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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
Interesting inequalities
sqing   1
N a minute ago by pooh123
Source: Own
Let $ a,b>0 $. Prove that
$$\frac{ab-1} {ab(a+b+2)} \leq \frac{1} {8}$$$$\frac{2ab-1} {ab(a+b+1)} \leq 6\sqrt 3-10$$
1 reply
sqing
Today at 4:24 AM
pooh123
a minute ago
Inequality with rational function
MathMystic33   4
N 2 minutes ago by ytChen
Source: Macedonian Mathematical Olympiad 2025 Problem 2
Let \( n > 2 \) be an integer, \( k > 1 \) a real number, and \( x_1, x_2, \ldots, x_n \) be positive real numbers such that \( x_1 \cdot x_2 \cdots x_n = 1 \). Prove that:

\[
\frac{1 + x_1^k}{1 + x_2} + \frac{1 + x_2^k}{1 + x_3} + \cdots + \frac{1 + x_n^k}{1 + x_1} \geq n.
\]
When does equality hold?
4 replies
MathMystic33
May 13, 2025
ytChen
2 minutes ago
Combinatorics: Easy or tough?
AlexLewandowski   3
N 9 minutes ago by FrancoGiosefAG
Source: Mexican MO 2000
A board $n$×$n$ is coloured black and white like a chessboard. The following steps are permitted: Choose a rectangle inside the board (consisting of entire cells)whose side lengths are both odd or both even, but not both equal to $1$, and invert the colours of all cells inside the rectangle. Determine the values of $n$ for which it is possible to make all the cells have the same colour in a finite number of such steps.

3 replies
AlexLewandowski
Feb 6, 2017
FrancoGiosefAG
9 minutes ago
D1033 : A problem of probability for dominoes 3*1
Dattier   1
N 14 minutes ago by Dattier
Source: les dattes à Dattier
Let $G$ a grid of 9*9, we choose a little square in $G$ of this grid three times, we can choose three times the same.

What the probability of cover with 3*1 dominoes this grid removed by theses little squares (one, two or three) ?
1 reply
Dattier
May 15, 2025
Dattier
14 minutes ago
inequality with roots
luci1337   2
N 26 minutes ago by Nguyenhuyen_AG
Source: an exam in vietnam
let $a,b,c\geq0: a+b+c=1$
prove that $\Sigma\sqrt{a+(b-c)^2}\geq\sqrt{3}$
2 replies
luci1337
May 18, 2025
Nguyenhuyen_AG
26 minutes ago
Very hard
steven_zhang123   0
44 minutes ago
Source: An article
Given a positive integer \(n\) and a positive real number \(m\), non-negative real numbers \(a_1, a_2, \cdots, a_n\) satisfy \(\sum_{i=1}^n a_i = m\). Define \(N\) as the number of elements in the following collection of subsets:

$$
\left\{ I \subseteq \{1, 2, \cdots, n\} : \prod_{i \in I} a_i \geq 1 \right\}.
$$
Find the maximum possible value of \(N\).
0 replies
steven_zhang123
44 minutes ago
0 replies
Interesting Inequality
lbh_qys   1
N an hour ago by nexu
Let $ a, b, c$ be real numbers such that $ (3a-2b-c)(3b-2c-a)(3c-2a-b)\neq 0 $ and $ a + b + c = 3 . $ Prove that
$$ \left( \frac{1}{3a - 2b - c} + \frac{1}{3b - 2c - a} + \frac{1}{3c - 2a - b} \right)^2 + a^2 + b^2 + c^2 \geq 3 + 3\sqrt{\frac 27}$$
1 reply
lbh_qys
2 hours ago
nexu
an hour ago
Goofy geometry
giangtruong13   1
N an hour ago by nabodorbuco2
Source: A Specialized School's Math Entrance Exam
Given the circle $(O)$, from $A$ outside the circle, draw tangents $AE,AF$ ($E,F$ are tangential points) and secant $ABC$ ($B,C$ lie on circle $O$, $B$ is between $A$ and $C$). $OA$ intersects $EF$ at $H$; $I$ is midpoint of $BC$. The line crossing $I$, paralleling with $CE$, intersects $EF$ at $D$. $CD$ intersects $AE$ at $K$. Let $N$ lie inside the triangle $FBC$ such that: $AF$=$AN$. From $N$ draw chords $BQ$, $RC$, $FP$ on circle $(O)$. Prove that: $PRQ$ is a isosceles triangle
1 reply
giangtruong13
Yesterday at 4:23 PM
nabodorbuco2
an hour ago
Rainbow vertices
goodar2006   3
N an hour ago by Sina_Sa
Source: Iran 3rd round 2012-Combinatorics exam-P1
We've colored edges of $K_n$ with $n-1$ colors. We call a vertex rainbow if it's connected to all of the colors. At most how many rainbows can exist?

Proposed by Morteza Saghafian
3 replies
goodar2006
Sep 20, 2012
Sina_Sa
an hour ago
Geometry hard problem.
noneofyou34   1
N an hour ago by User21837561
In a circle of radius R, three chords of length R are given. Their ends are joined with segments to
obtain a hexagon inscribed in the circle. Show that the midpoints of the new chords are the vertices of
an equilateral triang
1 reply
noneofyou34
2 hours ago
User21837561
an hour ago
Prove that two different boards can be obtained
hectorleo123   2
N an hour ago by alpha31415
Source: 2014 Peru Ibero TST P2
Let $n\ge 4$ be an integer. You have two $n\times n$ boards. Each board contains the numbers $1$ to $n^2$ inclusive, one number per square, arbitrarily arranged on each board. A move consists of exchanging two rows or two columns on the first board (no moves can be made on the second board). Show that it is possible to make a sequence of moves such that for all $1 \le i \le n$ and $1 \le j \le n$, the number that is in the $i-th$ row and $j-th$ column of the first board is different from the number that is in the $i-th$ row and $j-th$ column of the second board.
2 replies
hectorleo123
Sep 15, 2023
alpha31415
an hour ago
Inspired by lbh_qys
sqing   2
N 3 hours ago by JARP091
Source: Own
Let $ a, b $ be real numbers such that $ (a-3)(b-3)(a-b)\neq 0 $ and $ a + b =6 . $ Prove that
$$ \left( \frac{a}{b - 3} + \frac{b}{3 - a} + \frac{3}{a - b} \right)^2 + 2(a^2 + b^2 )\geq54$$Equality holds when $ (a,b)=\left(\frac{3}{2},\frac{9}{2}\right)$
$$\left( \frac{a+1}{b - 3} + \frac{b+1}{3 - a} + \frac{4}{a - b} \right)^2 + 2(a^2 + b^2 )\geq 60$$Equality holds when $ (a,b)=\left(3-\sqrt 3,3+\sqrt 3\right)$
$$  \left( \frac{a+3}{b - 3} + \frac{b+3}{3 - a} + \frac{6}{a - b} \right)^2 + 2\left(a^2 + b^2\right)\geq 72$$Equality holds when $ (a,b)=\left(3-\frac{3}{\sqrt 2},3+\frac{3}{\sqrt 2}\right)$
2 replies
sqing
3 hours ago
JARP091
3 hours ago
Interesting Inequality
lbh_qys   4
N 3 hours ago by lbh_qys
Given that \( a, b, c \) are pairwise distinct $\mathbf{real}$ numbers and \( a + b + c = 9 \), find the minimum value of

\[
\left( \frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b} \right)^2  + a^2 + b^2 + c^2.
\]
4 replies
lbh_qys
Today at 5:42 AM
lbh_qys
3 hours ago
Sum of complex fractions is an integer
Miquel-point   2
N 3 hours ago by cazanova19921
Source: Romanian IMO TST 1981, Day 3 P4
Let $n\geqslant 3$ be a fixed integer and $\omega=\cos\dfrac{2\pi}n+i\sin\dfrac{2\pi}n$.
Show that for every $a\in\mathbb{C}$ and $r>0$, the number
\[\sum\limits_{k=1}^n \dfrac{|a-r\omega^k|^2}{|a|^2+r^2}\]is an integer. Interpet this result geometrically.

Octavian Stănășilă
2 replies
Miquel-point
Apr 6, 2025
cazanova19921
3 hours ago
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
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#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
Y by
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
1541 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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