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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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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
collinearity as a result of perpendicularity and equality
parmenides51   2
N 11 minutes ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1996 OMM P6
In a triangle $ABC$ with $AB < BC < AC$, points $A' ,B' ,C'$ are such that $AA' \perp BC$ and $AA' = BC, BB' \perp  CA$ and $BB'=CA$, and $CC' \perp AB$ and $CC'= AB$, as shown on the picture. Suppose that $\angle AC'B$ is a right angle. Prove that the points $A',B' ,C' $ are collinear.
2 replies
parmenides51
Jul 28, 2018
FrancoGiosefAG
11 minutes ago
∑(a-b)(a-c)/(2a^2 + (b+c)^2) >= 0
Zhero   23
N 12 minutes ago by Rayvhs
Source: ELMO Shortlist 2010, A2
Let $a,b,c$ be positive reals. Prove that
\[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq 0. \]

Calvin Deng.
23 replies
Zhero
Jul 5, 2012
Rayvhs
12 minutes ago
An inequality
Rushil   13
N 19 minutes ago by JARP091
Source: Indian RMO 1994 Problem 8
If $a,b,c$ are positive real numbers such that $a+b+c = 1$, prove that \[ (1+a)(1+b)(1+c) \geq 8 (1-a)(1-b)(1-c) . \]
13 replies
Rushil
Oct 25, 2005
JARP091
19 minutes ago
3 var inequality
JARP091   6
N 27 minutes ago by JARP091
Source: Own
Let \( x, y, z \in \mathbb{R}^+ \). Prove that
\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]without using the Rearrangement Inequality or Chebyshev's Inequality.
6 replies
JARP091
Today at 8:54 AM
JARP091
27 minutes ago
Helplooo
Bet667   1
N 34 minutes ago by Lil_flip38
Let $ABC$ be an acute angled triangle.And altitudes $AD$ and $BE$ intersects at point $H$.Let $F$ be a point on ray $AD$ such that $DH=DF$.Circumcircle of $AEF$ intersects line $BC$ at $K$ and $L$ so prove that $BK=BL$
1 reply
Bet667
an hour ago
Lil_flip38
34 minutes ago
Cyclic sum of 1/(a+1/b+1)
v_Enhance   22
N 39 minutes ago by Rayvhs
Source: ELMO Shortlist 2013: Problem A2, by David Stoner
Prove that for all positive reals $a,b,c$,
\[\frac{1}{a+\frac{1}{b}+1}+\frac{1}{b+\frac{1}{c}+1}+\frac{1}{c+\frac{1}{a}+1}\ge \frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}. \]Proposed by David Stoner
22 replies
v_Enhance
Jul 23, 2013
Rayvhs
39 minutes ago
xf(x + xy) = xf(x) + f(x^2)f(y)
orl   14
N 39 minutes ago by jasperE3
Source: MEMO 2008, Team, Problem 5
Determine all functions $ f: \mathbb{R} \mapsto \mathbb{R}$ such that
\[ x f(x + xy) = x f(x) + f \left( x^2 \right) f(y) \quad  \forall  x,y \in \mathbb{R}.\]
14 replies
orl
Sep 10, 2008
jasperE3
39 minutes ago
i am not abel to prove or disprove
frost23   0
an hour ago
Source: made on my own
let $a_1a_1a_2a_2.............a_na_n$ be a perfect square then is it true that it must be of the form
$10^{2(n-1)}\cdot7744$
0 replies
frost23
an hour ago
0 replies
Beautiful Number Theory
tastymath75025   34
N an hour ago by Adywastaken
Source: 2022 ISL N8
Prove that $5^n-3^n$ is not divisible by $2^n+65$ for any positive integer $n$.
34 replies
tastymath75025
Jul 9, 2023
Adywastaken
an hour ago
Hard Functional Equation in the Complex Numbers
yaybanana   1
N an hour ago by jasperE3
Source: Own
Find all functions $f:\mathbb {C}\rightarrow \mathbb {C}$, s.t :

$f(xf(y)) + f(x^2+y) = f(x+y)x + f(f(y))$

for all $x,y \in \mathbb{C}$
1 reply
yaybanana
Apr 9, 2025
jasperE3
an hour ago
Find all numbers
Rushil   11
N an hour ago by frost23
Source: Indian RMO 1994 Problem 3
Find all 6-digit numbers $a_1a_2a_3a_4a_5a_6$ formed by using the digits $1,2,3,4,5,6$ once each such that the number $a_1a_2a_2\ldots a_k$ is divisible by $k$ for $1 \leq k \leq 6$.
11 replies
1 viewing
Rushil
Oct 25, 2005
frost23
an hour ago
Factorial Divisibility
Aryan-23   46
N an hour ago by anudeep
Source: IMO SL 2022 N2
Find all positive integers $n>2$ such that
$$ n! \mid \prod_{ p<q\le n, p,q \, \text{primes}} (p+q)$$
46 replies
Aryan-23
Jul 9, 2023
anudeep
an hour ago
IMO Shortlist 2011, G4
WakeUp   128
N 2 hours ago by Mathandski
Source: IMO Shortlist 2011, G4
Let $ABC$ be an acute triangle with circumcircle $\Omega$. Let $B_0$ be the midpoint of $AC$ and let $C_0$ be the midpoint of $AB$. Let $D$ be the foot of the altitude from $A$ and let $G$ be the centroid of the triangle $ABC$. Let $\omega$ be a circle through $B_0$ and $C_0$ that is tangent to the circle $\Omega$ at a point $X\not= A$. Prove that the points $D,G$ and $X$ are collinear.

Proposed by Ismail Isaev and Mikhail Isaev, Russia
128 replies
WakeUp
Jul 13, 2012
Mathandski
2 hours ago
IMO Shortlist 2011, G2
WakeUp   29
N 2 hours ago by AylyGayypow009
Source: IMO Shortlist 2011, G2
Let $A_1A_2A_3A_4$ be a non-cyclic quadrilateral. Let $O_1$ and $r_1$ be the circumcentre and the circumradius of the triangle $A_2A_3A_4$. Define $O_2,O_3,O_4$ and $r_2,r_3,r_4$ in a similar way. Prove that
\[\frac{1}{O_1A_1^2-r_1^2}+\frac{1}{O_2A_2^2-r_2^2}+\frac{1}{O_3A_3^2-r_3^2}+\frac{1}{O_4A_4^2-r_4^2}=0.\]

Proposed by Alexey Gladkich, Israel
29 replies
WakeUp
Jul 13, 2012
AylyGayypow009
2 hours ago
Sum of whose elements is divisible by p
nntrkien   45
N May 10, 2025 by Jupiterballs
Source: IMO 1995, Problem 6, Day 2, IMO Shortlist 1995, N6
Let $ p$ be an odd prime number. How many $ p$-element subsets $ A$ of $ \{1,2,\dots,2p\}$ are there, the sum of whose elements is divisible by $ p$?
45 replies
nntrkien
Aug 8, 2004
Jupiterballs
May 10, 2025
Sum of whose elements is divisible by p
G H J
Source: IMO 1995, Problem 6, Day 2, IMO Shortlist 1995, N6
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huashiliao2020
1292 posts
#33 • 1 Y
Y by cubres
Ha! The answer confirmation's so simple I knew there had to be a combinatorial way! So I'm not going to post some rouf solution.

By intuition we feel like our answer would be close to 1/p(2pCp) since it's around that expected value 1 in p subsets to be divisible by p; but since there are TWO elements same mod p, it kind of messes things up, while one distinct of each would help greatly. So we partition the sets into $\{1, 2, \dots, 2p\} = \{1, 2, \dots, p\} \cup \{p + 1, p + 2, \dots, 2p\} = M\cup N$; now, consider $M_k=\{m_1+k,m_2+k,...,m_m+k\}\forall k\ne p,m_1,m_2,...,m_m\in M$; do the similar thing for N. There are $p \left (\frac{1}{p} \binom{p}{k} \frac{1}{p} \binom{p}{p - k} \right) = \frac{1}{p} \binom{p}{k} \binom{p}{p - k}$ ways to choose some $N_i\equiv p-M_j\pmod p$; in particular, summing over all possible k simplifies to $\frac{1}{p} \left(\binom{2p}{p} - 2 \right)$; but since we also need to add set M and N, there are $2 + \frac{1}{p} \left(\binom{2p}{p} - 2 \right)$ many ways as our final answer. $\blacksquare$
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sixoneeight
1138 posts
#34 • 1 Y
Y by cubres
Goofy gen func rouf

Consider a generating function $$F(x,y) = (1+xy)(1+x^2y)(1+x^3y)\dots (1+x^{2p}y)$$This generating function describes the subsets, where the exponent of $x$ is the sum of the elements and the exponent of $y$ is the size. Therefore, we want the sum of the coefficients of $x^{kp}y^p$ for positive integers $k$.


We now apply a Roots of Unity Filter. The generating function containing only the terms $x^{kp}$ can be obtained by taking
\[
\frac{1}{p}\sum_{k=0}^{p-1}F(e^{\frac{ik\pi}{p}}x,y)
\]and we want the sum of the terms with exponent $y^p$ when evaluated at $x=1$. We write
\[
    (1+xy)(1+x^2y)\dots (1+x^{2p}y) =
    y^{2p}\left(-\frac1y-x\right)\left(-\frac1y-x^2\right)\left(-\frac1y-x^3\right)\dots \left(-\frac1y-x^{2p}\right)
\]Let $P_x(t)$ be a monic polynomial in $t$ with roots $x^i$ for $i=1,2,\dots 2p$. When $x\neq 1$ is a $p^\text{th}$ root of unity, $P_x(t) = (t^p-1)^2$ Then, we find that $$F(x,y) = y^{2p}P_x\left(-\frac1y\right)$$Plugging in $x=1$, we calculate that
\begin{align*}
[y^p]\frac{1}{p}\sum_{k=0}^{p-1}F(e^{\frac{ik\pi}{p}},y) &= [y^p]\frac{y^{2p}}{p}P_{e^\frac{ik\pi}{p}}\left(-\frac1y\right)\\
&= [y^p]\frac{y^{2p}}{p}\left((p-1)\left(\frac{-1}{y^p}-1\right)^2+\left(1+\frac{1}{y}\right)^{2p}\right)\\
&= [y^p]\frac{1}{p}\left((p-1)(y^p+1)^2 + (y+1)^{2p}\right)\\
&= \boxed{\frac{1}{p}\left(2(p-1)+\binom{2p}{p}\right)}
\end{align*}
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Spectator
657 posts
#35 • 4 Y
Y by OronSH, KevinYang2.71, KnowingAnt, cubres
Let $A(x,y)$ be the generating function
\[A(x,y) = (1+yx)(1+yx^2)\cdots(1+yx^{2p})\]We apply the roots of unity filter on $x$ to get
\[\frac{A(1,y)+A(w,y)+\cdots+A(w^{p-1},y)}{p} = \frac{(1+y)^{2p}+(p-1)(1+yw)\cdots(1+yw^{2p})}{p}\]We call this function on $y$, $B(y)$. Note that
\[(1+w)(1+w^2)\cdots(1+w^{p}) = 2\]Then, we apply the roots of unity filter on $y$ to get
\begin{align*}
    \frac{B(1)+B(w)+B(w^2)+\cdots B(w^{p-1})}{p} &= \frac{p+p\binom{2p}{p}+p+2^{2}(p-1)(p)}{p^2}
\end{align*}But, we need to subtract $2$ because it counts the empty set and the set with size $2p$. This gives us
\[\boxed{\frac{\binom{2p}{p}+2p-2}{p}}\]
This post has been edited 1 time. Last edited by Spectator, Nov 11, 2023, 3:50 PM
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GrantStar
821 posts
#36 • 1 Y
Y by cubres
Let $F(x,y)=\prod_{i=1}^{2p}(1+yx^i)$ be the gen func representing sums of subsets and their number of elements. Note that the answer is equal to \[\frac{1}{p^2}\left(\sum_{k=1}^p \sum_{k=1}^p F\left(e^{2\pi ij/p},e^{2\pi i k/p}\right)\right)-2\]by roots of unity filter, with the $-2$ coming from the empty set and $\{1,2,\dots,2p\}$ being included in this count. We thus compute this!!!
  • First, if $j=1,2,\dots,p-1$, then the sequence $j,2j,\dots, 2pj$ contains each residue modulo $p$ twice. Thus $j+k,2j+k,\dots, 2pj+k$ contains each residue twice. herefore, \[\sum_{k=1}^{p}F\left(e^{2\pi ij/p},e^{2\pi i k/p}\right)=p\prod_{k=1}^p \left(1+e^{2\pi i k/p}\right)^2=4p\]As \[\prod_{k=1}^p \left(1+e^{2\pi i k/p}\right)^2=\prod_{k=1}^p \left(-1-e^{2\pi i k/p}\right)^2=P(-1)^2=4\]in $P(x)=x^p-1$.
  • If $j=p$, then \[\sum_{k=1}^{p}F\left(1,e^{2\pi i k/p}\right)=p\sum_{k=1}^{p}F\left(1+e^{2\pi i k/p}\right)^{2p}\]By roots of unity filter on $(1+x)^{2p}$, we get that the above sum is $p\left(2+\binom{2p}{p}\right).$
Thus the total sum ignoring the division by $p^2$ and subtraction is \[p\left(2+\binom{2p}{p}\right)+4p(p-1)=p\binom{2p}{p}+4p^2-2p\]implying a final answer of \[\frac{p\binom{2p}{p}+4p^2-2p}{p^2}-2=\frac{\binom{2p}{p}-2}{p}+2.\]

Remark: N6 is crazy lol
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blueprimes
356 posts
#37 • 1 Y
Y by cubres
We will create a generating function $f(x, y)$, where the coefficient $c_{i, j}$ of $x^i y^j$ represents the number of subsets $A \subseteq \{1, 2, \dots, 2p \}$ where the sum of the elements of $A$ is $i$, while $|A| = j$. By considering how many numbers we extract from each individual residue class, it is not hard to find that
$$f(x, y) = \prod_{n = 0}^{p - 1} (1 + 2x^ny + x^{2n}y^2) = \prod_{n = 0}^{p - 1} (1 + x^ny)^2.$$We will use a double roots of unity filter to add all coefficients $c_{i, j}$ where $p \mid i, j$, then subtract $2$ to account for the cases when $j = 0, 2p$. Let $\omega = e^{2 \pi i / p}$. We want to evaluate $\frac{1}{p^2} \sum_{r = 0}^{p - 1} \sum_{s = 0}^{p - 1} f(\omega^r, \omega^s)$. When $r \ne 0$, $f(\omega^r, \omega^s) = \left[\prod_{n = 0}^{p - 1} (1 + \omega^n) \right]^2 = 2^2 = 4$. All cases belonging to the latter yield a total of $4p(p - 1)$. On the other hand, when $r = 0$, we have $\sum_{s = 0}^{p - 1} f(1, \omega^s) = \sum_{s = 0}^{p - 1} (1 + \omega^s)^{2p}$. Using the binomial theorem, only the terms when the exponent of $\omega$ is divisible by $p$ are left behind, and we obtain $p \left[\binom{2p}{p} + 2 \right]$. Our final answer is
$$\frac{4p(p - 1) + p \left[\binom{2p}{p} + 2 \right]}{p^2} - 2 = \frac{\binom{2p}{p} - 2}{p} + 2.$$
This post has been edited 1 time. Last edited by blueprimes, Apr 29, 2024, 1:24 AM
Reason: omega definition
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KnowingAnt
153 posts
#38 • 1 Y
Y by cubres
I think you only need one filter! We want to find the sum of the coefficients of $x^0y^p,x^py^p,x^{2p}y^p,\dots$ in
\[(1 + xy)(1 + x^2y)\dots(1 + x^{2p}y)\text{.}\]First fix $y$. Now let $\omega$ be a primitive $p$-th root of unity, we want the coefficient of $y^p$ in
\[\frac{P(1) + P(\omega) + \dots + P(\omega^{p - 1})}{p} = \frac{(1 + y)^{2p} + (p - 1)(1 + y^p)^2}{p}\]so the answer is
\[\frac1p\left(\binom{2p}{p} - 2\right) + 2\text{.}\]
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Mathandski
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#39 • 1 Y
Y by cubres
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Maximilian113
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#40 • 1 Y
Y by cubres
niceee :-D
Let $A(x, y) = (1+xy)(1+x^2y)(\cdots)(1+x^{2p}y).$ Clearly, we want the sum of the coefficients of the terms with $x$ of degree divisible by $p$ and $y$ having degree $p.$ Let $z=e^{2\pi i/p}.$ Then by Roots of Unity Filter we have $$\frac{1}{p} \sum^{p-1}_{k=0}A(z^k, y).$$However, for $k = 1, 2, \cdots, p-1,$ clearly the set $\{1, z, z^2, \cdots z^{p-1}\}$ is a permutation of $\{ z^k, z^{2k}, \cdots z^{pk} \},$ so $$A(z^k, y) = \left( \prod_{k=0}^{p-1} (1+z^ky) \right)^2 = \left( y^{2p} \prod_{k=0}^{p-1} (\frac{1}{y}+z^k) \right)^2 = y^{2p} \cdot \left( -\frac{1}{y^p}-1 \right)^2 = (y^p+1)^2.$$Hence, our sum equals $$\frac{1}{p} \left((p-1)(y^p+1)^2+(y+1)^{2p} \right).$$Now, we simply extract the coefficient of $y^p$ from here, and this is just $$\boxed{\frac{2p-2+\binom{2p}{p}}{p}}.$$
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golue3120
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#41 • 1 Y
Y by cubres
Here's the other genfunc solution.

Let $\textstyle\binom{n}{k}_q$ be the $q$-binomial coefficient and $\textstyle [n]_q=1+q+\dots+q^{n-1}=\frac{1-q^n}{1-q}$. Then it is well-known that
\[\sum_{\substack{S\subseteq\{1,2,\dots,2p\}\\|S|=p}}q^{\sum S}=q^{p(2p+1)}\binom{2p}{p}_q.\]
Let $\omega$ be a primitive $p$th root of unity. Then we have
\[\binom{2p}{p}_q=\frac{[2p]_q[2p-1]_q\dots[p+1]_q}{[p]_q[p-1]_q\dots[1]_q}=(1+q^p)\frac{[2p-1]_q\dots[p+1]_q}{[p-1]_q\dots[1]_q}.\]As we let $q\rightarrow\omega$, then the product cancels out, so we have $\textstyle\binom{2p}{p}_\omega=1+\omega^p=2$.

Hence by roots of unity filter, the desired result is $\textstyle\frac{1}{p}\sum_{i=0}^{n-1}\binom{n}{k}_{\omega^i}=\frac{\binom{2p}{p}-2}{p}+2$, since $\omega^0=1$ and all other powers are primitive $p$th roots of unity.
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smileapple
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#42 • 1 Y
Y by cubres
Define \[P(x,y)=\prod_{n=1}^{2p}(x+y^n).\]Note that if a subset $S\subseteq\{1,2,\dots,2p\}$ has $m$ elements that sum to $s$, the set $S$ will show up in the expansion of $P$ as $x^{2p-m}y^s$.

Now let $\zeta=e^{2\pi i/n}$. By roots of unity filter it suffices to find the coefficient $c_p$ of $x^p$ in the expansion of $Q(x)=\frac1p\sum_{n=0}^{p-1}P(x,\zeta^n)$. But note that $P(x,\zeta^n)$ is equal to $(x+1)^{2p}$ if $n=0$ and is equal to $(x^p+1)^2$ otherwise. Thus \[Q(x)=\frac{(x+1)^{2p}+(p-1)(x^p+1)^2}p,\]from which extracting $c_p$ yields \[c_p=\boxed{\frac{\binom{2p}p+2(p-1)}p},\]which is our answer. $\blacksquare$

Edit: 999th post?
This post has been edited 1 time. Last edited by smileapple, Feb 21, 2025, 1:35 AM
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eg4334
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#43 • 1 Y
Y by cubres
Lol what. Just take the generating function $(x+y)(x+y^2) \dots (x+y^{2p})$ where $x$ counts the number of elements we dont use and $y$ counts the exponent. We want $y$ to be a multiple of $p$, and $x$ to be $p$. To extract the first condition, take ROUF with a primitive $p$th root $\omega$. We get $\frac{(x+1)^{2p} + (p-1)((x+\omega)(x+\omega^2)\dots (x+\omega^p))^2}{p} = \frac{(x+1)^{2p} + (p-1)(x^p+1)^2}{p}$. Its obvious by binomial expansion that the answer from here is $\boxed{\frac{\binom{2p}{p} + 2(p-1)}{p}}$
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cubres
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#44
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lpieleanu
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#45 • 1 Y
Y by cubres
Solution
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Ilikeminecraft
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#46 • 1 Y
Y by cubres
We do ROUF on $(1 + xy)(1 + x^2y)\cdots(1 + x^{2p}y)$ and then subtract 2.

If $x = 1,$ it follows that we are just looking for the exponent of $y^{pk}$ and so it is $\binom{2p}p + 2$.

Now consider when $x = \omega^i$ for some $i > 0$ where $\omega$ is some primitive $p$-th root of unity.
\begin{align*}
    \sum_{i = 1}^{p - 1}\sum_{j = 0}^{p - 1} \prod_{k = 1}^{p} (1 + \omega^{k \cdot i + j})^2 & = \sum_{j = 0}^{p - 1}\sum_{i = 1}^{p - 1} \prod_{k = 0}^{p - 1}(w + \omega^{ki + j})^2 \\
    & = 4 (p-1)\sum_{j = 0}^{p - 1} \prod_{k = 1}^{p - 1}(1 + \omega^j)^2 \\
    & = 4(p - 1)p
\end{align*}Thus, to finish, we divide by $p^2$ and $-2$. This gives $\frac1{p}\left(\binom{2p}p - 2\right) + 2$
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Jupiterballs
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#47 • 1 Y
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