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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
Three circles are concurrent
Twoisaprime   23
N 3 minutes ago by Curious_Droid
Source: RMM 2025 P5
Let triangle $ABC$ be an acute triangle with $AB<AC$ and let $H$ and $O$ be its orthocenter and circumcenter, respectively. Let $\Gamma$ be the circle $BOC$. The line $AO$ and the circle of radius $AO$ centered at $A$ cross $\Gamma$ at $A’$ and $F$, respectively. Prove that $\Gamma$ , the circle on diameter $AA’$ and circle $AFH$ are concurrent.
Proposed by Romania, Radu-Andrew Lecoiu
23 replies
Twoisaprime
Feb 13, 2025
Curious_Droid
3 minutes ago
|a_i/a_j - a_k/a_l| <= C
mathwizard888   32
N 13 minutes ago by ezpotd
Source: 2016 IMO Shortlist A2
Find the smallest constant $C > 0$ for which the following statement holds: among any five positive real numbers $a_1,a_2,a_3,a_4,a_5$ (not necessarily distinct), one can always choose distinct subscripts $i,j,k,l$ such that
\[ \left| \frac{a_i}{a_j} - \frac {a_k}{a_l} \right| \le C. \]
32 replies
mathwizard888
Jul 19, 2017
ezpotd
13 minutes ago
Two lines meeting on circumcircle
Zhero   54
N 35 minutes ago by Ilikeminecraft
Source: ELMO Shortlist 2010, G4; also ELMO #6
Let $ABC$ be a triangle with circumcircle $\omega$, incenter $I$, and $A$-excenter $I_A$. Let the incircle and the $A$-excircle hit $BC$ at $D$ and $E$, respectively, and let $M$ be the midpoint of arc $BC$ without $A$. Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$. If $TI$ intersects $\omega$ again at $S$, prove that $SI_A$ and $ME$ meet on $\omega$.

Amol Aggarwal.
54 replies
Zhero
Jul 5, 2012
Ilikeminecraft
35 minutes ago
Help me this problem. Thank you
illybest   3
N an hour ago by jasperE3
Find f: R->R such that
f( xy + f(z) ) = (( xf(y) + yf(x) )/2) + z
3 replies
illybest
Today at 11:05 AM
jasperE3
an hour ago
line JK of intersection points of 2 lines passes through the midpoint of BC
parmenides51   4
N an hour ago by reni_wee
Source: Rioplatense Olympiad 2018 level 3 p4
Let $ABC$ be an acute triangle with $AC> AB$. be $\Gamma$ the circumcircle circumscribed to the triangle $ABC$ and $D$ the midpoint of the smallest arc $BC$ of this circle. Let $E$ and $F$ points of the segments $AB$ and $AC$ respectively such that $AE = AF$. Let $P \neq A$ be the second intersection point of the circumcircle circumscribed to $AEF$ with $\Gamma$. Let $G$ and $H$ be the intersections of lines $PE$ and $PF$ with $\Gamma$ other than $P$, respectively. Let $J$ and $K$ be the intersection points of lines $DG$ and $DH$ with lines $AB$ and $AC$ respectively. Show that the $JK$ line passes through the midpoint of $BC$
4 replies
parmenides51
Dec 11, 2018
reni_wee
an hour ago
AGM Problem(Turkey JBMO TST 2025)
HeshTarg   3
N an hour ago by ehuseyinyigit
Source: Turkey JBMO TST Problem 6
Given that $x, y, z > 1$ are real numbers, find the smallest possible value of the expression:
$\frac{x^3 + 1}{(y-1)(z+1)} + \frac{y^3 + 1}{(z-1)(x+1)} + \frac{z^3 + 1}{(x-1)(y+1)}$
3 replies
HeshTarg
2 hours ago
ehuseyinyigit
an hour ago
Shortest number theory you might've seen in your life
AlperenINAN   8
N 2 hours ago by HeshTarg
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
8 replies
AlperenINAN
Yesterday at 7:51 PM
HeshTarg
2 hours ago
Squeezing Between Perfect Squares and Modular Arithmetic(JBMO TST Turkey 2025)
HeshTarg   3
N 2 hours ago by BrilliantScorpion85
Source: Turkey JBMO TST Problem 4
If $p$ and $q$ are primes and $pq(p+1)(q+1)+1$ is a perfect square, prove that $pq+1$ is a perfect square.
3 replies
HeshTarg
3 hours ago
BrilliantScorpion85
2 hours ago
A bit too easy for P2(Turkey 2025 JBMO TST)
HeshTarg   0
2 hours ago
Source: Turkey 2025 JBMO TST P2
Let $n$ be a positive integer. Aslı and Zehra are playing a game on an $n \times n$ chessboard. Initially, $10n^2$ stones are placed on the squares of the board. In each move, Aslı chooses a row or a column; Zehra chooses a row or a column. The number of stones in each square of the chosen row or column must change such that the difference between the number of stones in a square with the most stones and a square with the fewest stones in that same row or column is at most 1. For which values of $n$ can Aslı guarantee that after a finite number of moves, all squares on the board will have an equal number of stones, regardless of the initial distribution?
0 replies
HeshTarg
2 hours ago
0 replies
D1030 : An inequalitie
Dattier   0
2 hours ago
Source: les dattes à Dattier
Let $0<a<b<c<d$ reals, and $n \in \mathbb N^*$.

Is it true that $a^n(b-a)+b^n(c-b)+c^n(d-c) \leq \dfrac {d^{n+1}}{n+1}$ ?
0 replies
Dattier
2 hours ago
0 replies
Long and wacky inequality
Royal_mhyasd   0
2 hours ago
Source: Me
Let $x, y, z$ be positive real numbers such that $x^2 + y^2 + z^2 = 12$. Find the minimum value of the following sum :
$$\sum_{cyc}\frac{(x^3+2y)^3}{3x^2yz - 16z - 8yz + 6x^2z}$$knowing that the denominators are positive real numbers.
0 replies
Royal_mhyasd
2 hours ago
0 replies
Vietnamese national Olympiad 2007, problem 4
hien   16
N 3 hours ago by de-Kirschbaum
Given a regular 2007-gon. Find the minimal number $k$ such that: Among every $k$ vertexes of the polygon, there always exists 4 vertexes forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.
16 replies
hien
Feb 8, 2007
de-Kirschbaum
3 hours ago
2n^2+4n-1 and 3n+4 have common powers
bin_sherlo   4
N 3 hours ago by CM1910
Source: Türkiye 2025 JBMO TST P5
Find all positive integers $n$ such that a positive integer power of $2n^2+4n-1$ equals to a positive integer power of $3n+4$.
4 replies
bin_sherlo
Yesterday at 7:13 PM
CM1910
3 hours ago
An interesting geometry
k.vasilev   19
N 3 hours ago by Ilikeminecraft
Source: All-Russian Olympiad 2019 grade 10 problem 4
Let $ABC$ be an acute-angled triangle with $AC<BC.$ A circle passes through $A$ and $B$ and crosses the segments $AC$ and $BC$ again at $A_1$ and $B_1$ respectively. The circumcircles of $A_1B_1C$ and $ABC$ meet each other at points $P$ and $C.$ The segments $AB_1$ and $A_1B$ intersect at $S.$ Let $Q$ and $R$ be the reflections of $S$ in the lines $CA$ and $CB$ respectively. Prove that the points $P,$ $Q,$ $R,$ and $C$ are concyclic.
19 replies
k.vasilev
Apr 23, 2019
Ilikeminecraft
3 hours ago
integer functional equation
ABCDE   149
N Yesterday at 1:49 AM by ezpotd
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 $x,y\in\mathbb{Z}$.
149 replies
ABCDE
Jul 7, 2016
ezpotd
Yesterday at 1:49 AM
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
479 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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