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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
My Unsolved Problem
ZeltaQN2008   3
N a minute ago by lolsamo
Let $\triangle ABC$ satisfy $AB<AC$. The circumcircle $(O)$ and the incircle $(I)$ of $\triangle ABC$ are tangent to the sides $AC,AB$ at $E,F$, respectively. The line $BI$ meets $EF$ at $M$ and intersects $AC$ at $P$, while the line $BO$ meets $CM$ at $Q$. Construct the common external tangent $\ell$ (different from $BC$) to the incircles of the triangles $PBC$ and $QBC$. Show that $\ell$ is parallel to the line $PQ$.
3 replies
ZeltaQN2008
Today at 11:03 AM
lolsamo
a minute ago
Problem 4
blug   1
N a minute ago by CHESSR1DER
Source: Czech-Polish-Slovak Junior Match 2025 Problem 4
Three non-negative integers are written on the board. In every step, the three numbers $(a, b, c)$ are being replaced with $a+b, b+c, c+a$. Find the biggest number of steps, after which the number $111$ will appear on the board.
1 reply
blug
2 hours ago
CHESSR1DER
a minute ago
Nice concurrency
navi_09220114   2
N 2 minutes ago by navi_09220114
Source: TASIMO 2025 Day 1 Problem 2
Four points $A$, $B$, $C$, $D$ lie on a semicircle $\omega$ in this order with diameter $AD$, and $AD$ is not parallel to $BC$. Points $X$ and $Y$ lie on segments $AC$ and $BD$ respectively such that $BX\parallel AD$ and $CY\perp AD$. A circle $\Gamma$ passes through $D$ and $Y$ is tangent to $AD$, and intersects $\omega$ again at $Z\neq D$. Prove that the lines $AZ$, $BC$ and $XY$ are concurrent.
2 replies
+1 w
navi_09220114
Today at 11:42 AM
navi_09220114
2 minutes ago
Find all integers
velmurugan   2
N 23 minutes ago by Assassino9931
Source: Titu and Dorin Book Problem
Find all positive integers $(x,n)$ such that $x^n + 2^n + 1$ is a divisor of $x^{n+1} + 2^{n+1} +1 $ .
2 replies
velmurugan
Jul 30, 2015
Assassino9931
23 minutes ago
Functional inequality
Jackson0423   1
N 31 minutes ago by CHESSR1DER
Show that there does not exist a function \( f : \mathbb{R}^+ \to \mathbb{R}^+ \) such that for all positive real numbers \( x, y \),
\[
f^2(x) \geq f(x+y)\left(f(x) + y\right).
\]
1 reply
Jackson0423
3 hours ago
CHESSR1DER
31 minutes ago
Problem 2
blug   1
N 39 minutes ago by atdaotlohbh
Source: Czech-Polish-Slovak Junior Match 2025 Problem 2
Find all triangles that can be divided into congruent right-angled isosceles triangles with side lengths $1, 1, \sqrt{2}$.
1 reply
blug
2 hours ago
atdaotlohbh
39 minutes ago
SOLVE IN NATURAL
Pirkuliyev Rovsen   2
N an hour ago by Assassino9931
Source: kolmogorov-2014
Solve in $ N$ the equation: $x{\cdot }y!+2y{\cdot }x!=z!$
2 replies
Pirkuliyev Rovsen
Sep 17, 2023
Assassino9931
an hour ago
Problem 3
blug   2
N an hour ago by blug
Source: Czech-Polish-Slovak Junior Match 2025 Problem 3
In a triangle $ABC$, $\angle ACB=60^{\circ}$. Points $D, E$ lie on segments $BC, AC$ respectively. Points $K, L$ are such that $ADK$ and $BEL$ are equlateral, $A$ and $L$ lie on opposite sides of $BE$, $B$ and $K$ lie on the opposite siedes of $AD$. Prove that
$$AE+BD=KL.$$
2 replies
blug
2 hours ago
blug
an hour ago
equal angles starting with a parallelogram with perpenducular
parmenides51   3
N an hour ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1994 OMM P3
$ABCD$ is a parallelogram. Take $E$ on the line $AB$ so that $BE = BC$ and $B$ lies between $A$ and $E$. Let the line through $C$ perpendicular to $BD$ and the line through $E$ perpendicular to $AB$ meet at $F$. Show that $\angle DAF = \angle BAF$.
3 replies
parmenides51
Jul 29, 2018
FrancoGiosefAG
an hour ago
An important lemma of isogonal conjugate points
buratinogigle   5
N an hour ago by trigadd123
Source: Own
Let $P$ and $Q$ be two isogonal conjugate with respect to triangle $ABC$. Let $S$ and $T$ be two points lying on the circle $(PBC)$ such that $PS$ and $PT$ are perpendicular and parallel to bisector of $\angle BAC$, respectively. Prove that $QS$ and $QT$ bisect two arcs $BC$ containing $A$ and not containing $A$, respectively, of $(ABC)$.
5 replies
+1 w
buratinogigle
Mar 23, 2025
trigadd123
an hour ago
A factorial equals sum of factorials
Ege_Saribass   1
N an hour ago by Ege_Saribass
Source: Own
Let $n$ be given positive integer. Suppose that there is only one positive integer $m$ which satisfies the equation
$$m! = a_1! + a_2! + a_3! + \dots + a_n!$$for some positive integers $a_1, a_2, a_3, \dots, a_n$.Then find all possible values of the positive integer $n$.
1 reply
Ege_Saribass
an hour ago
Ege_Saribass
an hour ago
Long and wacky inequality
Royal_mhyasd   3
N an hour ago by Royal_mhyasd
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.
3 replies
Royal_mhyasd
May 12, 2025
Royal_mhyasd
an hour ago
y^2 = x^3 + 2x^2 + 2x + 1
pokmui9909   10
N an hour ago by Assassino9931
Source: KJMO 2023 P1
Find all integer pairs $(x, y)$ such that $$y^2 = x^3 + 2x^2 + 2x + 1.$$
10 replies
pokmui9909
Nov 4, 2023
Assassino9931
an hour ago
Integer polynomial commutes with sum of digits
cjquines0   44
N an hour ago by Shreyasharma
Source: 2016 IMO Shortlist N1
For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geq 2016$, the integer $P(n)$ is positive and $$S(P(n)) = P(S(n)).$$
Proposed by Warut Suksompong, Thailand
44 replies
cjquines0
Jul 19, 2017
Shreyasharma
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
5003 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
157 posts
#12 • 1 Y
Y by LLL2019
Hilarious problem, I love it :D
Solution
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joshualiu315
2534 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
157 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
3448 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
113 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
113 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
658 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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