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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Thursday 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
Thursday 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
Sharygin 2025 CR P18
Gengar_in_Galar   5
N 9 minutes ago by hectorleo123
Source: Sharygin 2025
Let $ABCD$ be a quadrilateral such that the excircles $\omega_{1}$ and $\omega_{2}$ of triangles $ABC$ and $BCD$ touching their sides $AB$ and $BD$ respectively touch the extension of $BC$ at the same point $P$. The segment $AD$ meets $\omega_{2}$ at point $Q$, and the line $AD$ meets $\omega_{1}$ at $R$ and $S$. Prove that one of angles $RPQ$ and $SPQ$ is right
Proposed by: I.Kukharchuk
5 replies
Gengar_in_Galar
Mar 10, 2025
hectorleo123
9 minutes ago
BMO 2025
GreekIdiot   10
N 14 minutes ago by tranducphat
Does anyone have the problems? They should have finished by now.
10 replies
GreekIdiot
Apr 27, 2025
tranducphat
14 minutes ago
Infinitely many numbers of a given form
Stefan4024   19
N an hour ago by cursed_tangent1434
Source: EGMO 2016 Day 2 Problem 6
Let $S$ be the set of all positive integers $n$ such that $n^4$ has a divisor in the range $n^2 +1, n^2 + 2,...,n^2 + 2n$. Prove that there are infinitely many elements of $S$ of each of the forms $7m, 7m+1, 7m+2, 7m+5, 7m+6$ and no elements of $S$ of the form $7m+3$ and $7m+4$, where $m$ is an integer.
19 replies
Stefan4024
Apr 13, 2016
cursed_tangent1434
an hour ago
Very easy case of a folklore polynomial equation
Assassino9931   1
N an hour ago by iamnotgentle
Source: Bulgaria EGMO TST 2025 P6
Determine all polynomials $P(x)$ of odd degree with real coefficients such that $P(x^2 + 2025) = P(x)^2 + 2025$.
1 reply
Assassino9931
2 hours ago
iamnotgentle
an hour ago
Process on scalar products and permutations
Assassino9931   2
N an hour ago by Assassino9931
Source: RMM Shortlist 2024 C1
Fix an integer $n\geq 2$. Consider $2n$ real numbers $a_1,\ldots,a_n$ and $b_1,\ldots, b_n$. Let $S$ be the set of all pairs $(x, y)$ of real numbers for which $M_i = a_ix + b_iy$, $i=1,2,\ldots,n$ are pairwise distinct. For every such pair sort the corresponding values $M_1, M_2, \ldots, M_n$ increasingly and let $M(i)$ be the $i$-th term in the list thus sorted. This denes an index permutation of $1,2,\ldots,n$. Let $N$ be the number of all such permutations, as the pairs run through all of $S$. In terms of $n$, determine the largest value $N$ may achieve over all possible choices of $a_1,\ldots,a_n,b_1,\ldots,b_n$.
2 replies
Assassino9931
3 hours ago
Assassino9931
an hour ago
Square problem
Jackson0423   2
N an hour ago by Jackson0423
Construct a square such that the distances from an interior point to the vertices (in clockwise order) are
1,7,8,4 respectively.
2 replies
Jackson0423
Yesterday at 4:08 PM
Jackson0423
an hour ago
IMO Shortlist Problems
ABCD1728   2
N an hour ago by ABCD1728
Source: IMO official website
Where can I get the official solution for ISL before 2005? The official website only has solutions after 2006. Thanks :)
2 replies
1 viewing
ABCD1728
Yesterday at 12:44 PM
ABCD1728
an hour ago
Estimate on number of progressions
Assassino9931   1
N an hour ago by BlizzardWizard
Source: RMM Shortlist 2024 C4
Let $n$ be a positive integer. For a set $S$ of $n$ real numbers, let $f(S)$ denote the number of increasing arithmetic progressions of length at least two all of whose terms are in $S$. Prove that, if $S$ is a set of $n$ real numbers, then
\[ f(S) \leq \frac{n^2}{4} + f(\{1,2,\ldots,n\})\]
1 reply
Assassino9931
3 hours ago
BlizzardWizard
an hour ago
2^x+3^x = yx^2
truongphatt2668   10
N an hour ago by MittenpunktpointX9
Prove that the following equation has infinite integer solutions:
$$2^x+3^x = yx^2$$
10 replies
truongphatt2668
Apr 22, 2025
MittenpunktpointX9
an hour ago
find the radius of circumcircle!
jennifreind   1
N an hour ago by ricarlos
In $\triangle \rm ABC$, $  \angle \rm B$ is acute, $\rm{\overline{BC}} = 8$, and $\rm{\overline{AC}} = 3\rm{\overline{AB}}$. Let point $\rm D$ be the intersection of the tangent to the circumcircle of $\triangle \rm ABC$ at point $\rm A$ and the perpendicular bisector of segment $\rm{\overline{BC}}$. Given that $\rm{\overline{AD}} = 6$, find the radius of the circumcircle of $\triangle \rm BCD$.
IMAGE
1 reply
jennifreind
Yesterday at 2:12 PM
ricarlos
an hour ago
A folklore polynomial game
Assassino9931   1
N 2 hours ago by YaoAOPS
Source: RMM Shortlist 2024 A1, also Bulgaria Regional Round 2016, Grade 12
Fix a positive integer $d$. Yael and Ziad play a game as follows, involving a monic polynomial of degree $2d$. With Yael going first, they take turns to choose a strictly positive real number as the value of one of the coecients of the polynomial. Once a coefficient is assigned a value, it cannot be chosen again later in the game. So the game
lasts for $2d$ rounds, until Ziad assigns the final coefficient. Yael wins if $P(x) = 0$ for some real
number $x$. Otherwise, Ziad wins. Decide who has the winning strategy.
1 reply
Assassino9931
3 hours ago
YaoAOPS
2 hours ago
Game on board with gcd and lcm
Assassino9931   0
2 hours ago
Source: Bulgaria EGMO TST 2025 P3
On the board are written $n \geq 2$ positive integers with least common multiple $K$ and greatest common divisor $1$. It is known that $K$ is not a perfect square and is not among the initially written numbers. Two players $A$ and $B$ play the following game, taking turns alternatingly, with $A$ starting first. In a move the player has to write a number which has not been written so far, by taking two distinct integers $a$ and $b$ from the board and write LCM$(a,b)$ or LCM$(a,b)$/$a$. The player who writes $1$ or $K$ loses. Who has a winning strategy?
0 replies
Assassino9931
2 hours ago
0 replies
Popular children at camp with algebra and geometry
Assassino9931   0
3 hours ago
Source: RMM Shortlist 2024 C3
Fix an odd integer $n\geq 3$. At a maths camp, there are $n^2$ children, each of whom selects
either algebra or geometry as their favourite topic. At lunch, they sit at $n$ tables, with $n$ children
on each table, and start talking about mathematics. A child is said to be popular if their favourite
topic has a majority at their table. For dinner, the students again sit at $n$ tables, with $n$ children
on each table, such that no two children share a table at both lunch and dinner. Determine the
minimal number of young mathematicians who are popular at both mealtimes. (The minimum is across all sets of topic preferences and seating arrangements.)
0 replies
1 viewing
Assassino9931
3 hours ago
0 replies
Triangles in dissections
Assassino9931   0
3 hours ago
Source: RMM Shortlist 2024 C2
Fix an integer $n\geq 3$ and let $A_1A_2\ldots A_n$ be a convex polygon in the plane. Let $\mathcal{M}$ be the set of all midpoints $M_{i,j}$ of segments $A_iA_j$ where $i\neq j$. Assume that all of these midpoints are distinct, i.e. $\mathcal{M}$ consists of $\frac{n(n-1)}{2}$ elements. Dissect the polygon $M_{1,2}M_{2,3}\ldots M_{n,1}$ into triangles so that the following hold:

(1) The intersection of every two triangles (interior and boundary) is either empty or a common
vertex or a common side.
(2) The vertices of all triangles lie in M (not all points in M are necessarily used).
(3) Each side of every triangle is of the form $M_{i,j}M_{i,k}$ for some pairwise distinct indices $i,j,k$.

Prove that the total number of triangles in such a dissection is $3n-8$.
0 replies
Assassino9931
3 hours ago
0 replies
Number theory or function ?
matematikator   15
N Apr 23, 2025 by YaoAOPS
Source: IMO ShortList 2004, algebra problem 3
Does there exist a function $s\colon \mathbb{Q} \rightarrow \{-1,1\}$ such that if $x$ and $y$ are distinct rational numbers satisfying ${xy=1}$ or ${x+y\in \{0,1\}}$, then ${s(x)s(y)=-1}$? Justify your answer.

Proposed by Dan Brown, Canada
15 replies
matematikator
Mar 18, 2005
YaoAOPS
Apr 23, 2025
Number theory or function ?
G H J
Source: IMO ShortList 2004, algebra problem 3
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matematikator
110 posts
#1 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Does there exist a function $s\colon \mathbb{Q} \rightarrow \{-1,1\}$ such that if $x$ and $y$ are distinct rational numbers satisfying ${xy=1}$ or ${x+y\in \{0,1\}}$, then ${s(x)s(y)=-1}$? Justify your answer.

Proposed by Dan Brown, Canada
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grobber
7849 posts
#2 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Where did you find this problem? It is realy interesting.

I believe the answer to be 'yes', and I think I have a way of constructing the function, but it's only verified empirically, and I haven't fixed my ideas well enough to provide a full proof. Here's what I thought we could do:

It suffices to determine the function on the positive rationals, of course, so let's restrict our attention to those. Construct the Stern-Brocot tree step by step. At each step, to each leaf you attach a leaf to the left and a leaf to the right. Label the leaf to the left with $-1$ and the leaf to the right with $1$. It seems to work.
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Myth
4464 posts
#3 • 2 Y
Y by Adventure10, Mango247
We have $s(x)=s(1+x)=-s(1/x)$. I see a direct way to use continuous fractions here, since having these rules we can obtain $s(x)$ from $s(1)$. :?
Did I say something stupid?
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matematikator
110 posts
#4 • 2 Y
Y by Adventure10, Mango247
Solution: Let x=a/b be a positive rational, where a,b are coprime positive integers. Consider the squence of consecutive remainders given by euclidean algorithm for the ordered pairs (a,b). if u mod v denotes the least non negative remainder of u modulo v, this sequence can be written as
r_0=a, r_1=b, r_2=a mod b, ... , r_(i+1)=r_(i-1) mod r_i, ... , r_n=1, r_(n+1)=0 (1)

the index n=n(x) of the least nonzero remainder r_n=1 is uniquily determined by x, so
t(x)=(-1)^n(x) is a well-defined function from the positive rationals into {-1,1}. Now define s: Q-->{-1,1} by

s(x)=t(x) for x in Q, x>0 or s(x)=-t(-x) for x in Q, x<0 or s(0)=1

We prove that s has the desired properties. Let x and y be distinct rational numbers.
* If x+y=0, let x>0 and y<0 (x,y are nonzero). Then , by definition, s(x)=t(x),
s(y)=-t(-y)=-t(x), hence s(x)s(y)=-1.
**If xy=1 then x and y are of the same sign and neither equals 1 or -1. Suppose first
that x=a/b >0, y=b/a >0, with a,b coprime positive integers, one may also assume
that a>b. Euclidean algorithm for (a,b) starts a,b, (a mod b), ... .On the other hand,
the algorithm for pair (b,a) gives b,a,b,(a mod b)m,... .
Because a>b implies r_2=(b mod a)=b. Each term in (1) depends only on the previous
two, so the sequence for x has length by 1 less than the sequence for y, that is,
n(y)=n(x)+1. Hence t(y)=-t(x), and so s(y)=-s(x), as needed. Now the case of x,y
follows from the definition...
**If x+y=1 then ....

by the same logic it'll be true...
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Myth
4464 posts
#5 • 2 Y
Y by Adventure10, Mango247
We see that $t(x)$ is a number of subfractions in the continuous fraction representing $x$.
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matematikator
110 posts
#6 • 2 Y
Y by Adventure10, Mango247
absolutely...
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Raúl
69 posts
#7 • 2 Y
Y by Adventure10, Mango247
I knew it's imoral relive topic but I have a question: we can prove that $ s(1) = 0$ if we use $ s(x) = - s(\frac {1}{x})\Leftrightarrow s(1) = - s(1)\Leftrightarrow s(1) = 0$ contradiction.

I dont really understand :maybe:
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bearkiller
37 posts
#8 • 2 Y
Y by Adventure10, Mango247
x and y are distinct rational numbers if you use s(x) =-s(1/x) for x=1 then we have x=1=1/1=1/x=y.
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Wolstenholme
543 posts
#9 • 7 Y
Y by bcp123, jt314, JasperL, meowme, Adventure10, Mango247, Sedro
Every positive rational number has a unique continued fraction of the form $ a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{\ddots\frac{1}{a_{m - 1} + \frac{1}{a_m}}}}} $ for some $ m \in \mathbb{N} $ where $ a_i \in \mathbb{N} $ for all $ i $. From now on this will be denoted as $ [a_0; a_1, a_2, \dots, a_m] $.

For every $ a \in \mathbb{Q}^{+} $, let $ s(a) = (-1)^m $. Then let $ s(0) = -1 $ and for all $ b \in \mathbb{Q} \setminus \{0\} $ let $ s(-b) = -s(b) $. I claim that this $ s $ satisfies the desired properties. To prove this, consider two rational numbers $ x < y $.

$ 1) $ If $ x + y = 0 $ then by definition $ s(x)s(y) = -1 $ as desired.

$ 2) $ If $ x + y = 1 $ and $ x = 0 $ then by definition $ s(x)s(y) = -1 $ as desired since $ s(1) = 1 $.

$ 3) $ If $ x + y = 1 $ and $ x < 0 $ then to prove that $ s(x)s(y) = -1 $ it suffices to show that $ s(-x) = s(y) $. Let $ -x = [a_0; a_1, a_2, \dots, a_m] $. Then $ y = [a_0 + 1; a_1, a_2, \dots, a_m] $ and so $ s(-x) = s(y) = (-1)^m $ as desired.

$ 4) $ If $ x + y = 1 $ and $ x > 0 $ we can let $ y = [0; 1, a_2, \dots, a_m] $ so that $ x = [0; a_2 + 1, a_3, \dots, a_m] $ so $ s(y) = (-1)^m $ and $ s(x) = (-1)^{m - 1} $ which implies that $ s(x)s(y) = -1 $ as desired.

$ 5) $ If $ xy = 1 $, we can assume WLOG that $ 0 < x < 1 $. Then letting $ x = [0; a_1, a_2, \dots, a_m] $ we have that $ y = [a_1; a_2, a_3, \dots, a_m] $ so $ s(x) = (-1)^m $ and $ s(y) = (-1)^{m - 1} $ which implies that $ s(x)s(y) = -1 $ as desired.

This implies that my claim is true and so we are done. Note that this also implies that Grobber's idea about the Stern-Brocot tree works as well.
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dgrozev
2463 posts
#10 • 1 Y
Y by Adventure10
Note also that this function $s$ is unique up to multiplication by $-1$. Indeed let $x\in\mathbb{Q}^{+}$ and $x=[a_0;a_1,\ldots,a_m]$. Using $s(x+1)=s(x)$ and $s(x)=-s(1/x)$ it's easily obtained $s(x)=(-1)^m s(1)$.
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math_pi_rate
1218 posts
#11 • 3 Y
Y by AlastorMoody, amar_04, AmirKhusrau
Fake alg, pure NT :furious:. Anyway, here's my solution: For two integers $0<p<q$, let $f(p,q)$ denote the number of steps taken in the Euclidean Algorithm to find $\gcd(p,q)$. In other words, if we let $b_0=q,$ $b_1=p$ and write the system of equations $$b_r=a_{r+1}b_{r+1}+b_{r+2} \text{ such that } 0<b_{r+1}<b_r \quad \forall r \in \{0,1, \dots ,n\} \text{ and } b_{n+2}=0$$then $f(p,q)=n,$ where $n \geq 0$. Then we claim that the following function works-
The ProGawd Function wrote:
  1. $s(n)=1$ for $n \in \mathbb{N},$ and $s(-m)=-1$ for $m \in \mathbb{N} \cup \{0\}.$
    $\text{ }$
  2. $s \left(\frac{-1}{2} \right)=1.$
    $\text{ }$
  3. $s(1+x)=s(x)$ for $x \neq \frac{-1}{2},0.$
    $\text{ }$
  4. For integers $p,q$ with $0<p<q$, $$s \left (\frac{p}{q} \right)=(-1)^{f(p,q)+1}$$
It's easy to see that this uniquely defines the image of all rationals. Now we show that this indeed satisfies all the given properties.
  • First we prove $s(x)s(-x)=-1$ when $x \neq 0$. For $x \in \mathbb{N}$ and $x=\pm \frac{1}{2}$, this easily follows from the definition, so we can ignore these cases. WLOG assume $x>0$ and suppose $x \in (n-1,n)$ for some $n \in \mathbb{N}$ with $2x \neq 1$. Then from property 3 of our function, we have $s(x)=s(x-(n-1))$ and $s(-x)=s((n-1)-x)$. Since $0<x-(n-1)<1,$ so it suffices to show that $s(x) \cdot s(-x)=-1$ for $x \in (0,1)$.

    Take $x=\frac{p}{q}$ for $0<p<q$, and first suppose $p< \frac{q}{2}$. We have $$s \left( \frac{-p}{q} \right)=s \left(1-\frac{p}{q} \right)=(-1)^{f(q-p,q)+1}$$But, as $q-p>\frac{q}{2}>p$, so the first step of our Euclidean Algorithm while finding $\gcd(q-p,q)$ will be $q=(q-p)+p$. So $f(q-p,q)=1+f(p,q-p)$. Also, using $q>2p$, we get $f(p,q-p)=f(p,q)$. Thus, we have $$s \left( \frac{-p}{q} \right)=(-1)^{f(q-p,q)+1}=(-1)^{1+f(p,q)+1}=-s \left( \frac{p}{q} \right)$$Thus, for $x \in \left(0, \frac{1}{2} \right),$ we have $s(x)s(-x)=-1$. Finally, if $x \in \left(\frac{1}{2},1 \right),$ then $s(x)=s(x-1)$ and $s(-x)=s(1-x)$. Since $0<1-x<\frac{1}{2}$, then the previous result gives $s(1-x)s(-(1-x))=-1$. Combined with the above equalities, we get $s(x)s(-x)=-1$. Thus, the result is true for all $x \in \mathbb{Q}$.
    $\text{ }$
  • Now we show that $s(x)s(1-x)=-1$ for $x \neq \frac{1}{2}$. For $x=1$, the result is true by definition. And for $x \neq 1,\frac{1}{2}$, by the above result, we have $s(1-x)=-s(x-1)=-s((x-1)+1),$ which directly gives the desired property.
    $\text{ }$
  • Finally we show that $s(x)s \left(\frac{1}{x} \right)=-1$ when $x \neq 0,\pm 1$. Since $s(-x)=-s(x)$ (proved before), so it suffices to show the result for positive $x$. Also, WLOG we can take $0<x<1$, and write $x=\frac{p}{q}$ with $p<q$ and $p,q \in \mathbb{N}$. Then $$s \left(\frac{1}{x} \right)=s \left( \frac{q}{p} \right)=s \left( \frac{q}{p} -1\right)= \dots =s \left( \frac{q \pmod{p}}{p} \right)=(-1)^{f(q \pmod{p},p)+1}$$But, while finding $\gcd(q\pmod{p},p)$ we simply start from the second step in the Euclidean Algorithm to find $\gcd(p,q)$. So we have $f(q \pmod{p},p)+1=f(p,q)$. This gives us $$s \left(\frac{1}{x} \right)=(-1)^{f(q \pmod{p},p)+1}=(-1)^{f(p,q)}=-s \left(\frac{p}{q} \right)=-s(x)$$Thus, we have $s(x)s(-x)=-1$ as desired. $\blacksquare$
This post has been edited 1 time. Last edited by math_pi_rate, Mar 29, 2020, 8:57 AM
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blacksheep2003
1081 posts
#12
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Solution
This post has been edited 1 time. Last edited by blacksheep2003, Oct 13, 2020, 10:29 PM
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yayups
1614 posts
#13 • 5 Y
Y by Mathematicsislovely, TechnoLenzer, Mango247, Mango247, richrow12
The answer is yes, and we construct such a function. First, note that the only condition on $s(0)$ is that $s(0)=-s(1)$, which we can manually enforce at the end. Thus, we assume that the inputs to $s$ are nonzero rationals.

Write $x=a/b$ where $\gcd(a,b)=1$ and $b>0$. We set \[s(x) = \begin{cases}1 &\text{if } (a^{-1}\mod b)\le b/2 \\ -1 &\text{if } (a^{-1}\mod b)>b/2.\end{cases}\]From this definition, it is clear that $s(x)= - s(-x)$ and $s(x)=-s(1-x)$, so it suffices to show that $s(x)=-s(1/x)$. Since we know $s$ is odd, it suffices in fact to show $s(x)=-s(1/x)$ for $x>0$.

Lemma: If $a,b\ge 1$ are relatively prime positive integers, then \[a\cdot(a^{-1}\mod b) + b\cdot(b^{-1}\mod a) = ab+1.\]
Proof: Indeed, letting $T$ be the left side of the above equation, we see that $T\equiv 1\pmod{a}$ and $T\equiv 1\pmod{b}$, so $T\equiv 1\pmod{ab}$ since $\gcd(a,b)=1$.

Furthermore, we have \[a+b\le T\le a(b-1)+b(a-1)<2ab,\]so we must in fact have $T=ab+1$, as desired. $\blacksquare$

The above lemma implies that $(a^{-1}\mod b)\le b/2$ if and only if $(b^{-1}\mod a)>a/2$, which implies that $s(a/b) = -s(b/a)$.

Remark: The condition is basically the same as ISL 2017 N8, which motivates the classification of $s$.
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awesomeming327.
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#14
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We define $n(x)$ on the positive rational numbers to be the number of terms in the continued fraction of $x$. Then let $s(x)=-1^{n(x)}$ and $s(-x)=-s(x)$. Then, clearly $s(x)s(-x)=-1$. If $xy=1$ and $x>1$ then $y = 0+\tfrac{1}{x}$ and therefore $n(y)=1+n(x)$, so $s(x)s(y)=-1$.

Suppose $x+y=1$. WLOG, $x > \tfrac{1}{2}$. If $x < 1$ then $x=[0;1,a,\dots]$ and $y=[0;1+a,\dots]$ where the $\dots$ is the same. If $x>1$ then $x=(-y)+1$ so they have the same number of terms in the continued fraction, thus $s(x)=s(-y)=-s(y)$ as desired.
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asdf334
7585 posts
#15 • 1 Y
Y by megarnie
The answer is yes. We construct $s$ directly.
First assume $s(0)=1$. For any $x>0$ we have
\[s(x)=-s(-x)=s(x+1)\]and we also have $s(1)=-1$. Hence for any $n\in \mathbb{N}$ we have $s(n)=-1$.
For each $q\ge 2$ (starting from the smallest), we now choose $s\left(\frac{p}{q}\right)$ for $p$ starting from $1$ and relatively prime to $q$ according to the following rules:
  • If $p<q$ then $s\left(\frac{p}{q}\right)=-s\left(\frac{q}{p}\right)$.
  • If $p>q$ then $s\left(\frac{p}{q}\right)=s\left(\frac{p-q}{q}\right)$.
Evidently all outputs for nonnegative inputs have been chosen with no problems.
Now we prove that for distinct positive rationals $x$ and $y$ the conditions hold. Importantly our selection of outputs satisfies $s(x)=s(x+1)$ as well. (The only restriction on $s(0)$ is $s(0)=-s(1)$ which clearly holds.)
  • If $xy=1$ then WLOG $x=\frac{p}{q}$ with $p<q$. From the first rule in the previous section we are fine.
  • Notice that $x+y=0$ cannot hold. If $x+y=1$ then write $x=\frac{a}{a+b}$ and $y=\frac{b}{a+b}$. Then
    \[s(x)=-s(1/x)=-s(1/x-1)=-s(b/a)\]\[s(y)=-s(1/y)=-s(1/y-1)=-s(a/b)\]thus $s(x)s(y)=-1$.

Now we write $s(-x)=-s(x)$ for positive $x$.
  • If $x$ and $y$ are distinct, negative and satisfy $xy=1$ we are okay: $s(-x)s(-y)=-1\implies s(x)s(y)=-1$.
  • If $x$ and $y$ are distinct and satisfy $x+y=0$ we are okay by definition.
  • If $x+y=1$ and WLOG $x$ is negative while $y$ is positive, then
    \[s(y)=s(y-1)=-s(1-y)=-s(x)\]and we are okay.

All cases are exhausted and the construction works. $\blacksquare$
This post has been edited 1 time. Last edited by asdf334, Jun 2, 2024, 5:01 PM
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YaoAOPS
1537 posts
#16 • 1 Y
Y by NicoN9
This problem gets no big booms.


we claim the answer is yes. We first define a function $f: \mathbb{Q}^+ \times \mathbb{Q}^+ \to \{+1, -1\}$ for coprime nonzero inputs as follows: let $f(1, 1) = +1, f(1, a) = -1, f(a, 1) = +1$ for all $a \ne 1$. Then we define recursively
\[
	f(a+b,b) = f(a,a+b) = f(a,b).
\]This is well defined as $f$ is defined on input by Euclidean algorithm, and $f(a,b)$ is defined uniquely by whichever of $a, b$ is smaller.

Now note that $f(a, b) = -f(b, a)$ follows since $f(1, a) = -f(a, 1)$ recursively with the Euclidean algorithm. Furthermore, for $a < b$, $f(b-a, b) = f(b-a, a) = f(b,a) = -f(b,a) = -f(a,b)$ $(\heartsuit)$.

We now extend $f$ to negative coprime inputs by defining $f(-a, b) = f(a, -b) = -f(a, b)$. By $(\heartsuit)$ and by the definition, it follows that for $b \ne 1$
\[
	f(a, b) = f(a+b, b), f(b, a) = f(b, a+b)
\]holds for all $a$. We finally define $s\left(\frac{a}{b}\right) = f(a, b)$ for coprime $a, b$. Then $f(a,b) = -f(b,a)$ holds over the extension for the same reasons, $f(a,b) = -f(-a,b)$ holds by definition, and $f(a,b) = -f(b-a,b)$ holds since
\[
	f(b-a,b) = f(-a,b) = -f(a,b)
\]so we are done.
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