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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
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
Yesterday at 11:16 PM
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
Inequality with 3 variables and a special condition
Nuran2010   7
N a minute ago by sqing
Source: Azerbaijan Al-Khwarizmi IJMO TST 2024
For positive real numbers $a,b,c$ we have $3abc \geq ab+bc+ca$.
Prove that:

$\frac{1}{a^3+b^3+c}+\frac{1}{b^3+c^3+a}+\frac{1}{c^3+a^3+b} \leq \frac{3}{a+b+c}$.

Determine the equality case.
7 replies
Nuran2010
Apr 29, 2025
sqing
a minute ago
D1024 : Can you do that?
Dattier   1
N 9 minutes ago by Dattier
Source: les dattes à Dattier
Let $x_{n+1}=x_n^2+1$ and $x_0=1$.

Can you calculate $\sum\limits_{i=1}^{2^{2025}} x_i \mod 10^{30}$?
1 reply
Dattier
Apr 29, 2025
Dattier
9 minutes ago
4-var inequality
RainbowNeos   3
N 11 minutes ago by RainbowNeos
Given $a,b,c,d>0$, show that
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]
3 replies
RainbowNeos
Yesterday at 9:31 AM
RainbowNeos
11 minutes ago
Hard inequality
ys33   0
13 minutes ago
Let $a, b, c, d>0$. Prove that
$\sqrt[3]{ab}+ \sqrt[3]{cd} < \sqrt[3]{(a+b+c)(b+c+d)}$.
0 replies
1 viewing
ys33
13 minutes ago
0 replies
4 lines concurrent
Zavyk09   7
N 18 minutes ago by bin_sherlo
Source: Homework
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $BH, CH$ intersect $(O)$ again at $K, L$ respectively. Lines through $H$ parallel to $AB, AC$ intersects $AC, AB$ at $E, F$ respectively. Point $D$ such that $HKDL$ is a parallelogram. Prove that lines $KE, LF$ and $AD$ are concurrent at a point on $OH$.
7 replies
Zavyk09
Apr 9, 2025
bin_sherlo
18 minutes ago
Generalized mirror problem
Taha1381   8
N 22 minutes ago by Lemmas
Source: Iranian second round/day1/problem1
We have a rectangle with it sides being a mirror.A light Ray enters from one of the corners of the rectangle and after being reflected several times enters to the opposite corner it started.Prove that at some time the light Ray passed the center of rectangle(Intersection of diagonals.)
8 replies
Taha1381
May 2, 2019
Lemmas
22 minutes ago
4 variables with quadrilateral sides 2
mihaig   5
N 28 minutes ago by mihaig
Source: Own
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove
$$\left(a+b+c+d-2\right)^2+8\geq3\left(abc+abd+acd+bcd\right).$$
5 replies
mihaig
Apr 29, 2025
mihaig
28 minutes ago
Consecutive sum of integers sum up to 2020
NicoN9   1
N 30 minutes ago by Mathzeus1024
Source: Japan Junior MO Preliminary 2020 P2
Let $a$ and $b$ be positive integers. Suppose that the sum of integers between $a$ and $b$, including $a$ and $b$, are equal to $2020$.
All among those pairs $(a, b)$, find the pair such that $a$ achieves the minimum.
1 reply
NicoN9
4 hours ago
Mathzeus1024
30 minutes ago
IMO 2023 P2
799786   91
N 40 minutes ago by ND_
Source: IMO 2023 P2
Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.
91 replies
799786
Jul 8, 2023
ND_
40 minutes ago
Geometry
VicKmath7   9
N 42 minutes ago by tilya_TASh
Source: 8th European Mathematical Cup 2019 Junior Q3
Let $ABC$ be a triangle with circumcircle $\omega$. Let $l_B$ and $l_C$ be two lines through the points $B$ and $C$, respectively, such that $l_B  \parallel  l_C$. The second intersections of $l_B$ and $l_C$ with $\omega$ are $D$ and $E$, respectively. Assume that $D$ and $E$ are on the same side of $BC$ as $A$. Let $DA$ intersect $l_C$ at $F$ and let $EA$ intersect $l_B$ at $G$. If $O$, $O_1$ and $O_2$ are circumcenters of the triangles $ABC$, $ADG$ and $AEF$, respectively, and $P$ is the circumcenter of the triangle $OO_1O_2$, prove that $l_B  \parallel  OP \parallel l_C$.

Proposed by Stefan Lozanovski, Macedonia
9 replies
VicKmath7
Dec 26, 2019
tilya_TASh
42 minutes ago
2017 CNMO Grade 11 P5
minecraftfaq   3
N an hour ago by CovertQED
Source: 2017 China Northern MO, Grade 11, Problem 5
Length of sides of regular hexagon $ABCDEF$ is $a$. Two moving points $M,N$ moves on sides $BC,DE$, satisfy that $\angle MAN=\frac{\pi}{3}$. Prove that $AM\cdot AN-BM\cdot DN$ is a definite value.
3 replies
minecraftfaq
Feb 24, 2020
CovertQED
an hour ago
Inspired by Nice inequality
sqing   0
an hour ago
Source: Own
Let $  a,b,c >0  $. Show that
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1\right)^2 \geq \frac{16}{k+3}\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}+k\right)$$Where $ k\geq 1.$
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1\right)^2 \geq  \frac{b}{a}+\frac{c}{b}+\frac{a}{c}+13$$$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1\right)^2 \geq \frac{16}{5}\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}+2\right)$$
0 replies
sqing
an hour ago
0 replies
Mmo 9-10 graders P5
Bet667   5
N an hour ago by User141208
Let $a,b,c,d$ be real numbers less than 2.Then prove that $\frac{a^3}{b^2+4}+\frac{b^3}{c^2+4}+\frac{c^3}{d^2+4}+\frac{d^3}{a^2+4}\le4$
5 replies
Bet667
Apr 3, 2025
User141208
an hour ago
3 var inequality
sqing   2
N an hour ago by sqing
Source: Own
Let $ a,b,c>0 ,\frac{a}{b} +\frac{b}{c} +\frac{c}{a} \leq 2\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right). $ Prove that
$$a+b+c+2\geq abc$$Let $ a,b,c>0 , a^3+b^3+c^3\leq 2(ab+bc+ca). $ Prove that
$$a+b+c+2\geq abc$$
2 replies
sqing
Apr 30, 2025
sqing
an hour ago
Equal pairs in continuous function
CeuAzul   16
N Apr 12, 2025 by Ilikeminecraft
Let $f(x)$ be an continuous function defined in $\text{[0,2015]},f(0)=f(2015)$
Prove that there exists at least $2015$ pairs of $(x,y)$ such that $f(x)=f(y),x-y \in \mathbb{N^+}$
16 replies
CeuAzul
Aug 6, 2018
Ilikeminecraft
Apr 12, 2025
Equal pairs in continuous function
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G H BBookmark kLocked kLocked NReply
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CeuAzul
994 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $f(x)$ be an continuous function defined in $\text{[0,2015]},f(0)=f(2015)$
Prove that there exists at least $2015$ pairs of $(x,y)$ such that $f(x)=f(y),x-y \in \mathbb{N^+}$
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CeuAzul
994 posts
#2 • 2 Y
Y by Adventure10, Mango247
Anyone???
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hutu683
112 posts
#4 • 2 Y
Y by Adventure10, Mango247
Bump for solution. :trampoline:
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me9hanics
375 posts
#5 • 2 Y
Y by Adventure10, Mango247
I never took calculus (yet) so I can't write it down algebraicly but the solution is noticing that there will be at least 1 $x$, for which $f(x)=f(x+1)$, atleast 1 $x$, for which $f(x)=f(x+2)$... and you can notice that by moving the function by $k$ ($k=1,2,...,2015$) units, as long as $k \leq 2015$ there shall be a common point
This post has been edited 1 time. Last edited by me9hanics, Aug 27, 2018, 2:31 PM
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hutu683
112 posts
#6 • 2 Y
Y by Adventure10, Mango247
misinnyo wrote:
I never took calculus (yet) so I can't write it down algebraicly but the solution is noticing that there will be at least 1 $x$, for which $f(x)=f(x+1)$, atleast 1 $x$, for which $f(x)=f(x+2)$... and you can notice that by moving the function by $k$ ($k=1,2,...,2015$) units, as long as $k \leq 2015$ there shall be a common point

I don’t think you are right. If $f(0)=f(2015)=0$, and $f(x)>0$ when $0< x\leqslant 1007$, $f(x)<0$ when $2015>x>1007$, then there will be no solution when $k=2014$.
This post has been edited 1 time. Last edited by hutu683, Aug 27, 2018, 2:46 PM
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hutu683
112 posts
#8 • 2 Y
Y by Adventure10, Mango247
Sorry for the previous wrong solution.(deleted)
Prove by induction on $n$ (2015).
Sketch:
First show that there is some $t\in [0, n-1]$ such that $f(t)=f(t+1)$.
Then denote $g(x)=f(x)$ when $0\leqslant x\leqslant t$ and $g(x)=f(x+1)$ when $t\leqslant x\leqslant n-1$, and refer to the case of $n-1$.
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CyclicISLscelesTrapezoid
372 posts
#10 • 2 Y
Y by CeuAzul, ihatemath123
Replace $2015$ with positive integer $n$. We do induction on $n$, with the base case of $n=1$ obvious. The following claim lets us do induction.

Claim: There exists a real number $a$ with $0 \le a \le n-1$ such that $f(a+1)=f(a)$.

Proof: Suppose that there doesn't exist such $a$, and assume WLOG that $f(1)>f(0)$. If $f(2)<f(1)$, then consider $f(x)$ and $g(x)=f(x+1)$. Since $f(0)<g(0)$ and $f(1)>g(1)$, there exists $a \in (0,1)$ such that $f(a)=g(a)$ by the intermediate value theorem. Thus, we have $f(2)>f(1)$. Similarly, we have $f(2)<f(3),\ldots,f(n-1)<f(n)$, so \[f(0)<f(1)<\cdots<f(n).\]However, $f(n)=f(0)$, a contradiction, so such $a$ must exist. $\square$

Now, we can induct. Choose $a$ such that $f(a)=f(a+1)$, and consider the function \[f'(x)=\begin{cases}f(x) & x \le a \\ f(x+1) & x>a\end{cases}.\]This function is continuous and $f'(n-1)=f'(0)$, so there exist at least $n-1$ pairs $(x,y)$ such that $f'(x)=f'(y)$ and $x-y$ is a positive integer. Suppose that $x'=x$ if $x \le a$ and $x'=x+1$ if $x>a$, and define $y'$ similarly. Notice that each pair $(x,y)$ converts to a pair $(x',y')$ such that $f(x')=f(y')$ and $x'-y'$ is a positive integer, so we have $n-1$ pairs. We also have the pair $(a,a+1)$, so there are at least $n$ pairs and we are done. $\square$
This post has been edited 2 times. Last edited by CyclicISLscelesTrapezoid, Sep 5, 2023, 9:26 PM
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IAmTheHazard
5001 posts
#11
Y by
A very similar variation of this problem was apparently a China IMO camp problem. It's possible to solve the Chinese problem by specifically utilizing the fact that $f$ in that problem is a polynomial and solutions need not lie in $[0,n]$, but there's a totally valid solution without.
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Ritwin
156 posts
#12 • 1 Y
Y by LLL2019
Hilarious problem, I love it :D
Solution
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joshualiu315
2533 posts
#13
Y by
We will prove this using strong induction. The base case $n=1$ is vacuous so we move towards the inductive step. Suppose our claim holds for all integers $n$ from $1$ to $k-1$.


Claim: There exists some value of $t \in [0, n-1]$ such that $p(t) = p(t+1)$.

Proof: Assume for the sake of contradiction that there is no such $t$. Consider the set $\{p(1)-p(0), p(2)-p(1), \dots, p(n)-p(n-1)\}$. Evidently, none of the elements in the set can be $0$, but the cumulative sum is $0$. This means that at some point, the elements will change sign. Denote $f(x) = p(x)-p(x-1)$; $f(x)$ is continuous and changes sign at some point, so it must equal $0$ at some point by the Intermediate Value Theorem, contradiction. $\square$


Consider the function $q(x)$ such that $q(x)=p(x)$ when $x = [0,t)$ and $q(x)=p(x+1)$ when $x = [t+1,n-1]$. From the inductive claim, we know that $q(x)$ has at least $n-1$ pairs. Simply translate the pairs back to $p(x)$ and include $(t,t+1)$ to get the desired result. $\square$
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Martin2001
147 posts
#14
Y by
Let $f(x)=p(x)-p(x-1).$ Note that
$$\sum_{i=1}^{i=16}f(i)=0.$$We are trying to prove that there exists an $x$ such that $f(x)=0.$ Note that if there isn't an integer $i$ that works, then
at least one of $f(i)$ will be positive and at least one will be negative. Thus, by Intermediate Value Theorem, there has to exist such an $i.$
Delete the interval $[i-1, i].$ Note that the ends of the graph are the same, so the graph is still continuous, and that all $x-y$ are either unchanged or the value is increased by $1.$ Then simply continue as before $n-1$ more times, until you can't anymore. We succesefully have found $n$ different pairs, so we're done.$\blacksquare$
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ihatemath123
3446 posts
#15 • 2 Y
Y by peace09, cosmicgenius
oops i overcomplicated this :stretcher: i still think this solution is pretty nice, hopefully its not a fakesolve ?

For integers $k$ with $0 \leq k \leq n-1$, define $g_k : [0, 1) \to \mathbb R$ as $g_k (x) = f(x+k)$. Now plot these functions all at once; each intersection point counts as one of the pairs $(x,y)$ with $x-y \in \mathbb N$ such that $f(x)=f(y)$. We will show there are at least $n-1$ intersection points (none of the intersection points will count the pair $(x,y) = (0,n)$, so this is the $n$th pair).
[asy]
unitsize(0.6cm);

path p = (0,0)..(1,2)..(2,3)..(3,2.5)..(4,1.5)..(5,1)..(6,2)..(7,1)..(8,0);

draw(shift((12,0))*p, mediumred+linewidth(1.5));
draw(shift((10,0))*p, deepgreen+linewidth(1.5));
draw(shift((8,0))*p, royalblue+linewidth(1.5));
draw(shift((6,0))*p, orange+linewidth(1.5));
fill((0,-0.5)--(12,-0.5)--(12,4.5)--(0,4.5)--cycle, white);
fill((14,-0.5)--(20.1, -0.5)--(20.1,4.5)--(14, 4.5)--cycle, white);
draw((12,-0.5)--(12,4.5), black+linewidth(1.5));
draw((14,-0.5)--(14,4.5), black+linewidth(1.5));

draw(p, black+linewidth(1.5));
label("becomes", (9.5,1), fontsize(9));
label("$1$", (1,-1), fontsize(9)+mediumred);
draw(brace((0.1,-0.3),(1.9,-0.3),-0.4),mediumred+linewidth(1.5));
label("$2$", (3,-1), fontsize(9)+deepgreen);
draw(brace((2.1,-0.3),(3.9,-0.3),-0.4),deepgreen+linewidth(1.5));
label("$3$", (5,-1), fontsize(9)+royalblue);
draw(brace((4.1,-0.3),(5.9,-0.3),-0.4),royalblue+linewidth(1.5));
label("$4$", (7,-1), fontsize(9)+orange);
draw(brace((6.1,-0.3),(7.9,-0.3),-0.4),orange+linewidth(1.5));
[/asy]
If any $i$ and $j$ exist for which $g_i (0) = g_j (0)$, we can nudge $f$ by a negligible amount at certain points so that this is no longer the case – evidently, this only decreases the number of intersection points. Note that, now, $g_0 (0), g_1 (0), \dots, g_{n-1} (0)$ is a permutation of $g_0 (1), g_1 (1), \dots, g_{n-1} (1)$. (Well, $g_i(1)$ is not defined, but we can take the limiting value.) Each inversion in this permutation corresponds to an intersection point, so it suffices to prove the following lemma:

Claim: A permutation of $1, 2, \dots, n $ with one component has at least $n-1$ inversions.
Proof: Call an element of the permutation a "gamechanger" if it is greater than all the elements to its left. Let the gamechangers be $y_1 < y_2 <  \cdots < y_k$, and let their indices be $x_1 < x_2 < \cdots < x_k$. Clearly, we have $x_{i+1} < y_{i}$ for all $i$ – in particular, $y_1 > 1$. As such, each gamechanger $y_i$ comes before $y_{i-1}+1, y_{i-2}+1, \dots, y_i-1$. If $i > 1$, this gamechanger also comes before some number from $1$ to $y_{i-1} - 1$ since there are only $x_i - 1 < y_{i-1} - 1$ numbers before the gamechanger in the permutation.

Altogether: the first gamechanger is the larger element of $y_1-1$ inversions while for $i > 1$, the gamechanger $y_i$ is the larger element of at least $y_i - y_{i-1}$ elements. Since the largest game changer is $n$, we have a total of at least $n-1$ inversions.
This post has been edited 3 times. Last edited by ihatemath123, Aug 18, 2024, 1:35 AM
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CyclicISLscelesTrapezoid
372 posts
#16
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We solve the following problem (half of the solution is copied from my previous post in this thread):
Generalization of China TST Quiz 2001/1/1 wrote:
Let $a$ be a positive real number, and let $f \colon [0,a] \to \mathbb{R}$ be a continuous function with $f(0)=f(a)$. Find the minimum possible number of pairs of real numbers $(x,y)$ such that $f(x)=f(y)$ and $x-y$ is a positive integer.

The answer is $a$ if $a$ is an integer and $0$ otherwise.

If $a$ is an integer, then $f(x)=\tfrac{a}{2}-|x-\tfrac{a}{2}|$ works: for a positive integer $n \le a$, the only pair $(x,y)$ such that $f(x)=f(y)$ and $x-y=n$ is $(\tfrac{a+n}{2},\tfrac{a-n}{2})$, so there are $a$ pairs. If $a$ is not an integer, then construct $f$ by shifting a function with period $1$ by a nonconstant linear function:
\[f(x)=\cos(2\pi x)-1+\frac{1-\cos(2\pi a)}{a}x.\]Notice that $f(0)=f(a)=0$. For a positive integer $n$, we have $f(x+n)-f(x)=\tfrac{1-\cos(2\pi a)}{a}n>0$, so there are $0$ pairs.

It remains to prove that there are at least $a$ pairs if $a$ is an integer.

Claim: There exists a real number $b$ with $0 \le b \le a-1$ such that $f(b+1)=f(b)$.

Proof: Assume WLOG that $f(1) \ge f(0)$, and let $g(x)=f(x+1)$. Since $f(0)=f(a)$, there exists $n \in \{1,2,\ldots,a-1\}$ such that $f(n-1) \le f(n)$ and $f(n) \ge f(n+1)$. This means $f(n-1) \le g(n-1)$ and $f(n) \ge g(n)$, so there exists $b \in [n-1,n]$ such that $f(x)=g(x)$ by the intermediate value theorem, as desired. $\square$

Now, we induct on $a$. For the base case, the pair $(1,0)$ works for $a=1$. For the inductive step, choose $b$ such that $f(b)=f(b+1)$, and consider the function
\[f_1(x)=\begin{cases} f(x) & x \le b \\ f(x+1) & x>b \end{cases}.\]This function is continuous and $f_1(a-1)=f_1(0)$, so there exist at least $a-1$ pairs $(x,y)$ such that $f_1(x)=f_1(y)$ and $x-y$ is a positive integer. Suppose that $x'=x$ if $x \le b$ and $x'=x+1$ if $x>b$, and define $y'$ similarly. Notice that each pair $(x,y)$ converts to a pair $(x',y')$ such that $f(x')=f(y')$ and $x'-y'$ is a positive integer, so we have $a-1$ pairs by the inductive hypothesis. We also have the pair $(b,b+1)$, so there are at least $a$ pairs and we are done. $\blacksquare$
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quantam13
112 posts
#17
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Wow what a problem, i loved it :)

I prove the desired result with strond induction, while replacing 2015 with $n$

Claim: There is some value $t \in [0, n-1]$ such that $p(t+1)=p(t)$

Proof: Assume FTSOC that there is no such value. Upon considering the set $\{p(1)-p(0), p(2)-p(1), \cdots, p(n)-p(n-1)\}$, we get that none of the elements of the set can be zero, but by the condition the total sum is 0, and hence some time in the list it switches signs. But if we consider the function $q(x)=p(x+1)-p(x)$, we get a contradiction by the intermediate value theorem.

Now by the claim we take a value $r$ such that $f(r+1)=f(r)$ and remove it from the interval $[0,n]$, and consider the two remaining intervals together which finishes by strong induction
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quantam13
112 posts
#18
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Oopsies one small error in mine, we musnt use strong inductionand instead shift the part of $f$ above $r+1$ down and induct, and then all the pairs will be preserved and then we shift back to get the extra pair $(r,r+1)$, which solves the problem, noting that the shifting preserves continuity as $f(r+1)=f(r)$
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Likeminded2017
391 posts
#19
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We prove this works for any $n,$ by induction. First, if $n=1$ then $p(0)=p(1)$ so we are done. Suppose this holds until some $n=i.$ For $i+1,$ Let $q(x)=p(x+1),$ and let $r(x)=p(x+1)-p(x).$ If $r(x)$ is always zero then we are done. If not, observe $\int_{0}^{i} r(x)=0,$ so if $r$ is positive at some point, it must also be negative at some point. By the IVT, there must be some point $t$ where $r(t)$ is $0.$ Suppose this point is $t.$ Then $p(t)=p(t+1).$ Now consider
\[s(x)=
\begin{cases}
    p(x) & x \in (-\infty,t] \\
    p(x+1) & x \in (t, +\infty)
\end{cases}
\]this function is clearly continuous as $p(x)$ is continuous and $p(t)=p(t+1).$ Then $s(0)=p(0)=p(n)=s(n-1)$ so there are $i$ such pairs in function $s.$ If they exist in function $s,$ they must exist in function $p$ too. Therefore adding the extra $(t+1,t)$ pair yields a total of $i+1$ pairs, and we are done by induction.
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Ilikeminecraft
611 posts
#20
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We show this for any continuous function $g$. We can induct. If $n = 1,$ this fact is obviously true.

Now, let $k\in\{0, \dots, n - 1\}$ denote the integer such that $f(k) > f(k - 1), f(x) > f(k + 1)$ or the inequalities are swapped. By extremal value theorem, this must exist. By IVT, there exists $t\in[0, j]$ such that $f(t) = f(t + 1).$

Define $\overline{g}\colon[0, n - 1]\to\mathbb R$ to be a continuous function such that \[\overline{g}(x) = \begin{cases*}
    g(x) & x < t \\
    g(x - 1) & g > t
\end{cases*}\]This is obviously continuous. By inductive hypothesis, this also finishes.
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