y=f(x)$ in $(1)$, then we get $$f(x-f(f(x)))=f(f(x))-f(f(x))-1=-1 \enspace{........ (2)}$$So by (2), choose $y=x-f(f(x))$ gives $f(y)=-1$, so we have $$f(x+1)=f(x-f(y))=f(f(x))-f(y)-1=f(f(x))$$Applying this successively we obtain $$f(x+k)=f(f(x+k-1))=f(f(f(x+k-2)))=\cdots =f^{(k+1)}(x)\enspace\text{for all $x\in \mathbb{Z}, k\in\mathbb{N}$}\enspace{........ (3)}$$With this $(1)$ rewrites to $$f(x-f(y))=f(x+1)-f(y)-1$$
Now we claim that $f(-2)=-1$. Take $x=f(y)$, then using $(3)$ we get $$f(0)=f(f(y)-f(y))=f(f(y)+1)-f(y)-1=f(f(f(y)))-f(y)-1=f(y+2)-f(y)-1$$so take $y=-2$, $f(0)=f(0)-f(2)-1\Rightarrow f(-2)=-1$.

So if we set $f(0)=a$, then $$f(y+2)-f(y)=a+1\enspace{........ (4)}$$Since $f(-2)=-1$, by induction, $f(2m)=a+m(a+1)$ for all $m\in \mathbb{Z}$. Likewise, if we set $f(1)=b$, then $f(2m+1)=b+m(a+1)$ for all $m\in \mathbb{Z}$. In particular, $f(2m+1)-f(2m)=b-a$ for all $m$. Now we give a method to determine $b$ in terms of $f(4)$.

Set $x=b+1, y=1$, then $b=f(1)=f((b+1)-f(1))=f(b+2)-f(1)-1$, but $f(b+2)=f(f(f(b)))=f(f(f(f(1))))=f(4)$, so $b=f(4)-f(1)-1$. So we have $$b=\frac{f(4)-1}{2}\enspace{........ (5)}$$We split into $2$ cases.

Case 1: $a\neq-1$.

In view of $(3)$, if $a\neq -1$, and if $f(p)=f(p+k)$ for some $p\in \mathbb{Z}, k\in \mathbb{N}$, then for all $\ell\in \mathbb{N}$, we obtain $$f(p+\ell)=f^{(\ell+1)}(p)=f^{(\ell+1)}(p+k)=f(p+k+\ell)$$In particular, $f(p)=f(p+k)=f(p+2k)\Rightarrow k(a+1)=f(p+2k)-f(p)=0$, clearly a contradiction. So $f$ is injective.

Take $x=y=2m$, we get $f(2m-a-m(a+1))=f(2m-f(2m))=f(2m+1)-f(2m)-1=b-a-1$, but if $a\neq 1$, then $2m-a-m(a+1)$ takes more than $1$ value, which is impossible since $f$ is injective. So $a=1$, and we get $f(2m)=2m+1$ for all $m\in \mathbb{Z}$. Now, in view of $(5)$, since $f(4)=5$, $b=\frac{f(4)-1}{2}=2\Rightarrow f(2m+1)=2+m(a+1)=2m+2$. So $f(n)=n+1$ for all $n\in \mathbb{Z}$, which had been verified to be a solution.

Case 2: $a=-1$

Then $f(2m)=f(0)=-1$, and $f(2m+1)=f(1)=b$ for all $m\in \mathbb{Z}$. Then again in view of $(5)$, we get $b=\frac{f(4)-1}{2}=-1$, so $f(2m+1)=b=-1$, so $f(n)=-1$ for all $n\in \mathbb{Z}$, which is also verified to be a solution.

So the only solutions are $f(n)=-1$ for all $n\in \mathbb{Z}$, and $f(n)=n+1$ for all $n\in \mathbb{Z}$. Q.E.D
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adamov1
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#11 • 6 Y
Y by Mobashereh, wheisenberg, centslordm, Adventure10, Mango247, H_Taken
Let $P(x,y)$ be the assertion. $P(x,f(x))$ gives that there exists $a$ such that $f(a)=-1$. $P(x,a)$ gives
\[f(x+1)=f(f(x))\]$P(f(x)-1,x)$ gives
\[f(-1)=f(f(f(x)-1))-f(x)-1=f(f(x))-f(x)-1=f(x+1)-f(x)-1\longrightarrow f(x+1)-f(x)=f(-1)+1\]Thus $f$ is linear, so either $f$ is constant or injective. If $f$ is constant, it must clearly be identically $-1$, which works, and if it is injective we have that $x+1=f(x)$ which also works.
This post has been edited 2 times. Last edited by adamov1, Jul 8, 2016, 4:22 AM
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gavrilos
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#12 • 2 Y
Y by Adventure10, Mango247
Hello.

This was also problem 3 in 2016 Greece Team Selection Test.
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rkm0959
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#13 • 5 Y
Y by B.J.W.T, E.A.K, Pluto1708, Adventure10, Mango247
We show $f$ is linear.
$P(x,f(x))$ shows that there exists an $u$ such that $f(u)=-1$.
$P(x,u)$ shows that $f(x+1)=f(f(x))$. This gives us $f(x-f(y))=f(x+1)-f(y)-1$.
Now $x=f(y)-1$ gives $f(-1)=f(f(y))-f(y)-1=f(y+1)-f(y)-1$, so $f$ is linear as required.
Now set $f(x)=ax+b$ and solve. $a(x+1)+b=a(ax+b)+b$, so $a=a^2$ and $a+b=ab+b$.
This gives $a=1, b=1$ or $a=0$. If $a=0$, we easily see that $f(x) \equiv -1$.
Therefore, the solution set is $f(x)=x+1$ and $f(x)=-1$. $\blacksquare$
This post has been edited 1 time. Last edited by rkm0959, Jul 9, 2016, 4:42 AM
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DrMath
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#14 • 1 Y
Y by Adventure10
Let $P(x,y)$ denote the given assertion.

By $P(x, f(x))$ we have $f(x-f(f(x)))=-1$. Thus there is a value $y$ such that $f(y)=-1$. Taking $P(x,y)$ with $f(y)=-1$ gives $f(x+1)=f(f(x))$.

Suppose $f$ is one to one. Then we instantly get that $f(x)=x+1$.

Else, suppose $f(a)=f(b)$ for $a, b$ distinct. We claim this gives $f(x)=-1$ for all $x$. Note $P(a,y)$ and $P(b,y)$ gives $f$ is periodic. Let the period be $p$, and suppose for the sake of contradiction we can take $y$ to be a value such that $f(y)\neq -1$. Then $P(x-1, y)$ gives $f(x-1-f(y))=f(x)-f(y)-1$. Thus, if $r$ is in the range of $f$, so is $r-f(y)-1$. Thus, for some integer $k$, $r$ and $r-kp$ are both in the range of $f$. But take $f(y)=r$ and $f(y')=r -kp$. $P(x,y)$ and $P(x, y')$ gives our contradiction, as $f$ has period $p$. Thus if $f$ is periodic then $f(x)=-1$ for all $x$.
This post has been edited 1 time. Last edited by DrMath, Jul 9, 2016, 6:38 AM
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mathmoGJ
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#15 • 2 Y
Y by Adventure10, Mango247
Was also Problem 2 on the UK Team selection test 2.
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hnkevin42
226 posts
#16 • 1 Y
Y by Adventure10
This will be long.

We have the following basic pieces of information.

Setting $(x, y) \rightarrow (f(0), 0)$ yields $f(f(f(0))) = 2f(0) + 1. \quad (\textbf{I})$
Setting $(x, y) \rightarrow (0, f(0))$ yields $f(-f(f(0))) = -1. \quad (\textbf{II})$
Via $\textbf{(II)}$, setting $(x, y) \rightarrow (x, -f(f(0)))$ yields $f(x + 1) = f(f(x)). \quad (\textbf{III})$

We then attack with the following lemmas.

Lemma 1: For every integer $n$, we have $f(n(f(0) + 1)) = (n + 1)f(0) + n$.
Proof: We induct both ways, starting with the obviously true base case of $n = 0$ and going in the negative and positive direction.
We go in the positive direction. If $n$ is not negative, setting $(x, y) \rightarrow ((n + 1)f(0) + n, n(f(0) + 1))$ yields, using $\textbf{(III)}$,
$$f(0) = f(f((n + 1)f(0) + n)) - (n + 1)f(0) - n - 1$$$$\implies f((n + 1)f(0) + (n + 1)) = f((n + 1)(f(0) + 1)) = (n + 2)f(0) + (n + 1)$$which completes the inductive step. In the negative direction, setting $(x, y) \rightarrow (n(f(0) + 1) - 1, 0)$ yields, again using $\textbf{(III)}$,
$$f((n - 1)f(0) + (n - 1)) = f(f(n(f(0) + 1) - 1)) - f(0) - 1$$$$\implies f((n - 1)(f(0) + 1)) = (n + 1)f(0) + n - f(0) - 1 = nf(0) + (n - 1)$$which completes the inductive step, proving Lemma 1.

Lemma 2: We either have $f(0) = 1$ or $f(0) = -1$.
Proof: Note that by $(\textbf{III})$, we have $f(f(0)) = f(1)$. Since $f(f(x)) = f(x + 1)$ for all integers $x$, we have that $f$ is periodic, for all $x \ge \min(f(0), 1)$ if $f(0) \ne 1$. For these $x$, $f$ is bounded. But from Lemma 1, $f$ must be unbounded for positive $x$ and for $f(0) \ne -1$. This is because, if $f(0)$ is nonnegative, both $n(f(0) + 1)$ and $f(n(f(0) + 1))$ are positive and strictly increasing as $n$ grows in the positive integers. Likewise, if $f(0) < -1$, both $n(f(0) + 1)$ and $f(n(f(0) + 1))$ are positive and strictly increasing as $n$ gets more negative, so either way, $f$ is unbounded in the positive integers if $f(0) \ne -1$. Since $f$ is periodic and unbounded in the positive integers for all $f(0)$ not equal to $1$ or $-1$, we cannot have $f(0)$ be any other value than $1$ or $-1$.

END LEMMA: The solutions, and the only solutions, are $f(x) = x + 1$ and $f(x) = -1$.
Proof: Setting $(x, y) \rightarrow (0, 0)$ yields $(\star) f(-f(0)) = f(f(0)) - f(0) - 1 = f(1) - f(0) - 1$ via $\textbf{(III)}$. Then setting $(x, y) \rightarrow (f(1) - 1, -f(0))$ yields via $\textbf{(III)}$, $$f(f(0)) = f(f(f(1) - 1)) - f(1) + f(0) \implies f(f(1)) = 2f(1) - f(0).$$We know $f(f(1)) = f(f(f(0))) = 2f(0) + 1$ from $\textbf{(I)}$ and $\textbf{(III)}$, so then $2f(1) = 3f(0) + 1$.

If $f(0) = 1$, then $f(1) = 2$ and, with $(\star)$, $f(-1) = 0$. Then, plugging in $(x, y) \rightarrow (x, -1)$ yields that for all integers $x$, we need $f(x) = f(f(x)) - 1 \implies f(x + 1) = f(x) + 1$. From here we get that $f(0) = 1$ leads to the one solution $f(x) = x + 1$.

If $f(0) = -1$, then $f(1) = -1$ and, with $(\star)$, $f(2) = -1$. Then, plugging in $(x, y) \rightarrow (x, -1)$ yields that for all integers $x$, we need $f(x + 1) = f(f(x))$. A quick induction starting with $x = 1$ yields that for all positive integers $x$, $f(x) = -1$. Now suppose $f(k) = a \ne -1$ for any $k < -1$. Setting $(x, y) \rightarrow (x, -1)$ yields that for all integers $x$, we need $f(x - a) = f(x + 1) - a - 1$. If we choose a positive integer $x = k > |a|$, then we have $k - a, k + 1$ are positive integers and $f(k - a) = f(k + 1) - a + 1 \implies -1 = -1 - a + 1$, which is a contradiction since we said $a \ne 1$. Thus, we have the one solution $f(x) = -1$ for all $x$.

Since both solutions easily work when plugged back into the original equation, The answer is $$\boxed{f(x) = x + 1, f(x) = -1.}$$
This post has been edited 2 times. Last edited by hnkevin42, Jul 12, 2016, 12:49 PM
Reason: Copy error
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tenplusten
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#17 • 2 Y
Y by Adventure10, Mango247
Good problem.

Claim 1: There exist $a\in R$ such that $f(a)=-1$.
Proof:Just see $P(x,f(x))$

Claim 2: $f(f(x))=f(x+1)$
Proof: $P(x,a)$ $\implies$ $f(x+1)=f(f(x))$

Claim 3: $f(f(f(x)))-f(x)=f(0)+1$
Proof: Just see $P(f(x),x)$

Claim 4: $f(x+2)=f(x)+f(0)+1$
Proof: Using "Claim 2" we get $f(f(f(x)))=f(f(x+1))=f(x+2)$
So $f(f(f(x)))-f(x)=f(0)+1=f(x+2)-f(x)$.

Since we got $f(x+2)=f(x)+f(0)+1$
Easy to show that $f(x)=x+1$ and $f\equiv-1$
So solutions are $f(x)=x+1$ for all $x\in Z$ and $f\equiv -1$
This post has been edited 6 times. Last edited by tenplusten, Feb 6, 2017, 3:42 PM
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MSTang
6012 posts
#18 • 3 Y
Y by adityaguharoy, Adventure10, Mango247
Claim 1: $-1$ is in the range of $f$.

Proof: Taking $y=f(x)$ gives $f(x-f(f(x)) = -1$ for all $x$. $\spadesuit$

Claim 2: Over all $x$, the quantity $f(f(x)) - f(x)$ is constant.

Proof. Let $a, b$ be integers. Taking $x=f(a)+f(b)$ and $y=a$ gives \[f(f(b)) = f(f(f(a)+f(b))) - f(a) - 1.\]Taking $x=f(a)+f(b)$ and $y=b$ gives \[f(f(a)) = f(f(f(a)+f(b))) - f(b) - 1.\]Comparing the two equations gives $f(f(a)) - f(a) = f(f(b)) - f(b)$, as claimed. $\spadesuit$

Finisher. Let $f(f(x)) - f(x) = k$ for some constant $k$. Then the given equation becomes \[f(x-f(y)) = f(x) - f(y) + (k-1)\]for all $x, y$. Using Claim 1, choose $y$ in the above equation such that $f(y) = -1$, giving \[f(x+1) = f(x) + k\]for all $x$. Since $f$ has domain $\mathbb{Z}$, we conclude that $f$ is of the form $f(x) = kx + c$ for some constants $k$ and $c$. In the original equation, we need \[k(x-ky-c)+c = k(kx+c) + c - ky - c - 1\]or \[kx - k^2y - ck + c = k^2x + ck - ky - 1.\]Comparing $x$ coefficients gives $k = k^2$, so $k \in \{0, 1\}$. If $k=0$, then $c = -1$, giving the solution $f(x) = -1$ for all $x$. If $k=1$, then $c = 1$, giving the solution $f(x) = x + 1$ for all $x$. It is easy to check that both these functions work. $\square$
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adityaguharoy
4657 posts
#19 • 2 Y
Y by Adventure10, Mango247
MSTang wrote:
Claim 1: $-1$ is in the range of $f$.

Proof: Taking $y=f(x)$ gives $f(x-f(f(x)) = -1$ for all $x$. $\spadesuit$

Claim 2: Over all $x$, the quantity $f(f(x)) - f(x)$ is constant.

Proof. Let $a, b$ be integers. Taking $x=f(a)+f(b)$ and $y=a$ gives \[f(f(b)) = f(f(f(a)+f(b))) - f(a) - 1.\]Taking $x=f(a)+f(b)$ and $y=b$ gives \[f(f(a)) = f(f(f(a)+f(b))) - f(b) - 1.\]Comparing the two equations gives $f(f(a)) - f(a) = f(f(b)) - f(b)$, as claimed. $\spadesuit$

Finisher. Let $f(f(x)) - f(x) = k$ for some constant $k$. Then the given equation becomes \[f(x-f(y)) = f(x) - f(y) + (k-1)\]for all $x, y$. Using Claim 1, choose $y$ in the above equation such that $f(y) = -1$, giving \[f(x+1) = f(x) + k\]for all $x$. Since $f$ has domain $\mathbb{Z}$, we conclude that $f$ is of the form $f(x) = kx + c$ for some constants $k$ and $c$. In the original equation, we need \[k(x-ky-c)+c = k(kx+c) + c - ky - c - 1\]or \[kx - k^2y - ck + c = k^2x + ck - ky - 1.\]Comparing $x$ coefficients gives $k = k^2$, so $k \in \{0, 1\}$. If $k=0$, then $c = -1$, giving the solution $f(x) = -1$ for all $x$. If $k=1$, then $c = 1$, giving the solution $f(x) = x + 1$ for all $x$. It is easy to check that both these functions work. $\square$

It is somewhat similar to the solution I had for this problem.
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