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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 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
chat gpt
fuv870   6
N a minute ago by jkim0656
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
6 replies
2 viewing
fuv870
3 hours ago
jkim0656
a minute ago
Unsolved Diophantine(I think)
Nuran2010   2
N 3 minutes ago by ohiorizzler1434
Find all solutions for the equation $2^n=p+3^p$ where $n$ is a positive integer and $p$ is a prime.(Don't get mad at me,I've used the search function and did not see a correct and complete solution anywhere.)
2 replies
Nuran2010
Mar 14, 2025
ohiorizzler1434
3 minutes ago
Functional Equations Marathon March 2025
Levieee   0
10 minutes ago
1. before posting another problem please try your best to provide the solution to the previous solution because we don't want a backlog of many problems
2.one is welcome to send functional equations involving calculus (mainly basic real analysis type of proofs) as long it is of the form $\text{"find all functions:"}$
0 replies
1 viewing
Levieee
10 minutes ago
0 replies
Another NT FE
nukelauncher   60
N 23 minutes ago by pi271828
Source: ISL 2019 N4
Find all functions $f:\mathbb Z_{>0}\to \mathbb Z_{>0}$ such that $a+f(b)$ divides $a^2+bf(a)$ for all positive integers $a$ and $b$ with $a+b>2019$.
60 replies
nukelauncher
Sep 22, 2020
pi271828
23 minutes ago
Perfect Squares, Infinite Integers and Integers
steven_zhang123   4
N 37 minutes ago by ohiorizzler1434
Source: China TST 2001 Quiz 5 P1
For which integer \( h \), are there infinitely many positive integers \( n \) such that \( \lfloor \sqrt{h^2 + 1} \cdot n \rfloor \) is a perfect square? (Here \( \lfloor x \rfloor \) denotes the integer part of the real number \( x \)?
4 replies
steven_zhang123
Yesterday at 12:06 PM
ohiorizzler1434
37 minutes ago
another geometry problem with sharky-devil point
anyone__42   11
N 40 minutes ago by zhenghua
Source: The francophone mathematical olympiads P1
Let $ABC$ be an acute triangle with $AC>AB$, Let $DEF$ be the intouch triangle with $D \in (BC)$,$E \in (AC)$,$F \in (AB)$,, let $G$ be the intersecttion of the perpendicular from $D$ to $EF$ with $AB$, and $X=(ABC)\cap (AEF)$.
Prove that $B,D,G$ and $X$ are concylic
11 replies
anyone__42
Jun 27, 2020
zhenghua
40 minutes ago
IMO Shortlist 2011, Algebra 5
orl   18
N an hour ago by mathfun07
Source: IMO Shortlist 2011, Algebra 5
Prove that for every positive integer $n,$ the set $\{2,3,4,\ldots,3n+1\}$ can be partitioned into $n$ triples in such a way that the numbers from each triple are the lengths of the sides of some obtuse triangle.

Proposed by Canada
18 replies
orl
Jul 11, 2012
mathfun07
an hour ago
Line Perpendicular to Euler Line
tastymath75025   55
N an hour ago by ohiorizzler1434
Source: USA TSTST 2017 Problem 1, by Ray Li
Let $ABC$ be a triangle with circumcircle $\Gamma$, circumcenter $O$, and orthocenter $H$. Assume that $AB\neq AC$ and that $\angle A \neq 90^{\circ}$. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ in $\triangle ABC$, respectively. Let $P$ be the intersection of line $MN$ with the tangent line to $\Gamma$ at $A$. Let $Q$ be the intersection point, other than $A$, of $\Gamma$ with the circumcircle of $\triangle AEF$. Let $R$ be the intersection of lines $AQ$ and $EF$. Prove that $PR\perp OH$.

Proposed by Ray Li
55 replies
tastymath75025
Jun 29, 2017
ohiorizzler1434
an hour ago
Foot from vertex to Euler line
cjquines0   31
N an hour ago by pUssydestroyer777
Source: 2016 IMO Shortlist G5
Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $ABC$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $AB$ and $AC$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $BC$, and let $M$ be the midpoint of $BC$. Prove that the circumcentre of triangle $XSY$ is equidistant from $P$ and $M$.
31 replies
1 viewing
cjquines0
Jul 19, 2017
pUssydestroyer777
an hour ago
Inequality => square
Rushil   12
N 2 hours ago by ohiorizzler1434
Source: INMO 1998 Problem 4
Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.
12 replies
Rushil
Oct 7, 2005
ohiorizzler1434
2 hours ago
p + q + r + s = 9 and p^2 + q^2 + r^2 + s^2 = 21
who   28
N 2 hours ago by asdf334
Source: IMO Shortlist 2005 problem A3
Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.
28 replies
who
Jul 8, 2006
asdf334
2 hours ago
H not needed
dchenmathcounts   44
N 3 hours ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
3 hours ago
IZHO 2017 Functional equations
user01   51
N 3 hours ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
3 hours ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 3 hours ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
3 hours ago
IMO ShortList 2003, number theory problem 8
orl   10
N Jan 5, 2024 by IAmTheHazard
Source: IMO ShortList 2003, number theory problem 8
Let $p$ be a prime number and let $A$ be a set of positive integers that satisfies the following conditions:

(i) the set of prime divisors of the elements in $A$ consists of $p-1$ elements;

(ii) for any nonempty subset of $A$, the product of its elements is not a perfect $p$-th power.

What is the largest possible number of elements in $A$ ?
10 replies
orl
Oct 4, 2004
IAmTheHazard
Jan 5, 2024
IMO ShortList 2003, number theory problem 8
G H J
Source: IMO ShortList 2003, number theory problem 8
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orl
3647 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $p$ be a prime number and let $A$ be a set of positive integers that satisfies the following conditions:

(i) the set of prime divisors of the elements in $A$ consists of $p-1$ elements;

(ii) for any nonempty subset of $A$, the product of its elements is not a perfect $p$-th power.

What is the largest possible number of elements in $A$ ?
Attachments:
This post has been edited 1 time. Last edited by djmathman, May 27, 2018, 3:55 PM
Reason: adjusted wording according to https://anhngq.files.wordpress.com/2010/07/imo-2003-shortlist.pdf
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orl
3647 posts
#2 • 2 Y
Y by Adventure10, Mango247
Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions :)
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Zhero
2043 posts
#3 • 4 Y
Y by huricane, mijail, Adventure10, Mango247
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v_Enhance
6857 posts
#4 • 8 Y
Y by SK_pi3145, Mathematicsislovely, mijail, PIartist, guptaamitu1, HamstPan38825, Ru83n05, starchan
Solution from Twitch Solves ISL:

Let $D = p-1$. Then the question (thinking in terms of the exponents) can be phrased as follows:

What's the largest multiset of vectors in ${\mathbb F}_p^{D}$ such that no nonempty subset has zero sum?

We claim the answer is $D \cdot (p-1)$. A construction is given by taking
  • $p-1$ copies of the vector $\left< 1, 0, \dots, 0\right>$;
  • $p-1$ copies of the vector $\left< 0, 1, \dots, 0\right>$;
  • \dots;
  • $p-1$ copies of the vector $\left< 0, 0, \dots, 1 \right>$.
To show it's best possible, suppose we have vectors $v_1$, \dots, $v_N$ with coordinates given as \[ v_k = \left< v_{k,1}, v_{k,2}, \dots, v_{k,D} \right> 	\qquad k = 1, 2, \dots, N.  \]Then, we construct the polynomial \[ 	F(X_1, \dots, X_N) 	= \prod_{i=1}^D 	\left[ 1 - \left( \sum_{k=1}^N X_k v_{k,j} \right)^{p-1} \right] 	- \prod_{i=1}^N (1-X_i) \]If we imagine the $X_i \in \{0,1\}$ as Bernoulli random variables indicating whether $v_k$ is used in a set or not, then $F(X_1, \dots, X_N) \neq 0$ exactly when the $X_i$'s equal to $1$ correspond to a nonempty subset of the vectors which have vanishing sum.
Now assume for contradiction $N < D \cdot (p-1)$. Then the largest degree term is given in the latter product; it is exactly $(-1)^{N+1} X_1 X_2 \dots X_N$. So if we quote Alon's combinatorial nullstellensatz, it now follows that there is a choice of $X_i \in \{0,1\}$ such that $F(X_1, \dots, X_N) \neq 0$ which is a contradiction.
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Math_Is_Fun_101
159 posts
#5 • 2 Y
Y by Derfu37, guptaamitu1
Solution with derfu37 and a hint to use Chevalley's from Al3jandro0000.

We claim the answer is $(p-1)^2$. This is achievable by taking the set
\[ A = \{ q_i^{pj+1}\vert(i,j)\in\{1,\ldots,p-1\}^2 \} \]where $\{q_i\}_{i=1}^{\infty}$ is the sequence of primes. Now we show $|A|>(p-1)^2$ is impossible. Let $n=|A|$. Note that by condition (i) we can represent the elements of $A$ as vectors in $\mathbb F_p^{p-1}$. Let these be
\[ v_i = (e_{i1},\ldots,e_{i(p-1)})\text{ for } 1\le i\le n. \]Now, we define $f_j\colon\mathbb F_p^n\rightarrow\mathbb F_p^n$ as
\[ f_j(x_1,\ldots,x_n) = \sum_{i=1}^n x_i^{p-1}e_{ji}. \]Consider the system of equations $f_j(x_1,\ldots,x_n)=0$ for $1\le j\le p-1$. Note that $(0,\ldots,0)$ is a trivial solution. Suppose for the sake of contradiction that $n>(p-1)^2$. Then, we have
\[ n>(p-1)^2=(p-1)\cdot(p-1)=\sum_{j=1}^{p-1}\deg(f_j). \]Thus, quoting Chevalley's Theorem there must exist a nontrivial solution to the system. However, since $x_i^{p-1}\in{0,1}$, this nontrivial solution specifies a nonempty subset of $A$ for which the product is a perfect $p-$th power, yielding the desired contradiction. $\blacksquare$
This post has been edited 3 times. Last edited by Math_Is_Fun_101, Oct 13, 2021, 7:51 AM
Reason: Mispelled Chevalley's
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Leo.Euler
577 posts
#6
Y by
One can easily rephrase this problem using the prime factorizations of the elements of $A$; this asks for the size of the largest multiset of vectors of $\mathbb{F}_p^{p-1}$ where no nonempty subset has a sum of zero.

The answer is $(p-1)^2$; this can be obtained by taking\[\bigcup_{\textbf{x}\in \mathbb{F}_p^{p-1} \\ |\textbf{x}|=1}\{\textbf{x} \ \text{repeated} \ p-1 \ \text{times}\}.\]Assume for contradiction that there is a solution of vectors $v_1, \ldots, v_N$ with $N>p-1$ that works, where we write\[ v_i = (e_{i1},\ldots,e_{i(p-1)}) \]for each $i$. Now define\[ f_k(x_1,\ldots,x_N) = \sum_{i=1}^N x_i^{p-1}e_{ki} \]for all $1 \le k \le p-1$. Since $(0, \ldots, 0)$ is a common root of all $f_k$, the Chevalley-Warning theorem guarantees a nontrivial root common to all $f_k$. By Fermat's Little Theorem, each $x_i^{p-1}$ is either $0$ or $1$, so there is a nonempty subset of $\{v_1, \ldots, v_N\}$ with a zero sum by using this nontrivial root, a contradiction. Hence $(p-1)^2$ is the size of the largest multiset of vectors of $\mathbb{F}_p^{p-1}$ where no nonempty subset has a sum of zero, and we are done.
This post has been edited 4 times. Last edited by Leo.Euler, Apr 11, 2023, 3:30 PM
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HamstPan38825
8846 posts
#7
Y by
Chevalley-Warning essentially nukes this.

The answer is $(p-1)^2$. For construction, pick primes $q_1, q_2, \cdots, q_{p-1}$, and use $q_1, q_1^{p+1}, \cdots, q_1^{1+p(p-2)}$ and similarly for the other $q_i$'s. This obviously satisfies both conditions.

For the proof of maximality, set the $q_i$ to be the same and choose $k = (p-1)^2 + 1$ distinct elements from $A$, which we will label $x_1$ through $ x_k$. For each $j$, let $e_{ij}$ be the exponent of $q_i$ in $x_j$.

Now, for the polynomials $$f_i(X_1, \cdots, X_k) = X_1^{p-1}e_{i1} + X_2^{p-1}e_{i2} + \cdots + X_k^{p-1} e_{ik},$$by Chevalley-Warning there exists a nontrivial solution to the system $$f_1(x_1, x_2, \cdots, x_k) \equiv f_2(x_1, x_2, \cdots, x_k) \equiv \cdots \equiv 0 \pmod p.$$Pick a subset $B$ of all nonzero $x_i$ in this set that satisfy the condition. Then by definition and Fermat's Little Theorem it follows that $\sum_{j \in B} e_{ij} \equiv 0 \pmod p$ for each $i$, thus the product of the elements in $B$ is a perfect $p$th power, contradiction.
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CANBANKAN
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#8
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Claim: If $x_j = (x_{j,1},\cdots, x_{j,n}) \in (\mathbb{Z}/p\mathbb{Z})^n$ ($j\in \{1,\cdots,K\}$) such that for all $e_1, \cdots, e_K \in \{0,1\}$ (not all zero), then $\sum e_jx_j \ne 0$. Then $K\le n(p-1)$ (this bound is optimal because I can have $p-1$ copies of $(\underbrace{0, \cdots, 0}_{j}, 1, \underbrace{0, \cdots, 0}_{n-1-j})$ for $j=0,\cdots,n-1$)

Proof: Assume $K\ge n(p-1)+1$. Consider the polynomial

$$P(f_1, \cdots, f_K) = \prod\limits_{j=1}^n (1-(\sum_{k=1}^K f_kx_{k,j})^{p-1})$$
If $\sum e_jx_j \ne 0 \iff$ there exists $t$ such that $\sum_k e_k x_{k,t} \ne 0$. Then $1-(\sum_k e_k x_{k,t})^{p-1}=0$ in $\mathbb{Z}/p\mathbb{Z}$. Hence $P(f_1,\cdots,f_K)=0$ for all $f_1,\cdots,f_k \in \{0,1\}$ that are not all zero, and $P(0,\cdots,0)=1$.

We first flatten the polynomial i.e. replace $f_k^j$ with $f_k$ when $j\ge 2$ (i.e. we take this polynomial and take its remainder mod $f_1^2-f_1, f_2^2-f_2, \cdots, f_K^2-f_K$). Call this polynomial $\tilde P$. Now note $$\tilde P + (1-f_1)(1-f_2)\cdots (1-f_K)$$vanishes in $\{0,1\}^K$, but one can show that $$v\colon [K] \to (\mathbb{Z}/p\mathbb{Z})^{2^K}, v_S = ( \prod\limits_{j\in S} f_j )_{(f_1,\cdots,f_k)\in \{0,1\}^K}$$are linearly independent in $\{0,1\}^K$ (or we induct by seeing this as a linear poly in $f_1$ namely $R(f_2,\cdots,f_K) f_1 + Q(f_2,\cdots,f_K)$ and since this poly is 0 when $f_1\in \{0,1\}$, $R(f_2,\cdots,f_K), Q(f_2,\cdots,f_K)$ satisfy the problem conditions for $K-1$) so $$\tilde P + (1-f_1)(1-f_2)\cdots (1-f_K) \equiv 0$$
This is absurd since $\deg \tilde P < K$
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YaoAOPS
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#9
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Yet another C-W nuke solution

Claim: $|A| = (p-1)^2$ is achieveable.
Proof. Let $p_n$ denote the $n$th prime for $1 \le n \le p - 1$. Then $p-1$ occurences of each $p_n$ suffices. $\blacksquare$

Claim: $(p-1)^2$ is maximal.
Proof. FTSOC assume there exists some $A$ of size $n = p^2 - 2p$. Enumerate $A$ as $a_{1}, a_{2}, \dots, a_n$ and the prime divisors as $p_1, p_2, \dots, p_n$. For each $i$, let $a_i = p_1^{c_{i1}}p_2^{c_{i2}}\dots p_n^{c_{in}}$.
Construct the $p-1$ polynomials $f_1, f_2, \dots f_{p-1}$ in ${\mathbb F}_p$ such that \[ f_i(x_1, x_2, \dots, x_n) = \sum_{j=1}^{n} c_{ij}x_j^{p-1}. \]Note that $\deg f_i = p-1$ and that $f_i$ is $0$ iff the elements of $A$ at its nonzero inputs have a product with $\nu_{p_i}$ divisible by $p$.
Since $n > \deg f_i \cdot (p-1) = (p-1)^2$, and $(0, 0, \dots, 0)$ is one of the common roots of every polynomial, by Chevalley-Warnining there exists another root, contradiction. $\blacksquare$
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Sourorange
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Let $D = p-1$. Then the question (thinking in terms of the exponents) can be phrased as follows:

What's the largest multiset of vectors in ${\mathbb F}_p^{D}$ such that no nonempty subset has zero sum?

We claim the answer is $D \cdot (p-1)$. A construction is given by taking
  • $p-1$ copies of the vector $\left< 1, 0, \dots, 0\right>$;
  • $p-1$ copies of the vector $\left< 0, 1, \dots, 0\right>$;
  • $......$;
  • $p-1$ copies of the vector $\left< 0, 0, \dots, 1 \right>$.
To show it's best possible, suppose we have vectors $v_1$, \dots, $v_N$ with coordinates given as \[ v_k = \left< v_{k,1}, v_{k,2}, \dots, v_{k,D} \right> 	\qquad k = 1, 2, \dots, N.  \]Then, we construct the polynomial \[ 	F(X_1, \dots, X_N) 	= \prod_{i=1}^D 	\left[ 1 - \left( \sum_{k=1}^N X_k v_{k,j} \right)^{p-1} \right] 	- \prod_{i=1}^N (1-X_i) \]If we imagine the $X_i \in \{0,1\}$ as Bernoulli random variables indicating whether $v_k$ is used in a set or not, then $F(X_1, \dots, X_N) \neq 0$ exactly when the $X_i$'s equal to $1$ correspond to a nonempty subset of the vectors which have vanishing sum.
Now assume for contradiction $N < D \cdot (p-1)$. Then the largest degree term is given in the latter product; it is exactly $(-1)^{N+1} X_1 X_2 \dots X_N$. So if we quote Alon's combinatorial nullstellensatz, it now follows that there is a choice of $X_i \in \{0,1\}$ such that $F(X_1, \dots, X_N) \neq 0$ which is a contradiction. $\square$
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IAmTheHazard
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#13
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The answer is $(p-1)^2$

Interpret each number as a $\mathbb{F}_p^{p-1}$ vector with entries corresponding to their $\nu_q$ modulo $p$: we are asked to find the largest multiset of vectors such that no nonempty subset of them sum to the $0$ vector. For a construction, we may take $p-1$ copies of each of the $p-1$ vectors with one entry equal to $1$ and the rest $0$.

I will now prove that $k:=|A|>(p-1)^2$ is impossible. To that end, suppose we had $k$ vectors $v_1,\ldots,v_k$ and the $j$-th element of $v_i$ was $a_{ij}$. Consider the following polynomial in $\mathbb{F}_p[x_1,\ldots,x_k]$:
$$P(x):=\prod_{i=1}^{k} (1-x_i)-\prod_{j=1}^{p-1} \left(1-\left(\sum_{i=1}^k a_{ij}x_i\right)^{p-1}\right).$$If we restrict $x_i \in \{0,1\}$ for all $i$, then choosing some subset of the vectors and summing them is equivalent to setting some subset of the $x_i$ to $1$ (and the rest $0$). If we choose the empty set, i.e. $x_i=0$ always, then the first and second products are both $1$. Otherwise, the first product clearly vanishes, and by hypothesis, since some entry of their sum is nonzero, Fermat's little theorem guarantees that the second product vanishes as well. Thus $P$ vanishes for any choice of $(x_1,\ldots,x_k) \in \{0,1\}^k$, but since $k>(p-1)^2$ it contains the maximal-degree term $(-1)^kx_1\ldots x_k$, which yields a contradiction by combinatorial nullstellensatz. $\blacksquare$
This post has been edited 1 time. Last edited by IAmTheHazard, Jan 8, 2024, 2:08 AM
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