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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
AD=BE implies ABC right
v_Enhance   118
N 14 minutes ago by Siddharthmaybe
Source: European Girl's MO 2013, Problem 1
The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$. The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$. Prove that, if $AD=BE$, then the triangle $ABC$ is right-angled.
118 replies
v_Enhance
Apr 10, 2013
Siddharthmaybe
14 minutes ago
Cyclic Quads and Parallel Lines
gracemoon124   17
N 15 minutes ago by Siddharthmaybe
Source: 2015 British Mathematical Olympiad?
Let $ABCD$ be a cyclic quadrilateral. Let $F$ be the midpoint of the arc $AB$ of its circumcircle which does not contain $C$ or $D$. Let the lines $DF$ and $AC$ meet at $P$ and the lines $CF$ and $BD$ meet at $Q$. Prove that the lines $PQ$ and $AB$ are parallel.
17 replies
gracemoon124
Aug 16, 2023
Siddharthmaybe
15 minutes ago
Problem 1 (First Day)
Valentin Vornicu   137
N 17 minutes ago by Siddharthmaybe
1. Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.
137 replies
Valentin Vornicu
Jul 12, 2004
Siddharthmaybe
17 minutes ago
Concentric Circles
MithsApprentice   62
N 18 minutes ago by Siddharthmaybe
Source: USAMO 1998
Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$.
62 replies
MithsApprentice
Oct 9, 2005
Siddharthmaybe
18 minutes ago
four points lie on a circle
pohoatza   77
N 23 minutes ago by Siddharthmaybe
Source: IMO Shortlist 2006, Geometry 2, AIMO 2007, TST 1, P2
Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic.

Proposed by Vyacheslev Yasinskiy, Ukraine
77 replies
pohoatza
Jun 28, 2007
Siddharthmaybe
23 minutes ago
Hard combi
EeEApO   6
N 30 minutes ago by aidan0626
In a quiz competition, there are a total of $100 $questions, each with $4$ answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?

Now, suppose my spouse and I move into a new home. Every year, we have twins. Starting at the age of $16$, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
6 replies
EeEApO
May 8, 2025
aidan0626
30 minutes ago
The familiar right angle from the orthocenter
buratinogigle   0
44 minutes ago
Source: Own, HSGSO P6
Let $ABC$ be a triangle inscribed in a circle $\omega$ with orthocenter $H$ and altitude $BE$. Let $M$ be the midpoint of $AH$. Line $BM$ meets $\omega$ again at $P$. Line $PE$ meets $\omega$ again at $Q$. Let $K$ be the orthogonal projection of $E$ on the line $BC$. Line $QK$ meets $\omega$ again at $G$. Prove that $GA\perp GH$.
0 replies
buratinogigle
44 minutes ago
0 replies
This question just asks if you can factorise 12 factorial or not
Sadigly   4
N an hour ago by NicoN9
Source: Azerbaijan Junior MO 2025 P1
A teacher creates a fraction using numbers from $1$ to $12$ (including $12$). He writes some of the numbers on the numerator, and writes $\times$ (multiplication) between each number. Then he writes the rest of the numbers in the denominator and also writes $\times$ between each number. There is at least one number both in numerator and denominator. The teacher ensures that the fraction is equal to the smallest possible integer possible.

What is this positive integer, which is also the value of the fraction?
4 replies
Sadigly
Friday at 7:34 AM
NicoN9
an hour ago
Japanese Olympiad
parkjungmin   2
N an hour ago by parkjungmin
It's about the Japanese Olympiad

I can't solve it no matter how much I think about it.

If there are people who are good at math

Please help me.
2 replies
parkjungmin
Yesterday at 6:51 PM
parkjungmin
an hour ago
Japanese high school Olympiad.
parkjungmin   0
an hour ago
It's about the Japanese high school Olympiad.

If there are any students who are good at math, try solving it.
0 replies
parkjungmin
an hour ago
0 replies
Divisibility NT
reni_wee   0
an hour ago
Source: Iran 1998
Suppose that $a$ and $b$ are natural numbers such that
$$p = \frac{b}{4}\sqrt{\frac{2a-b}{2a+b}}$$is a prime number. Find all possible values of $a$,$b$,$p$.
0 replies
reni_wee
an hour ago
0 replies
Six variables
Nguyenhuyen_AG   0
an hour ago
Let $a,\,b,\,c,\,x,\,y,\,z$ be six positive real numbers. Prove that
$$\frac{a}{b+c} \cdot \frac{y+z}{x} + \frac{b}{c+a} \cdot \frac{z+x}{y} + \frac{c}{a+b} \cdot \frac{x+y}{z} \geqslant 2+\sqrt{\frac{8abc}{(a+b)(b+c)(c+a)}}.$$
0 replies
Nguyenhuyen_AG
an hour ago
0 replies
Integration Bee Kaizo
Calcul8er   60
N an hour ago by Svyatoslav
Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!
60 replies
Calcul8er
Mar 2, 2025
Svyatoslav
an hour ago
Marginal Profit
NC4723   2
N 2 hours ago by Juno_34
Please help me solve this
2 replies
NC4723
Dec 11, 2015
Juno_34
2 hours ago
Putnam 2019 A5
hoeij   17
N Apr 19, 2025 by Ilikeminecraft
Let $p$ be an odd prime number, and let $\mathbb{F}_p$ denote the field of integers modulo $p$. Let $\mathbb{F}_p[x]$ be the ring of polynomials over $\mathbb{F}_p$, and let $q(x) \in \mathbb{F}_p[x]$ be given by $q(x) = \sum_{k=1}^{p-1} a_k x^k$ where $a_k = k^{(p-1)/2}$ mod $p$. Find the greatest nonnegative integer $n$ such that $(x-1)^n$ divides $q(x)$ in $\mathbb{F}_p[x]$.
17 replies
hoeij
Dec 10, 2019
Ilikeminecraft
Apr 19, 2025
Putnam 2019 A5
G H J
G H BBookmark kLocked kLocked NReply
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hoeij
35 posts
#1 • 1 Y
Y by Adventure10
Let $p$ be an odd prime number, and let $\mathbb{F}_p$ denote the field of integers modulo $p$. Let $\mathbb{F}_p[x]$ be the ring of polynomials over $\mathbb{F}_p$, and let $q(x) \in \mathbb{F}_p[x]$ be given by $q(x) = \sum_{k=1}^{p-1} a_k x^k$ where $a_k = k^{(p-1)/2}$ mod $p$. Find the greatest nonnegative integer $n$ such that $(x-1)^n$ divides $q(x)$ in $\mathbb{F}_p[x]$.
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hoeij
35 posts
#3 • 3 Y
Y by OmicronGamma, Guardiola, Adventure10
Answer: For $f(x) \in \mathbb{F}_p[x] - \{0\}$ let $m(f)$ denote its multiplicity at $x=1$ so $f(x) = (x-1)^{m(f)} g(x)$ for some $g(x) \in \mathbb{F}_p[x]$ with $g(1) \neq 0$. If $m(f) \not\equiv 0$ mod $p$, then $m(f') = m(f)-1$ by the product rule. We are asked to compute $m(q)$.

Let $f_n := \sum_{k=0}^{p-1} k^n x^k \in \mathbb{F}_p[x]$. Then $f_0 = x^{p-1}+\cdots+x^0 = (x^p-1)/(x-1) = (x-1)^{p-1}$ using the fact that $x^p-1 = (x-1)^p$ in $\mathbb{F}_p[x]$. Since $f_{n+1} = x f_n'$ it follows that $m(f_n) = m(f_0) - n$ for $n = 1,2,\ldots, m(f_0)$. In particular $m(q) = m(f_{(p-1)/2}) = m(f_0) - (p-1)/2 = (p-1)/2$.
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Anzoteh
126 posts
#4 • 3 Y
Y by Guardiola, Adventure10, Mango247
Kiran's solution uses the transformation $D^m(f)=xf'$, which is way neater. Here's a solution that repeats the above, i.e. $D^m(f)=f'$ and as usual we find the smallest $n$ with $D^n(f)(1)\neq 0$. Finding derivative term wise, get
$$
D^m(q)(1) = \sum_{k=1}^{p-1} k^{(p-1)/2}k(k-1)\cdots (k-m+1)
$$(some explanation needed as $k-m$ could go to 0. We can now write the above in terms of the following polynomial:
$$
s_m(x)=x^{(p-1)/2}x(x-1)\cdots (x-m+1)
=\sum_{k=1}^{m}b_kx^{(p-1)/2+k}
$$and
$$
D^m(q)(1) = \sum_{j=1}^{p-1}s_m(j)
=\sum_{k=1}^m b_k\sum_{x=1}^{p-1}x^{(p-1)/2+k}
$$Now for $k<\frac{p-1}{2}$ each term $\sum_{x=1}^{p-1}x^{(p-1)/2+k}=0$ (primitive root argument), which means the sum will be $0$ for $0\le m<\frac{p-1}{2}$. For $m=\frac{p-1}{2}$, only $k=m=\frac{p-1}{2}$ is left and
$$D^m(q)(1)=\sum_{k=1}^m b_k(-1)$$but $b_k$ is the coefficient of the monic polynomial $s_m(x)$, hence equal 1, so $D^m(q)(1)=-1\neq 0$ for $m=\frac{p-1}{2}$.
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hoeij
35 posts
#5 • 2 Y
Y by Adventure10, Mango247
Note that the exponent $(p-1)/2$ in the problem is irrelevant. If you replace it by another positive integer $n<p$ then the answer is simply $p-1-n$.
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TheStrangeCharm
290 posts
#6 • 2 Y
Y by Adventure10, Mango247
We first claim that for each $0\leq \ell < \frac{p - 1}{2}$, we have that
\[
(*) \sum_{k\in \mathbb{F}_p^{\times}}k^{\ell}a_k = 0.
\]Indeed, the above sum can be written as
\[
\sum_{k\in \mathbb{F}_p^{\times 2}}k^{\ell} - \sum_{k\in \mathbb{F}_p^{\times} - \mathbb{F}_p^{\times 2}}k^{\ell}.
\]But we have that
\[
\sum_{k\in \mathbb{F}_p^{\times 2}}k^{\ell} + \sum_{k\in \mathbb{F}_p^{\times} - \mathbb{F}_p^{\times 2}}k^{\ell} = \sum_{k\in \mathbb{F}_p^{\times}}k^{\ell} = 0.
\]To see this, take some $\alpha\in \mathbb{F}_p^{\times}$ with $\alpha^{\ell} \neq 1$, which is possible as $\ell < p - 1$, to see that
\[
\sum_{k\in \mathbb{F}_p^{\times}}k^{\ell} = \sum_{k\in \mathbb{F}_p^{\times}}(\alpha k)^{\ell} \implies (1 - \alpha^{\ell})\sum_{k\in \mathbb{F}_p^{\times}}k^{\ell} = 0 \implies \sum_{k\in \mathbb{F}_p^{\times}}k^{\ell} = 0.
\]Thus, we have that
\[
\sum_{k\in \mathbb{F}_p^{\times 2}}k^{\ell} - \sum_{k\in \mathbb{F}_p^{\times} - \mathbb{F}_p^{\times 2}}k^{\ell} = 2\sum_{k\in \mathbb{F}_p^{\times 2}}k^{\ell} = \sum_{m\in \mathbb{F}_p^{\times}}m^{2\ell} = 0,
\]as $2\ell < p - 1$, so we can use the trick of taking some $\alpha\in \mathbb{F}_p^{\times}$ with $\alpha^{2\ell}\neq 1$ as above, thus establishing $(*)$.

Now, return to our polynomial
\[
\sum_{k = 1}^{p - 1}a_kx^k. 
\]Note that this polynomial always vanishes at $x = 1$ by taking $(*)$ with $\ell = 0$. Suppose we take $\ell > 0$ derivatives of this expression and evaluate the result at $1$. The result will be
\[
\sum_{k = 1}^{p - 1}k(k - 1)\cdots k(k - \ell + 1)a_k. 
\]But we can expand out $k(k - 1)\cdots k(k - \ell + 1)$ as some polynomial in $k$ with integer coefficients and degree $\ell$. Thus, as long as $\ell < \frac{p - 1}{2}$, this expression will vanish by repeatedly applying (*). Thus, $(x - 1)^{\ell}$ divides our polynomial for $\ell < \frac{p - 1}{2}$. To see that $(x - 1)^{\frac{p - 1}{2}}$ does not divide our polynomial, it suffices to show that
\[
\sum_{k\in \mathbb{F}_p^{\times}}k^{\frac{p - 1}{2}}a_k \neq 0,
\]by expanding out $k(k - 1)\cdots k(k - \frac{p - 1}{2} + 1)$ as a polynomial in $k$ and applying $(*)$ to all $\ell < \frac{p - 1}{2}$. But this sum is just
\[
\sum_{k\in \mathbb{F}_p^{\times}}k^{p - 1} = p - 1\neq 0. \text{     }\square
\]
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DVDthe1st
341 posts
#7 • 2 Y
Y by JohnDoeSmith, Adventure10
Let $\chi(k) = k^{(p-1)/2} = \left(\frac{k}{p}\right)$. Then $q(x)^2 \equiv \sum_{k=0}^{p-1} (\chi\ast\chi)(k)x^k \pmod{x^p - 1}$. But it's fairly well known that $\chi \ast \chi$ is a non-zero constant.

(Note: $(\chi\ast\chi)(k) = \sum_{i\in \mathbb F_p} \chi(i)\chi(k-i) $.)
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hoeij
35 posts
#8 • 1 Y
Y by Adventure10
Short proof for A5: Let $f = 1+x+\cdots+x^{p-1} = (x-1)^{p-1}$ and denote $D(f) = xf'$. Then $q(x) = D^{(q-1)/2}(f)$. For $n=0,1,\ldots,p-1$, the multiplicity of $x-1$ in $D^n(f)$ is $p-1-n$
(because any time the multiplicity is not 0 mod p, it drops by 1 when you differentiate).
This post has been edited 1 time. Last edited by hoeij, Dec 10, 2019, 3:57 AM
Reason: added "because any time..."
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trumpeter
3332 posts
#9 • 1 Y
Y by Adventure10
The answer is $\boxed{\frac{p-1}{2}}$.

The only number theoretic fact necessary is that the $m$th moment of $\mathbb{F}_p$ for $m\not\equiv0\pmod{p-1}$ is $0$. There are a few different reasons this is true: (number theory) plug in a primitive root and use geometric series, (group theory) the sum is a multiple of the sum of a subgroup with zero sum, (algebra) use Newton sums on the polynomial $x^{p-1}-1$.

An immediate corollary is that for any polynomial that has $0$ as a root has degree $\leq p-2$, the sum of the polynomial over $\mathbb{F}_p$ is $0$.

Take the $j$th derivative; we get
\[
	q^{(j)}(1)=\sum_{a\in\mathbb{F}_p}a^{\frac{p-1}{2}}a^{\underline{j}}
\]where $x^{\underline{j}}=x(x-1)\cdots(x-j+1)$ is the falling factorial. When $j=0,\ldots,\frac{p-3}{2}$, this is $0$ by the corollary. When $j=\frac{p-1}{2}$, the corollary implies that
\[
	q^{(j)}(1)=\sum_{a\in\mathbb{F}_p}a^{p-1}=-1.
\]So the first $\frac{p-1}{2}-1$ derivatives of $q$ at $1$ (a root of $q$) are $0$, so $1$ has multiplicity $\frac{p-1}{2}$ in $q$.
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yayups
1614 posts
#10 • 4 Y
Y by pad, Vfire, Adventure10, Mango247
All statements in this solution will be over $\mathbb{F}_p[x]$. Let $q^{(m)}(x)$ denote the $m$th derivative of $q(x)$. We have the following well known lemma.

Lemma 1: Suppose we have an integer $1\le n\le p$. Then $(x-1)^n\mid q(x)$ if and only if $q^{(m)}(1)=0$ for all $m=0,1,\ldots,n-1$.

Proof: Use the division algorithm for polynomials repeatedly to write $q(x)$ in the form \[q(x)=b_0+b_1(x-1)+b_2(x-1)^2+\cdots+b_{p-1}(x-1)^{p-1}.\]In this setting, it's clear that $(x-1)^n\mid q(x)$ if and only if $b_m=0$ for all $m=0,1,\ldots,n-1$. Differentiating $m$ times, we see that $m!b_m=q^{(m)}(1)$, and since $m\le n-1\le p-1$, we thus have $b_m=0\iff q^{(m)}(1)=0$, so the result follows. $\blacksquare$

In our setting, we have \[q^{(m)}(1)=\sum_{k=m}^{p-1}k^{\frac{p-1}{2}}k(k-1)\cdots(k-m+1)=\sum_{k=0}^{p-1}k^{\frac{p-1}{2}}k(k-1)\cdots(k-m+1).\]To evaluate this, we have the following useful lemma.

Lemma 2: For any integer $r=1,2,\ldots,p-2$, we have \[\sum_{k=0}^{p-1}k^r=0.\]Recall that we are working over $\mathbb{F}_p$.

Proof: Let $g$ be a primitive root mod $p$. Then, the sum is \[\sum_{\ell=0}^{p-2}(g^\ell)^r=\sum_{\ell=0}^{p-2}=\frac{(g^r)^{p-1}-1}{g^r-1}=0,\]which works since $g^r-1$, due to $1\le r\le p-2$. $\blacksquare$

Lemma 2 clearly implies that $q^{(m)}(1)=0$ for $m=0,\ldots,\frac{p-1}{2}-1$. It also implies that \[q^{\left(\frac{p-1}{2}\right)}(1)=\sum_{k=0}^{p-1} k^{p-1}=p-1\ne 0,\]so the maximal $n$ such that $(x-1)^n\mid q(x)$ is $n=\boxed{\frac{p-1}{2}}$ by lemma 1.
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hoeij
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#11 • 2 Y
Y by Adventure10, Mango247
Trumpeter: The only number theoretic fact needed for this exercise is the Freshman's dream: $(x-1)^p = x^p - 1$. The rest is calculus: the formula for a geometric sum, the fact that $x d/dx$ applied to $x^k$ gives $k x^k$, and the fact that if a multiplicity of a root is not zero in our field then it decreases by 1 when you differentiate. Nothing else is needed for the proof.

Complete proof for A5: The geometric sum $f := 1+x+\cdots+x^{p-1}$ can be written as $f = (x^p-1)/(x-1) = (x-1)^{p-1}$. Starting with $f$ and its multiplicity $p-1$, keep applying $x d/dx$ until you reach $q(x)$ and its multiplicity.
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donot
1180 posts
#12 • 2 Y
Y by Adventure10, Mango247
It seems I'm late, but I had a more number-theoretic solution
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Alex2
168 posts
#13 • 1 Y
Y by Adventure10
Sol
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v_Enhance
6877 posts
#14 • 3 Y
Y by XbenX, Mathematicsislovely, v4913
Solution from Twitch Solves ISL:

The answer is $n = \frac{1}{2}(p-1)$ as claimed.

We use derivatives in the following way.
Claim: Define $q_0 = q$, and $q_{i+1} = x \cdot q_i'$. Suppose $n$ is such that $q_1$, \dots, $q_{n-1}$ has $x=1$ as a root, but $q_n$ does not have $x=1$ as a root. Then $n$ is the multiplicity of $x=1$ in $q$.
Proof. This follows from the fact that $q_{i+1}$ will have multiplicity of $x=1$ one less than in $q_i$. $\blacksquare$
On the other hand, we may explicitly compute \[ q_n(1) = \sum_{k=1}^{p-1} k^n \left( \frac kp \right) = 		\sum_{k \text{ qr}} k^n - 		\underbrace{2 \sum_{k=1}^{p-1} k^n}_{=0 \text{ for } n < p-1} \]Let $g$ be a primitive root modulo $p$. The first sum then equals \[ g^0 + g^{2n} + g^{4n} + \dots + g^{(p-3) n} \]which equals $\frac{p-1}{2}$ if $n = (p-1)/2$ but $\frac{g^{(p-1)n}-1}{g^{2n}-1} = 0$ otherwise.
Consequently, the answer is $n = \frac{1}{2}(p-1)$ as claimed.
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amuthup
779 posts
#15 • 1 Y
Y by SK_pi3145
I believe this is a new solution, can someone confirm that it works?

The answer is $\boxed{\frac{p-1}{2}}.$

Work in $\mathbb{F}_p,$ and let $f(n)=n^{(p-1)/2}.$ Additionally, define $S_k(n)$ inductively by $S_0(n)=1$ and $S_{k}(n)=\sum_{i=1}^{n}S_{k-1}(i).$

By synthetic division, it suffices to show the following. $$\sum_{i=1}^{p-1}S_{0}(i)f(i)=\sum_{i=1}^{p-1}S_{1}(i-1)f(i)=\sum_{i=1}^{p-1}S_{2}(i-2)f(i)=\dots=\sum_{i=1}^{p-1}S_{(p-3)/2}\left(i-\frac{p-1}{2}\right)f(i)=0,$$$$\sum_{i-1}^{p-1}S_{(p-1)/2}\left(i-\frac{p-1}{2}\right)f(i)\ne 0.$$To show that the sums on top evaluate to $0,$ induct from left to right; the base case follows from the fact that the number of nonzero QRs and QNRs are equal.

Now consider the $k$th sum from the left. By the inductive hypothesis, we may shift to obtain $\sum_{i=1}^{p-1}S_{k-1}(i)f(i).$ Therefore, since $S_{k-1}$ is a polynomial of degree $k-1,$ it suffices to show that $\sum_{i=1}^{p-1}i^{k-1}f(i)=0.$

This follows from Newton's sums on the polynomials $x^{(p-1)/2}\pm 1$ precisely when $k\ne\frac{p-1}{2},$ so we are done.
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HamstPan38825
8863 posts
#16
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Kind of a boring problem; you just keep taking derivatives.

The answer is $\frac{p-1}2$. Notice that including all terms with negative powers that evaluate to zero, the $n$th derivative of $q(x)$ is equal to $$q^(n')(x) =\sum_{k=1}^{p-1} k^{\frac{p-1}2} \cdot k(k-1)(k-2)\cdots(k-n+1) x^{k-n}.$$In particular, for $n < \frac{p-1}2$, we can always split this into sums of the form $c\sum_{k=1}^{p-1} k^i$ for some $c \in \mathbb Z$ and $0 \leq i \leq p-2$. All these sums evaluate to zero modulo $p$, hence $q^(n')(1) = 0$.

On the other hand, when $n = \frac{p-1}2$, the leading $k$ term has exponent $p-1$, and the sum $c\sum_{k=1}^{p-1} k^{p-1}$ evaluates to $-1$, and hence the process terminates. It follows by the definition of derivative that the multiplicity of $1$ in $q$ is $\frac{p-1}2$.
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john0512
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#17
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The answer is $n=\frac{p-1}{2}$.

Using the Legendre symbol, define $$P(x)=\sum_{k=1}^{p-1} \left ( \frac{k}{p}  \right )x^k.$$
The problem is asking to find the smallest positive integer $n$ such that
$$P^{(n)}(1)\not\equiv 0\pmod{p}.$$
Using the definition of $P(x)$, we can explictly write $P^{(n)}(x)$ as
$$P^{(n)}(1)=\sum_{QR}r(r-1)\dots(r-n+1)-\sum_{NQR}r(r-1)\dots(r-n+1).$$
However, since
$$\sum_{NQR}r(r-1)\dots(r-n+1)=\sum_r r(r-1)\dots(r-n+1)-\sum_{QR}r(r-1)\dots(r-n+1),$$we have
$$P^{(n)}(1)=2\sum_{QR}r(r-1)\dots(r-n+1)-\sum_r r(r-1)\dots(r-n+1)$$$$=\sum_{r}(r^2)(r^2-1)\dots (r^2-n+1)-\sum_r r(r-1)\dots(r-n+1).$$
Now, we will use the following lemma (well-known):
Quote:
The sum of $k^m$ over $k=1,2,\dots,p-1$ vanishes mod $p$ unless $p-1\mid m$.

When $n<\frac{p-1}{2}$, everything is degree less than $p-1$, and the constant terms are the same, so everything vanishes. However, when $n=\frac{p-1}{2}$, the first sum has a non-vanishing component $r^{p-1}$, so it will not be zero as everything else vanishes. Thus, $n=\frac{p-1}{2}$ fails and we are done.
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OronSH
1745 posts
#18 • 1 Y
Y by centslordm
Answer $\tfrac{p-1}2$.

Fix a primitive root $g$ so that $a_{g^i}=(-1)^i$. Then notice we can write $\sum_{k=1}^{p-1}a_kk^n$ as a sum of $q(1),q'(1),\dots,q^{(n)}(1)$. In particular to preserve degrees the last term is added once. Thus the first time $q^{(n)}(1)\ne 0$ is the same as the first time $\sum_{k=1}^{p-1}a_kk^n\ne 0$. However, the idea is that we can rewrite this as $\sum_{i=1}^{p-1}(-g^n)^i$ which equals $g^n\frac{(-g^n)^{p-1}-1}{g^n+1}$ so if this is $\ne 0$ then we must have $g^n=-1$ so $n=\frac{p-1}2$. In fact for this $n$ we get $\sum_{k=1}^{p-1}a_kk^n=\sum_{k=1}^{p-1}a_k^2=p-1\ne 0$ as desired.
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Ilikeminecraft
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#19
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Differentiate and multiply by $x$ $\frac{p - 1}{2} - 1$ times. We are left with \[P(x) = \sum_{k = 1}^{p - 1}\frac1{k}x^k\]. Plugging in $x = 1,$ we get that the sum is diviisble by $x - 1.$ Differentiate one more time and we get \[\sum_{k = 1}^{p - 1}x^k = \frac{x^k - 1}{x - 1}\]which obviously has 0 powers of $x - 1.$ This implies that there are exactly $\frac{p - 1}{2}$ factors of $x - 1$ in the original expression.
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