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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.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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 A Letter to MSM
Arr0w   23
N Sep 19, 2022 by scannose
Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
[list]
[*]Firstly, the case of $0^0$. It is usually regarded that $0^0=1$, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.

[*]What about $\frac{\infty}{\infty}$? The issue here is that $\infty$ isn't even rigorously defined in this expression. What exactly do we mean by $\infty$? Unless the example in question is put in context in a formal manner, then we say that $\frac{\infty}{\infty}$ is meaningless.

[*]What about $\frac{1}{0}$? Suppose that $x=\frac{1}{0}$. Then we would have $x\cdot 0=0=1$, absurd. A more rigorous treatment of the idea is that $\lim_{x\to0}\frac{1}{x}$ does not exist in the first place, although you will see why in a calculus course. So the point is that $\frac{1}{0}$ is undefined.

[*]What about if $0.99999...=1$? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
\begin{align*}
\sum_{n=1}^{\infty} \frac{9}{10^n}&=9\sum_{n=1}^{\infty}\frac{1}{10^n}=9\sum_{n=1}^{\infty}\biggr(\frac{1}{10}\biggr)^n=9\biggr(\frac{\frac{1}{10}}{1-\frac{1}{10}}\biggr)=9\biggr(\frac{\frac{1}{10}}{\frac{9}{10}}\biggr)=9\biggr(\frac{1}{9}\biggr)=\boxed{1}
\end{align*}
[*]What about $\frac{0}{0}$? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
[/list]
Hopefully all of these issues and their corollaries are finally put to rest. Cheers.

2nd EDIT (6/14/22): Since I originally posted this, it has since blown up so I will try to add additional information per the request of users in the thread below.

INDETERMINATE VS UNDEFINED

What makes something indeterminate? As you can see above, there are many things that are indeterminate. While definitions might vary slightly, it is the consensus that the following definition holds: A mathematical expression is be said to be indeterminate if it is not definitively or precisely determined. So how does this make, say, something like $0/0$ indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that $0/0$ is an indeterminate form.

But we need to more specific, this is still ambiguous. An indeterminate form is a mathematical expression involving at most two of $0$, $1$ or $\infty$, obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are
\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function $f:\mathbb{R}^{+}\cup\{0\}\rightarrow\mathbb{R}$ given by the mapping $x\mapsto \sqrt{x}$ is undefined for $x<0$. On the other hand, $1/0$ is undefined because dividing by $0$ is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context.

WHEN THE WATERS GET MUDDIED

So with this notion of indeterminate and undefined, things get convoluted. First of all, just because something is indeterminate does not mean it is not undefined. For example $0/0$ is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.

One example of making something undefined into something defined is the extended real number line, which we define as
\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let $-\infty\le a\le \infty$ for each $a\in\overline{\mathbb{R}}$ which means that via this order topology each subset has an infimum and supremum and $\overline{\mathbb{R}}$ is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In $\overline{\mathbb{R}}$ it is perfectly OK to say that,
\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,
\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define $\infty \times 0=0$ as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by $0$ is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as
\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, $\tilde{\infty}$ means complex infinity, since we are in the complex plane now. Here's the catch: division by $0$ is allowed here! In fact, we have
\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]where $\tilde{\infty}/\tilde{\infty}$ and $0/0$ are left undefined. We also have
\begin{align*}
z+\tilde{\infty}=\tilde{\infty}, \forall z\ne -\infty\\
z\times \tilde{\infty}=\tilde{\infty}, \forall z\ne 0
\end{align*}Furthermore, we actually have some nice properties with multiplication that we didn't have before. In $\mathbb{C}^*$ it holds that
\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]but $\tilde{\infty}-\tilde{\infty}$ and $0\times \tilde{\infty}$ are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with $\mathbb{C}^*$, by defining them as
\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]They behave in a similar way to the Riemann Sphere, with division by $0$ also being allowed with the same indeterminate forms (in addition to some other ones).
23 replies
Arr0w
Feb 11, 2022
scannose
Sep 19, 2022
k i Marathon Threads
LauraZed   0
Jul 2, 2019
Due to excessive spam and inappropriate posts, we have locked the Prealgebra and Beginning Algebra threads.

We will either unlock these threads once we've cleaned them up or start new ones, but for now, do not start new marathon threads for these subjects. Any new marathon threads started while this announcement is up will be immediately deleted.
0 replies
LauraZed
Jul 2, 2019
0 replies
k i Basic Forum Rules and Info (Read before posting)
jellymoop   368
N May 16, 2018 by harry1234
f (Reminder: Do not post Alcumus or class homework questions on this forum. Instructions below.) f
Welcome to the Middle School Math Forum! Please take a moment to familiarize yourself with the rules.

Overview:
[list]
[*] When you're posting a new topic with a math problem, give the topic a detailed title that includes the subject of the problem (not just "easy problem" or "nice problem")
[*] Stay on topic and be courteous.
[*] Hide solutions!
[*] If you see an inappropriate post in this forum, simply report the post and a moderator will deal with it. Don't make your own post telling people they're not following the rules - that usually just makes the issue worse.
[*] When you post a question that you need help solving, post what you've attempted so far and not just the question. We are here to learn from each other, not to do your homework. :P
[*] Avoid making posts just to thank someone - you can use the upvote function instead
[*] Don't make a new reply just to repeat yourself or comment on the quality of others' posts; instead, post when you have a new insight or question. You can also edit your post if it's the most recent and you want to add more information.
[*] Avoid bumping old posts.
[*] Use GameBot to post alcumus questions.
[*] If you need general MATHCOUNTS/math competition advice, check out the threads below.
[*] Don't post other users' real names.
[*] Advertisements are not allowed. You can advertise your forum on your profile with a link, on your blog, and on user-created forums that permit forum advertisements.
[/list]

Here are links to more detailed versions of the rules. These are from the older forums, so you can overlook "Classroom math/Competition math only" instructions.
Posting Guidelines
Update on Basic Forum Rules
What belongs on this forum?
How do I write a thorough solution?
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Mathcounts FAQ and resources
Mathcounts and how to learn

As always, if you have any questions, you can PM me or any of the other Middle School Moderators. Once again, if you see spam, it would help a lot if you filed a report instead of responding :)

Marathons!
Relays might be a better way to describe it, but these threads definitely go the distance! One person starts off by posting a problem, and the next person comes up with a solution and a new problem for another user to solve. Here's some of the frequently active marathons running in this forum:
[list][*]Algebra
[*]Prealgebra
[*]Proofs
[*]Factoring
[*]Geometry
[*]Counting & Probability
[*]Number Theory[/list]
Some of these haven't received attention in a while, but these are the main ones for their respective subjects. Rather than starting a new marathon, please give the existing ones a shot first.

You can also view marathons via the Marathon tag.

Think this list is incomplete or needs changes? Let the mods know and we'll take a look.
368 replies
jellymoop
May 8, 2015
harry1234
May 16, 2018
Is EGMO good for JMO Geometry Questions?
MathRook7817   2
N 4 hours ago by CatCatHead
Hi guys, I was just wondering if EGMO is a good book for JMO/AMO/olympiad level questions, or if there exists another olympiad geo book. Thanks!
2 replies
MathRook7817
Today at 3:05 AM
CatCatHead
4 hours ago
purple comet discussion
ConfidentKoala4   64
N Today at 3:11 AM by MathCosine
when can we discuss purple comet
64 replies
1 viewing
ConfidentKoala4
May 2, 2025
MathCosine
Today at 3:11 AM
2025 Math and AI 4 Girls Competition: Win Up To $1,000!!!
audio-on   70
N Today at 3:00 AM by stuffedmath
Join the 2025 Math and AI 4 Girls Competition for a chance to win up to $1,000!

Hey Everyone, I'm pleased to announce the dates for the 2025 MA4G Competition are set!
Applications will open on March 22nd, 2025, and they will close on April 26th, 2025 (@ 11:59pm PST).

Applicants will have one month to fill out an application with prizes for the top 50 contestants & cash prizes for the top 20 contestants (including $1,000 for the winner!). More details below!

Eligibility:
The competition is free to enter, and open to middle school female students living in the US (5th-8th grade).
Award recipients are selected based on their aptitude, activities and aspirations in STEM.

Event dates:
Applications will open on March 22nd, 2025, and they will close on April 26th, 2025 (by 11:59pm PST)
Winners will be announced on June 28, 2025 during an online award ceremony.

Application requirements:
Complete a 12 question problem set on math and computer science/AI related topics
Write 2 short essays

Prizes:
1st place: $1,000 Cash prize
2nd place: $500 Cash prize
3rd place: $300 Cash prize
4th-10th: $100 Cash prize each
11th-20th: $50 Cash prize each
Top 50 contestants: Over $50 worth of gadgets and stationary


Many thanks to our current and past sponsors and partners: Hudson River Trading, MATHCOUNTS, Hewlett Packard Enterprise, Automation Anywhere, JP Morgan Chase, D.E. Shaw, and AI4ALL.

Math and AI 4 Girls is a nonprofit organization aiming to encourage young girls to develop an interest in math and AI by taking part in STEM competitions and activities at an early age. The organization will be hosting an inaugural Math and AI 4 Girls competition to identify talent and encourage long-term planning of academic and career goals in STEM.

Contact:
mathandAI4girls@yahoo.com

For more information on the competition:
https://www.mathandai4girls.org/math-and-ai-4-girls-competition

More information on how to register will be posted on the website. If you have any questions, please ask here!


70 replies
audio-on
Jan 26, 2025
stuffedmath
Today at 3:00 AM
USAMO Medals
YauYauFilter   20
N Today at 2:38 AM by vincentwant
YauYauFilter
Apr 24, 2025
vincentwant
Today at 2:38 AM
Doubt on a math problem
AVY2024   14
N Today at 2:35 AM by Soupboy0
Solve for x and y given that xy=923, x+y=84
14 replies
AVY2024
Apr 8, 2025
Soupboy0
Today at 2:35 AM
Mass points question
Wesoar   0
Today at 2:27 AM
So I was working my way through mass points, and I found a rule that basically says:

"If transversal line EF crosses cevian AD in triangle ABC, you must split mass A into Mass ab and Mass ac. Could someone explain to me why this makes sense/why we couldn't just use mass A?
0 replies
Wesoar
Today at 2:27 AM
0 replies
What's the chance that two AoPS accounts generate with the same icon?
Math-lover1   16
N Today at 2:21 AM by martianrunner
So I've been wondering how many possible "icons" can be generated when you first create an account. By "icon" I mean the stack of cubes as the first profile picture before changing it.

I don't know a lot about how AoPS icons generate, so I have a few questions:
- Do the colors on AoPS icons generate through a preset of colors or the RGB (red, green, blue in hexadecimal form) scale? If it generates through the RGB scale, then there may be greater than $256^3 = 16777216$ different icons.
- Do the arrangements of the stacks of blocks in the icon change with each account? If so, I think we can calculate this through considering each stack of blocks independently.
16 replies
Math-lover1
May 2, 2025
martianrunner
Today at 2:21 AM
Easy number theory
britishprobe17   31
N Today at 2:20 AM by martianrunner
The number of factors from 2024 that are greater than $\sqrt{2024}$ are
31 replies
britishprobe17
Oct 16, 2024
martianrunner
Today at 2:20 AM
prime numbers
wpdnjs   109
N Today at 1:44 AM by ReticulatedPython
does anyone know how to quickly identify prime numbers?

thanks.
109 replies
wpdnjs
Oct 2, 2024
ReticulatedPython
Today at 1:44 AM
max number of candies
orangefronted   12
N Today at 1:27 AM by iwastedmyusername
A store sells a strawberry flavoured candy for 1 dollar each. The store offers a promo where every 4 candy wrappers can be exchanged for one candy. If there is no limit to how many times you can exchange candy wrappers for candies, what is the maximum number of candies I can obtain with 100 dollars?
12 replies
orangefronted
Apr 3, 2025
iwastedmyusername
Today at 1:27 AM
9 Have you participated in the MATHCOUNTS competition?
aadimathgenius9   43
N Today at 12:29 AM by Inaaya
Have you participated in the MATHCOUNTS competition before?
43 replies
aadimathgenius9
Jan 1, 2025
Inaaya
Today at 12:29 AM
How to get a 300+ on the NWEA MAP MATH test (URGENT)
nmlikesmath   16
N Today at 12:26 AM by Inaaya
I have 4 days till this test, I'm wondering how do I get a 300+ and what do I need to know, thank you.
16 replies
nmlikesmath
May 3, 2025
Inaaya
Today at 12:26 AM
Warning!
VivaanKam   18
N Today at 12:18 AM by Iwowowl253
This problem will try to trick you! :!:

18 replies
VivaanKam
Yesterday at 5:08 PM
Iwowowl253
Today at 12:18 AM
9 Was the 2025 AMC 8 harder or easier than last year?
Sunshine_Paradise   196
N Yesterday at 11:49 PM by giratina3
Also what will be the DHR?
196 replies
Sunshine_Paradise
Jan 30, 2025
giratina3
Yesterday at 11:49 PM
Scary Binomial Coefficient Sum
EpicBird08   43
N Apr 19, 2025 by Mathgloggers
Source: USAMO 2025/5
Determine, with proof, all positive integers $k$ such that $$\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k$$is an integer for every positive integer $n.$
43 replies
EpicBird08
Mar 21, 2025
Mathgloggers
Apr 19, 2025
Scary Binomial Coefficient Sum
G H J
G H BBookmark kLocked kLocked NReply
Source: USAMO 2025/5
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EpicBird08
1751 posts
#1 • 2 Y
Y by KevinYang2.71, cubres
Determine, with proof, all positive integers $k$ such that $$\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k$$is an integer for every positive integer $n.$
This post has been edited 2 times. Last edited by EpicBird08, Mar 21, 2025, 12:06 PM
Z K Y
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EpicBird08
1751 posts
#2 • 3 Y
Y by KevinYang2.71, Kingsbane2139, cubres
We claim that the answer is $\boxed{\text{all even positive integers}}.$

Proof that odd $k$ fail: Just take $n = 2$ and get $2^k + 2$ is divisible by $3,$ which implies $2^k \equiv 1 \pmod{3}$ This can only happen if $k$ is even.

Proof that even $k$ work: Substitute $n-1$ in place for $n$ in the problem; we then must prove that $$\frac{1}{n} \sum_{i=0}^{n-1} \binom{n-1}{i}^k$$is an integer for all positive integers $n.$ Call $n$ $k$-good if $n$ satisfies this condition. Letting $\Omega(n)$ denote the number of not necessarily distinct prime factors of $n,$ we will prove that all positive integers $n$ are $k$-good by induction on $\Omega(n).$ The base case is trivial since if $\Omega(n) = 0,$ then $n = 1,$ which vacuously works.

Now suppose that the result was true for all $n$ such that $\Omega(n) \le a.$ Let the prime factorization of $n$ be $p_1^{e_1} \cdot p_2^{e_2} \cdots p_m^{e_m},$ where $p_1, \dots, p_m$ are distinct primes and $e_1, \dots, e_m \in \mathbb{N}$ such that $e_1 + \dots + e_m = a+1.$ We will show that $$\sum_{i=0}^{n-1} \binom{n-1}{i}^k \equiv 0 \pmod{p_l^{e_l}}$$for all $1 \le l \le m,$ which finishes the inductive step by CRT.

We compute
\begin{align*}
\binom{n-1}{i} &= \frac{(n-1)!}{i! (n-i-1)!} \\
&= \frac{(n-1)(n-2)\cdots (n-i)}{i!} \\
&= \prod_{j=1}^i \frac{n-j}{j},
\end{align*}so $$\binom{n-1}{i}^k = \prod_{j=1}^i \left(\frac{n-j}{j}\right)^k.$$If $p_l \nmid j,$ then $p_l \nmid n-j$ as well since $p_l \mid n.$ Thus $j$ has a modular inverse modulo $p_l^{e_k},$ so we can say $$\left(\frac{n-j}{j}\right)^k \equiv \left(\frac{-j}{j}\right)^k \equiv (-1)^k \equiv 1 \pmod{p_l^{e_k}}$$because $k$ is even. Hence the terms in this product for which $p_l \nmid j$ contribute nothing to the binomial coefficient.

If $p_l \mid j,$ then we have $$\frac{n-j}{j} = \frac{\frac{n}{p_l} - \frac{j}{p_l}}{\frac{j}{p_l}}.$$Reindexing the product in terms of $\frac{j}{p_l},$ we get $$\binom{n-1}{i}^k \equiv \prod_{j'=1}^{\lfloor i/p_l \rfloor} \left(\frac{n/p_l - j'}{j'}\right)^k \equiv \binom{n/p_l - 1}{\lfloor i/p_l \rfloor}^k \pmod{p_l^{e_l}}.$$Therefore, plugging back into our sum gives $$\sum_{i=0}^{n-1} \binom{n-1}{i}^k \equiv p_l \sum_{i=0}^{n/p_l - 1} \binom{n/p_l - 1}{i}^k \pmod{p_l^{e_l}}.$$By the inductive hypothesis, the sum on the right-hand side is divisible by $p_l^{e_l - 1},$ so the entire sum is divisible by $p_l^{e_l}.$ This completes our inductive step.

Therefore, we have proven that all positive integers $n$ are $k$-good, and we are done.

Click to reveal hidden text
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KevinYang2.71
421 posts
#3 • 2 Y
Y by cubres, deduck
original statement says "for every positive integer $n$"

We claim that $k$ is $\boxed{\mathrm{even}}$.

From $n=2$ we get $3\mid 2+2^n$ so clearly $k$ is even.

Now suppose $k$ is even.

Claim 1. If $\alpha=\nu_2(n+1)$, then $2^\alpha$ divides
\[
\sum_{i=0}^n\binom{n}{i}^k.
\]Proof. We proceed by induction on $\alpha$ with the base case $\alpha=0$ trivial.

Assume the statement for $\alpha-1$. Let $n+1=:2^\alpha m$ and let us work in $\mathbb{Z}/2^\alpha\mathbb{Z}$. Note that if $r$ is even, $(r+1)^{-1}$ exists so
\[
\binom{n}{r+1}=\frac{n-r}{r+1}\binom{n}{r}=\frac{2^\alpha m-(r+1)}{r+1}\binom{n}{r}=-\binom{n}{r}.
\]We prove that $\binom{n}{2r}=(-1)^r\binom{\frac{n-1}{2}}{r}$ for $r=0,\,1,\,\ldots,\,\frac{n-1}{2}$ by induction on $r$ with the base case $r=0$ trivial.

Assume $\binom{n}{2r}=(-1)^r\binom{\frac{n-1}{2}}{r}$. We have
\begin{align*}
\binom{n}{2r+2}&=\frac{n-2r-1}{2r+2}\binom{n}{2r+1}\\
&=-\frac{\frac{n-1}{2}-r}{r+1}\binom{n}{2r}\\
&=(-1)^{r+1}\frac{\frac{n-1}{2}-r}{r+1}\binom{\frac{n-1}{2}}{r}\\
&=(-1)^{r+1}\binom{\frac{n-1}{2}}{r+1},
\end{align*}completing the induction step.

Thus,
\begin{align*}
\sum_{i=0}^n\binom{n}{i}^k&=\sum_{r=0}^{\frac{n-1}{2}}\left(\binom{n}{2r}+\binom{n}{2r+1}\right)^k\\
&=2\sum_{r=0}^{\frac{n-1}{2}}\binom{n}{2r}^k\\
&=2\sum_{r=0}^{\frac{n-1}{2}}\binom{n}{2r}^k\\
&=2\sum_{r=0}^{\frac{n-1}{2}}\left((-1)^r\binom{\frac{n-1}{2}}{r}\right)^k\\
&=2\sum_{r=0}^{\frac{n-1}{2}}\binom{\frac{n-1}{2}}{r}^k.
\end{align*}Since $\nu_2\left(\frac{n+1}{2}\right)=\alpha-1$, by the induction hypothesis with $n':=\frac{n-1}{2}$, $2^{\alpha-1}$ divides
\[
\sum_{r=0}^{n'}\binom{n'}{r}^k.
\]Thus
\[
\sum_{i=0}^n\binom{n}{i}^k=2\sum_{r=0}^{n'}\binom{n'}{r}^k=0,
\]as desired. $\square$

Odd $p$ case is the same except there is no $(-1)^r$ in the proof. Since all prime powers dividing $n+1$ divide $\sum_{i=0}^n\binom{n}{i}^k$, $n+1$ divides $\sum_{i=0}^n\binom{n}{i}^k$. $\square$

How many point dock for dropping ^k in Claim 1 but doing the odd $p$ case correctly (I wrote a Claim 2 that was basically identical to Claim 1 but modified for odd $p$). Also by dropping the ^k, I did not induct on $\alpha$ in Claim 1 because the last summation (wrongly) becomes $0$ directly.
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arfekete
257 posts
#4 • 1 Y
Y by cubres
Will I get docked if I just said it was sufficient to prove that the term is divisible by $p^{v_p{(n + 1)}}$ for any arbitrary $p | n + 1$ and didn't mention CRT? (I defined p-adic notation)
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Countmath1
180 posts
#5 • 1 Y
Y by cubres
this is the one thing i started on before i had to leave. i got that all odd k fail and k=2 works by vandermonde's + catalan. 0/21 day 2 baby
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balllightning37
389 posts
#7 • 1 Y
Y by cubres
Nice, solution was same as #2.

Do we get docked for not defining $v_p$? I was in a hurry...
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plang2008
337 posts
#8 • 1 Y
Y by cubres
Bruh I interpreted this as “For each positive integer $n$, find all positive integers $k$ such that this expression is a positive integer”

Then the answer should be all positive integers if $n + 1$ is a power of $2$ and all even positive integers otherwise
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Sleepy_Head
565 posts
#9 • 1 Y
Y by cubres
how many points for correct answer (no proof), odd $k$ doesn't work, and $k=2$ works?
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HamstPan38825
8859 posts
#10 • 1 Y
Y by cubres
nvm this solution is actually wrong you have to induct it :/
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bachkieu
136 posts
#11 • 1 Y
Y by cubres
dnw vp moment??
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Pear222
5 posts
#12 • 3 Y
Y by CertifiedNoob, andyloo666, cubres
sniped by @2above

We claim that the answer is $\boxed{\text{all even positive integers}}$. To show that odd doesn't work, just look at $n=2$.

The main part of the problem is proving that $n+1$ divides \[\sum_{i=0}^n \binom ni ^{2k}.\]Pick a prime $p$ dividing $n+1$, and let $\nu_p(n+1) = e \ge 1$ so that $p^e$ is the maximal power of $p$ dividing $n+1$. Let $n = mp^e-1$ with $\gcd(m,p) = 1$ The key claim is as follows:

Claim 1: For each $0 \le a < e$, we have that \[\sum_{i=0}^{\left\lfloor \frac{n}{p^a}\right\rfloor} \binom{n}{ip^a}^{2k} \equiv p\sum_{i=0}^{\left\lfloor \frac{n}{p^{a+1}}\right\rfloor} \binom{n}{ip^{a+1}}^{2k}\bmod{p^{e-a}}\]
Proof: Consider the values \[\binom{n}{0}^2, \binom{n}{p^a}^2, \dots, \binom{n}{(mp^{e-a}-1)p^a}^2.\]We claim that these values can be blocked into consecutive groups of $p$ such that the values in each block are equal modulo $p^{e-a}$. Specifically, for $p\nmid i+1$, we claim that \[\binom{n}{ip^a} \equiv \binom{n}{(i+1)p^a} \pmod {p^{e-a}}.\]Indeed, we have that
\begin{align*}
\binom{n}{(i+1)p^a}^2 &= \left(\frac{n}{1}\cdot \frac{n-1}{2} \cdots \frac{n+1-j}{j} \cdots \frac{n+1 - (i+1)p^a}{(i+1)p^a}\right)^2 \\
&= \binom{n}{ip^a}^2 \prod_{j = ip^a + 1}^{(i+1)p^a} \left(\frac{n+1}j - 1\right)^2
\end{align*}But for every $ip^a + 1 < j \le (i+1)p^a$, $\nu_p(j) \le a$ since $p\nmid i+1$. Therefore $\frac{n+1}{j} \equiv 0 \bmod{p^{e-a}}$, so the entire product is equal to $1\pmod{p^{e-a}}$; thus the subclaim is true. Therefore we have that
\begin{align*}
\sum_{i=0}^{\left\lfloor \frac{n}{p^a} \right\rfloor} \binom{n}{ip^a}^{2k} &=\sum_{i=0}^{\frac 1p\left\lfloor \frac{n}{p^a} \right\rfloor} \sum_{j= ip}^{ip+p-1} \left(\binom{n}{jp^a}^2\right)^k\\
& \equiv \sum_{i=0}^{\left\lfloor \frac{n}{p^{a+1}} \right\rfloor} p\left(\binom{n}{ip(p^a)}^2\right)^k \pmod{ p^{e-a}}\\
& \equiv p\sum_{i=0}^{\left\lfloor \frac{n}{p^{a+1}} \right\rfloor} \binom{n}{ip^{a+1}}^{2k}
\end{align*}and the claim is proven.

To finish, note that a quick induction implies that \[p^{e-a} \mid \sum_{i=0}^{\left\lfloor \frac{n}{p^a}\right\rfloor} \binom{n}{ip^a}^{2k}\]for all $0\le a \le e$. Taking $a = 0$ gives that \[p^e \mid \sum_{i=0}^{n} \binom ni^k\]for any prime power $p^e$ dividing $n+1$. This means that \[\frac 1{n+1} \sum_{i=0}^n \binom ni^{2k}\]is an integer for any $k\in \mathbb{N}$, so we are done.
This post has been edited 2 times. Last edited by Pear222, Apr 5, 2025, 8:22 PM
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pianoboy
320 posts
#13 • 1 Y
Y by cubres
I noticed (by taking powers over first 10 rows of Pascal triangle) that (n choose k)^4 = (n choose k)^2 mod (n+1). Is that always true?

If we prove that the problem is basically solved.
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YaoAOPS
1540 posts
#14 • 1 Y
Y by cubres
$n = 215, k = 54$ is a counterexample.
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NoSignOfTheta
1731 posts
#15 • 1 Y
Y by cubres
No its not
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YaoAOPS
1540 posts
#16 • 1 Y
Y by cubres
yes it is????????
Attachments:
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pianoboy
320 posts
#17 • 1 Y
Y by cubres
What ? This false conjecture with an absurdly high counterexample?
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OronSH
1730 posts
#18 • 3 Y
Y by centslordm, peppapig_, cubres
Answer is even $k$. Odd $k$ fail at $n=2$.

Let $m=n+1$ and set $\nu_p(m)=s$. The main idea is that \begin{align*}\binom{pm-1}{pi+j}&=\frac{(pm-1)(pm-2)\cdots(pm-pi-j)}{1\cdot 2\cdots(pi+j)}\\&=\left(\frac{pm-1}1\cdot\frac{pm-2}2\cdots\frac{pm-p+1}{p-1}\cdot\frac{pm-p-1}{p+1}\cdots\right)\cdot\frac{(m-1)(m-2)\cdots(m-i)}{1\cdot 2\cdots i}\\&\equiv\pm\binom{m-1}i\pmod{p^{s+1}}.\end{align*}Then \[\sum_{i=0}^{pm-1}\binom{pm-1}i^k\equiv p\sum_{i=0}^{m-1}\binom{m-1}i^k\pmod{p^{s+1}}\]so inducting on $\nu_p(m)$ works.
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OronSH
1730 posts
#19 • 1 Y
Y by cubres
pianoboy wrote:
What ? This false conjecture with an absurdly high counterexample?

$n=8,k=3$ is a counterexample
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NoSignOfTheta
1731 posts
#20 • 1 Y
Y by cubres
YaoAOPS wrote:
yes it is????????

ohhh I thought u were talking about the actual problem
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Ilikeminecraft
616 posts
#21 • 1 Y
Y by cubres
how many points will proving the case for $\operatorname{rad}(n) = n$(prime exponents are 1)
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awesomeguy856
7265 posts
#22 • 3 Y
Y by OronSH, bjump, cubres
OronSH wrote:
Answer is even $k$. Odd $k$ fail at $n=2$.

Let $m=n+1$ and set $\nu_p(m)=s$. The main idea is that \begin{align*}\binom{pm-1}{pi+j}&=\frac{(pm-1)(pm-2)\cdots(pm-pi-j)}{1\cdot 2\cdots(pi+j)}\\&=\left(\frac{pm-1}1\cdot\frac{pm-2}2\cdots\frac{pm-p+1}{p-1}\cdot\frac{pm-p-1}{p+1}\cdots\right)\cdot\frac{(m-1)(m-2)\cdots(m-i)}{1\cdot 2\cdots i}\\&\equiv\pm\binom{m-1}i\pmod{p^{s+1}}.\end{align*}Then \[\sum_{i=0}^{pm-1}\binom{pm-1}i^k\equiv p\sum_{i=0}^{m-1}\binom{m-1}i^k\pmod{p^{s+1}}\]so inducting on $\nu_p(m)$ works.

more like pmo
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krithikrokcs
148 posts
#23 • 1 Y
Y by cubres
if i wrote my solution backwards will i get points off?
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v_Enhance
6877 posts
#24 • 5 Y
Y by Curious_Droid, jkim0656, cubres, krithikrokcs, Sedro
The answer is all even $k$.
Let's abbreviate $S(n) \coloneq \binom n0^k + \dots + \binom nn^k$ for the sum in the problem.
Proof that even $k$ is necessary. Choose $n=2$. We need $3 \mid S(2) = 2+2^k$, which requires $k$ to be even.

Remark: It's actually not much more difficult to just use $n = p-1$ for prime $p$, since $\binom{p-1}{i} \equiv (-1)^i \pmod p$. Hence $S(p-1) \equiv 1 + (-1)^k + 1 + (-1)^k + \dots + 1 \pmod p$, and this also requires $k$ to be even. This special case is instructive in figuring out the proof to follow.

Proof that $k$ is sufficient. From now on we treat $k$ as fixed, and we let $p^e$ be a prime fully dividing $n+1$. The basic idea is to reduce from $n+1$ to $(n+1)/p$ by an induction.

Remark: Here is a concrete illustration that makes it clear what's going on. Let $p = 5$. When $n = p-1 = 4$, we have \[ S(4) = 1^k + 4^k + 6^k + 4^k + 1^k \equiv 1 + 1 + 1 + 1 + 1 \equiv 0 \pmod 5. \]When $n = p^2-1 = 24$, the $25$ terms of $S(24)$ in order are, modulo $25$, \begin{align*} S(24) &\equiv 1^k + 1^k + 1^k + 1^k + 1^k\\ &+ 4^k + 4^k + 4^k + 4^k + 4^k \\ &+ 6^k + 6^k + 6^k + 6^k + 6^k \\ &+ 4^k + 4^k + 4^k + 4^k + 4^k \\ &+ 1^k + 1^k + 1^k + 1^k + 1^k \\ &= 5(1^k + 4^k + 6^k + 4^k + 1^k). \end{align*}The point is that $S(24)$ has five copies of $S(4)$, modulo $25$.
To make the pattern in the remark explicit, we prove the following lemma on each individual binomial coefficient.
Lemma: Suppose $p^e$ is a prime power which fully divides $n+1$. Then \[ \binom{n}{i} \equiv \pm \binom{\frac{n+1}{p}-1}{\left\lfloor i/p \right\rfloor} \pmod{p^e}. \]Proof. [Proof of lemma] It's easiest to understand the proof by looking at the cases $\left\lfloor i/p \right\rfloor \in \{0,1,2\}$ first.
  • For $0 \le i < p$, since $n \equiv -1 \mod p^e$, we have \[ \binom{n}{i} = \frac{n(n-1) \dots (n-i+1)}{1 \cdot 2 \cdot \dots \cdot i} \equiv \frac{(-1)(-2) \dots (-i)}{1 \cdot 2 \cdot \dots \cdot i} \equiv \pm 1 \pmod{p^e}. \]
  • For $p \le i < 2p$ we have \begin{align*} \binom{n}{i} &\equiv \pm 1 \cdot \frac{n-p+1}{p} \cdot \frac{(n-p)(n-p-1) \dots (n-i+1)}{(p+1)(p+2) \dots i} \\ &\equiv \pm 1 \cdot \frac{\frac{n-p+1}{p}}{1} \cdot \pm 1 \\ &\equiv \pm \binom{\frac{n+1}{p}-1}{1} \pmod{p^e}. \end{align*}
  • For $2p \le i < 3p$ the analogous reasoning gives \begin{align*} \binom ni &\equiv \pm 1 \cdot \frac{n-p+1}{p} \cdot \pm 1 \cdot \frac{n-2p+1}{2p} \cdot \pm 1 \\ &\equiv \pm \frac{\left(\frac{n+1}{p}-1\right)\left( \frac{n+1}{p}-2 \right) }{1 \cdot 2} \\ &\equiv \pm \binom{\frac{n+1}{p}-1}{2} \pmod{p^e}. \end{align*}
\dots And so on. The point is that in general, if we write \[ \binom ni = \prod_{0 \le j \le i} \frac{n-(j-1)}{j} \]then the fractions for $p \nmid j$ are all $\pm 1 \pmod{p^e}$. So only considers those $j$ with $p \mid j$; in that case one obtains the claimed $\binom{\frac{n+1}{p}-1}{\left\lfloor i/p \right\rfloor}$ exactly (even without having to take modulo $p^e$). $\blacksquare$
From the lemma, it follows if $p^e$ is a prime power which fully divides $n+1$, then \[ S(n) \equiv p \cdot S\left( \frac{n+1}{p}-1 \right) \pmod{p^e} \]by grouping the $n+1$ terms (for $0 \le i \le n$) into consecutive ranges of length $p$ (by the value of $\left\lfloor i/p \right\rfloor$).
This post has been edited 1 time. Last edited by v_Enhance, Mar 21, 2025, 2:59 PM
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Mathandski
756 posts
#25 • 1 Y
Y by cubres
I initially fakesolved this problem writing a proof that was somewhat a convoluted way of saying $\binom{n}{i} \equiv \binom{-1}{i} \pmod{n}$. Realized this with 2 hours left and had to start over. 25AMO5 gave me the exact same feeling as 24JMO4. It took another 1:15 to solve correctly - easily the most stressful hour of my life. I measured my heart rate with roughly an hour left and it was at 50 beats / 20 seconds = 150 BPM.
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v_Enhance
6877 posts
#26 • 2 Y
Y by NaturalSelection, cubres
Realized this with 2 hours left and had to start over. In total it took 1:15 to solve correctly; the most stressful hour of my life. I measured my heart rate with 45 minutes left and it was at 50 beats / 20 seconds = 150 BPM
I remember that experience as a student too. In my case, the problem was USAMO 2014/4, but I only had 20 minutes to fix my wrong solution.
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peppapig_
281 posts
#27 • 15 Y
Y by YaoAOPS, OronSH, KnowingAnt, golue3120, centslordm, i3435, sixoneeight, EpicBird08, VicKmath7, Lam040208, Supercali, cubres, krithikrokcs, scannose, john0512
Haven't seen this solution yet! Pure manipulation, no induction on $p$.

We claim that the answer is all even $k$, odd $k$ dies to $n=2$.

For even $k$, let $k=2m$ for $m\in \mathbb{Z}^+$, note that
\[\binom{n}{i}^{2m}=\binom{n}{0}^{2m}+\left(\binom{n}{1}^{2m}-\binom{n}{0}^{2m}\right)+\left(\binom{n}{2}^{2m}-\binom{n}{1}^{2m}\right)+\dots+\left(\binom{n}{i}^{2m}-\binom{n}{i-1}^{2m}\right).\]Now, summing this up over all $1\le i\le n$, we have
\[\sum_{i=0}^{n}\binom{n}{i}^{2m}=(n+1)\binom{n}{0}^{2m}+\sum_{i=1}^{n}(n+1-i)\left(\binom{n}{i}^{2m}-\binom{n}{i-1}^{2m}\right).\]
It now suffices to show that
\[(n+1)\mid (n+1-i)\left(\binom{n}{i}^{2m}-\binom{n}{i-1}^{2m}\right),\]for all $1\le i\le n$. However, note that since $2m$ is even, we have that
\[(n+1-i)\left(\binom{n}{i}+\binom{n}{i-1}\right) \mid (n+1-i)\left(\binom{n}{i}^{2m}-\binom{n}{i-1}^{2m}\right).\]But we also have that
\[(n+1-i)\left(\binom{n}{i}+\binom{n}{i-1}\right)=(n+1-i)\binom{n+1}{i}=(n+1-i)\cdot \frac{(n+1)!}{(n+1-i)!i!}=(n+1)\cdot \frac{n!}{(n-i)!i!},\]which is just $(n+1)\binom{n}{i}$. This is clearly divisible by $n+1$, proving that
\[(n+1)\mid (n+1-i)\left(\binom{n}{i}^{2m}-\binom{n}{i-1}^{2m}\right),\]for all $1\le i\le n$.

Summing this over all $i$, this means that
\[(n+1)\mid \sum_{i=1}^{n}(n+1-i)\left(\binom{n}{i}^{2m}-\binom{n}{i-1}^{2m}\right),\]so
\[(n+1)\mid (n+1)\binom{n}{0}^{2m}+\sum_{i=1}^{n}(n+1-i)\left(\binom{n}{i}^{2m}-\binom{n}{i-1}^{2m}\right)=\sum_{i=0}^{n}\binom{n}{i}^{2m},\]as desired. Therefore all even $k$ work, completing our proof.
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ihatemath123
3446 posts
#28 • 6 Y
Y by peace09, Lhaj3, sixoneeight, cubres, biomathematics, megarnie
theres no way this problem hasnt already been posted somewhere in hso or math overflow 20 years ago or smth
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solasky
1566 posts
#29 • 1 Y
Y by cubres
v_Enhance wrote:
The answer is all even $k$.
Let's abbreviate $S(n) \coloneq \binom n0^k + \dots + \binom nn^k$ for the sum in the problem.
Proof that even $k$ is necessary. Choose $n=2$. We need $3 \mid S(2) = 2+2^k$, which requires $k$ to be even.

Remark: It's actually not much more difficult to just use $n = p-1$ for prime $p$, since $\binom{p-1}{i} \equiv (-1)^i \pmod p$. Hence $S(p-1) \equiv 1 + (-1)^k + 1 + (-1)^k + \dots + 1 \pmod p$, and this also requires $k$ to be even. This special case is instructive in figuring out the proof to follow.

Proof that $k$ is sufficient. From now on we treat $k$ as fixed, and we let $p^e$ be a prime fully dividing $n+1$. The basic idea is to reduce from $n+1$ to $(n+1)/p$ by an induction.

Remark: Here is a concrete illustration that makes it clear what's going on. Let $p = 5$. When $n = p-1 = 4$, we have \[ S(4) = 1^k + 4^k + 6^k + 4^k + 1^k \equiv 1 + 1 + 1 + 1 + 1 \equiv 0 \pmod 5. \]When $n = p^2-1 = 24$, the $25$ terms of $S(24)$ in order are, modulo $25$, \begin{align*} S(24) &\equiv 1^k + 1^k + 1^k + 1^k + 1^k\\ &+ 4^k + 4^k + 4^k + 4^k + 4^k \\ &+ 6^k + 6^k + 6^k + 6^k + 6^k \\ &+ 4^k + 4^k + 4^k + 4^k + 4^k \\ &+ 1^k + 1^k + 1^k + 1^k + 1^k \\ &= 5(1^k + 4^k + 6^k + 4^k + 1^k). \end{align*}The point is that $S(24)$ has five copies of $S(4)$, modulo $25$.
To make the pattern in the remark explicit, we prove the following lemma on each individual binomial coefficient.
Lemma: Suppose $p^e$ is a prime power which fully divides $n+1$. Then \[ \binom{n}{i} \equiv \pm \binom{\frac{n+1}{p}-1}{\left\lfloor i/p \right\rfloor} \pmod{p^e}. \]Proof. [Proof of lemma] It's easiest to understand the proof by looking at the cases $\left\lfloor i/p \right\rfloor \in \{0,1,2\}$ first.
  • For $0 \le i < p$, since $n \equiv -1 \mod p^e$, we have \[ \binom{n}{i} = \frac{n(n-1) \dots (n-i+1)}{1 \cdot 2 \cdot \dots \cdot i} \equiv \frac{(-1)(-2) \dots (-i)}{1 \cdot 2 \cdot \dots \cdot i} \equiv \pm 1 \pmod{p^e}. \]
  • For $p \le i < 2p$ we have \begin{align*} \binom{n}{i} &\equiv \pm 1 \cdot \frac{n-p+1}{p} \cdot \frac{(n-p)(n-p-1) \dots (n-i+1)}{(p+1)(p+2) \dots i} \\ &\equiv \pm 1 \cdot \frac{\frac{n-p+1}{p}}{1} \cdot \pm 1 \\ &\equiv \pm \binom{\frac{n+1}{p}-1}{1} \pmod{p^e}. \end{align*}
  • For $2p \le i < 3p$ the analogous reasoning gives \begin{align*} \binom ni &\equiv \pm 1 \cdot \frac{n-p+1}{p} \cdot \pm 1 \cdot \frac{n-2p+1}{2p} \cdot \pm 1 \\ &\equiv \pm \frac{\left(\frac{n+1}{p}-1\right)\left( \frac{n+1}{p}-2 \right) }{1 \cdot 2} \\ &\equiv \pm \binom{\frac{n+1}{p}-1}{2} \pmod{p^e}. \end{align*}
\dots And so on. The point is that in general, if we write \[ \binom ni = \prod_{0 \le j \le i} \frac{n-(j-1)}{j} \]then the fractions for $p \nmid j$ are all $\pm 1 \pmod{p^e}$. So only considers those $j$ with $p \mid j$; in that case one obtains the claimed $\binom{\frac{n+1}{p}-1}{\left\lfloor i/p \right\rfloor}$ exactly (even without having to take modulo $p^e$). $\blacksquare$
From the lemma, it follows if $p^e$ is a prime power which fully divides $n+1$, then \[ S(n) \equiv p \cdot S\left( \frac{n+1}{p}-1 \right) \pmod{p^e} \]by grouping the $n+1$ terms (for $0 \le i \le n$) into consecutive ranges of length $p$ (by the value of $\left\lfloor i/p \right\rfloor$).

Wait I proved the lemma in-contest but I didn’t realize you can just induct on that to finish :sob: I didn’t even write it down b/c I thought it was a dead end, oh well
This post has been edited 1 time. Last edited by solasky, Mar 21, 2025, 4:29 PM
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Mathandski
756 posts
#30 • 3 Y
Y by solasky, GrantStar, cubres
v_Enhance wrote:
Remark: Here is a concrete illustration that makes it clear what's going on. Let $p = 5$. When $n = p-1 = 4$, we have \[ S(4) = 1^k + 4^k + 6^k + 4^k + 1^k \equiv 1 + 1 + 1 + 1 + 1 \equiv 0 \pmod 5. \]When $n = p^2-1 = 24$, the $25$ terms of $S(24)$ in order are, modulo $25$, \begin{align*} S(24) &\equiv 1^k + 1^k + 1^k + 1^k + 1^k\\ &+ 4^k + 4^k + 4^k + 4^k + 4^k \\ &+ 6^k + 6^k + 6^k + 6^k + 6^k \\ &+ 4^k + 4^k + 4^k + 4^k + 4^k \\ &+ 1^k + 1^k + 1^k + 1^k + 1^k \\ &= 5(1^k + 4^k + 6^k + 4^k + 1^k). \end{align*}The point is that $S(24)$ has five copies of $S(4)$, modulo $25$.

These are the exact numbers I used to motivate my solve as well :O
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golue3120
57 posts
#31 • 5 Y
Y by GrantStar, OronSH, centslordm, cubres, MS_asdfgzxcvb
well, I guess I have to post this

Answer is all even $k$, necessity follows from setting $n=2$. Henceforth assume $k$ is even.

Lemma. Let $p^e$ be a prime power. Then $\textstyle (1+x)^{p^e}\equiv (1+x^p)^{p^{e-1}}\pmod p^e$.
Proof. When $e=1$, $\textstyle(1+x)^p=1+x^p+\sum_{i=1}^{p-1}\binom pix^i\equiv 1+x^p\pmod p$. Now we induct on $e$. Suppose $\textstyle (1+x)^{p^e}\equiv (1+x^p)^{p^{e-1}}\pmod p^e$. Then $\textstyle (1+x)^{p^e}=(1+x^p)^{p^{e-1}}+p^eQ$ where $Q$ is some integer polynomials. Raising both sides to the power of $p$, we have $\textstyle (1+x)^{p^{e+1}}=(1+x^p)^{p^e}+\text{terms divisible by }p^{e+1}$, as desired.

We now prove that for every positive integer $m$, prime $p$, and nonnegative integer $e$,
\[p^e\mid\sum_{i=0}^{mp^e-1}\binom{mp^e-1}i^k.\]
We induct on $e$. If $e=0$, this is trivial. Now suppose it holds for $e$. Working modulo $p^{e+1}$, we have
\[(1-x)^{mp^{e+1}-1}=\frac{(1-x)^{mp^{e+1}}}{1-x}=\frac{(1+(-x)^p)^{mp^e}}{1-x}=\frac{1+(-x)^p}{1-x}(1+(-x)^p)^{mp^e-1}=(1+(-x)^p)^{mp^e-1}\sum_{i=0}^{p-1}x^i.\]Thus by comparing coefficients, $\textstyle\binom{mp^{e+1}-1}{qp+r}\equiv\pm\binom{mp^e-1}{q}$ for $0\le q<mp^e$, $0\le r<p$. Therefore, modulo $p^{e+1}$,
\[\sum_{i=0}^{mp^{e+1}-1}\binom{mp^{e+1}-1}{i}^k=\sum_{q=0}^{mp^e-1}\sum_{r=0}^{p-1}\binom{mp^e-1}{q}=p\sum_{q=0}^{mp^e-1}\binom{mp^e-1}{q}.\]By the inductive hypothesis, the last sum is a multiple of $p^e$, hence the first sum is a multiple of $p^{e+1}$.
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Curious_Droid
35 posts
#32 • 2 Y
Y by peppapig_, cubres
Here is my solution which I stumbled upon after four hours of feverish grinding. I cannot make up my mind whether it is beautiful or just incredibly ugly. Scariest thing is, I still don't know if its correct :blush:

Lemma: If $\nu_p(c) = \nu_p(m) \ge 1$, then $p^k \mid {m-1 \choose c\cdot p^k -1}$ for any $k \ge 1$.

Proof: Kummers Theorem.

Corollary: Take positive integers $i, n$. Define $g = \gcd(i, n+1)$ and $i = g\cdot a\cdot b$ where $\gcd(b, n+1) = 1$ and $b$ is maximal. Then $a \mid {n \choose i-1}$.

Proof: Take one prime exponent $p^k \mid a$, where $k = \nu_p(a)$. By maximality of $b$, $p \mid n+1$. Obviously $p \nmid b$, and we must have $p \nmid \frac{n+1}{g}$. Thus $\nu_p(n+1) = \nu_p(g) = \nu_p\left(\frac{i}{p^k}\right)$, all $\ge 1$, and the desired follows by applying the Lemma. $\Box$

The case where $k$ is odd is easy, so we assume $k$ is even and show the desired conclusion.

Now consider the following process: Start with $n+1$ ones lined up in a row. Then on step $i$, multiply the central $n+1-2i$ terms by $\frac{n+1-i}{i}$. Obviously, we eventually construct the sequence of all binomial coefficients, and after step $i$, the sequence will be

$$1, {n \choose 1}, {n \choose 2}, \dots, {n \choose i-1}, \underbrace{{n \choose i}, \dots, {n \choose i}}_{\text{$n+1-2i$ copies}}, {n \choose i-1}, \dots, {n \choose 2}, {n \choose 1}, 1$$
Let $S_i$ be the sum of the $k$th powers of these terms after step $i$. As a result, $S_0 = n+1$. We claim that $S_i$ is invariant throughout the process.

Proof: First, note that $S_i - S_{i-1} = (n+1-2i)\left( {n \choose i}^k - {n \choose i-1}^k \right) =(n+1-2i){n \choose i-1}^k\left( \left(\frac{n+1-i}{i}\right)^k -1 \right) $. Now define $g = \gcd(i, n+1)$ and $i = g\cdot a\cdot b$ where $\gcd(b, n+1) = 1$ and $b$ is maximal. Further, let $n+1 = gd$. By the Corollary, $a \mid {n \choose i-1}$. Thus, we may set $X = \left(\frac{{n \choose i-1}}{a}\right)^k \in \mathbb Z$. Now
\begin{align*}
{n \choose i-1}^k\left( \left(\frac{n+1-i}{i}\right)^k -1 \right) &= X \cdot a^k \cdot \left( \left(\frac{gd-gab}{gab}\right)^k -1 \right)\\
&= X \cdot \left( \left(\frac{d-ba}{b}\right)^k -a^k \right)\\
&\equiv X \cdot \left( \left(\frac{-ba}{b}\right)^k -a^k \right)\\
&\equiv X \cdot \left( (-a)^k -a^k \right)\\
&\equiv 0 \pmod{d},
\end{align*}where division by $b$ was allowed because $\gcd(b, d) = 1$. Further, it is obvious that $g \mid n+1-2i$, so in conclusion, $n+1 = gd \mid S_i - S_{i-1}$. $\Box$

It is clear to see how we finish from here.

Note: Just a worse, more convoluted version of peppapig_'s solution.
This post has been edited 7 times. Last edited by Curious_Droid, Mar 22, 2025, 2:25 AM
Reason: clown
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john0512
4186 posts
#33 • 1 Y
Y by cubres
Clearly, $k$ is even as $n=2$ gives $3\mid 2+2^k$. Now, we show that all even $k$ work. Let $k=2m$.

Main Claim: If $p^r$ is a prime power such that $n\equiv -1\pmod{p^r}$, then
$${n\choose i}^{2m} \equiv {\lfloor n/p\rfloor \choose \lfloor i/p\rfloor}^{2m} \pmod{p^r}.$$
Consider the equation

$${n\choose i}=(\frac{n}{1})(\frac{n-1}{2})(\frac{n-2}{3})\dots(\frac{n-i+1}{i}).$$
Denote the "$k$th slot" as the fraction $\frac{n-k+1}{k}$. Considering just the slots that are multiples of $p$,

$$\frac{n-p+1}{p}\cdot \frac{n-2p+1}{2p}\cdots \frac{n-p\lfloor i/p \rfloor+1}{p\lfloor i/p\rfloor}$$$$=\frac{\lfloor n/p\rfloor}{1}\cdot \frac{\lfloor n/p\rfloor-1}{2}\cdots \frac{\lfloor n/p\rfloor-\lfloor i/p \rfloor +1}{\lfloor i/p\rfloor}$$$$={\lfloor n/p \rfloor \choose \lfloor i/p \rfloor}.$$
However, if $p\nmid k$, then the $k$th slot is
$$\frac{n-k+1}{k}\equiv \frac{-k}{k}\equiv -1\pmod{p^r},$$so if the exponent is even, the slots that are not multiples of $p$ do not affect the residue mod $p^r$ at all, which shows the claim.

Let $f(n)= \sum_{i=0}^n {n\choose i}^{2m}$. Then, if $n\equiv -1\pmod{p^r}$, then by the above claim,

$$f(n)={n\choose 0}^{2m} + {n\choose 1}^{2m}+\dots+{n\choose n}^{2m}$$$$\equiv {\lfloor n/p\rfloor \choose \lfloor 0/p\rfloor}^{2m}+{\lfloor n/p\rfloor \choose \lfloor 1/p\rfloor}^{2m}+\dots+{\lfloor n/p\rfloor \choose \lfloor n/p\rfloor}^{2m}$$$$\equiv p \left [ {\lfloor n/p \rfloor \choose 0}^{2m}+{\lfloor n/p \rfloor \choose 1}^{2m}+\dots+{\lfloor n/p \rfloor \choose \lfloor n/p \rfloor}^{2m}  \right ] \pmod{p^r}$$$$f(n) \equiv pf(\lfloor n/p \rfloor)\pmod{p^r}.$$
Finally we induct on the number of trailing $p-1$'s in the base $p$ representation of $n$ to show that $n\equiv -1\pmod{p^r}$ implies $p^r\mid f(n)$. If there is one trailing $p-1$, then clearly the above implies $p\mid f(n)$. Then, if $n$ has $r$ trailing $p-1$'s, then $\lfloor n/p \rfloor$ has $r-1$ trailing $p-1$'s. Thus, if $p^{r-1}\mid f(\lfloor n/p\rfloor)$, then $p^r\mid f(n)$, as desired.

Since $p^r\mid n+1$ implies $p^r\mid f(n)$, we are done.
This post has been edited 2 times. Last edited by john0512, Mar 22, 2025, 5:16 AM
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plang2008
337 posts
#34 • 2 Y
Y by cubres, megarnie
I misread it. The answer is all $k$ if $n + 1$ is a power of $2$ and all even $k$ otherwise.


Consider a prime $p \mid n + 1$, and let $a = \nu_p(n + 1)$. Notice that by definition we have $\binom ni = \prod_{j=1}^i \frac{n+1-j}{j}$. Since $p \mid n + 1$, we have $p \mid n + 1 - j$ if $p \mid j$ and $p\nmid n + 1 - j$ otherwise, so for each term, either both the numerator are divisible by $p$, or neither are. Let $M$ be the number of terms in the denominator that are not divisible by $p$.

For each term such that $p \nmid j$, we have $n + 1 - j \equiv -j$, so $\frac{n + 1 - j}{j} \equiv -1 \pmod {p^a}$. For each term such that $p \mid j$, we can divide out a $p$ from both the numerator and the denominator. Notice that what's left is simply $\binom{\lfloor n/p \rfloor}{\lfloor i/p \rfloor}$. Thus, we conclude that \[\boxed{\binom ni \equiv (-1)^M \binom{\lfloor n/p \rfloor}{\lfloor i/p \rfloor}}.\]

Proof that $k$ even works for all $p$: For $n = p - 1$, we clearly have \[\sum_{i=0}^n \binom ni^k \equiv \sum_{i=0}^n (-1)^{Mk} \equiv p \equiv 0 \pmod p.\]
Now consider $n = p(d + 1) - 1$ where $\nu_p(n) = a$, and suppose that $n = d$ satisfies the induction hypothesis for the prime $p$. Clearly $\nu_p(d + 1) = a - 1$. Then we have \[\sum_{i=0}^n \binom ni^k \equiv \sum_{i=0}^n (-1)^{Mk} \binom{\lfloor n/p \rfloor}{\lfloor i/p \rfloor}^k \equiv p\sum_{i=0}^d \binom di^k \pmod {p^a}\]By the induction hypothesis, $p^{a-1}$ divides the inner binomial sum, so since we are multiplying it by $p$, $p^a$ must divide $\sum_{i=0}^n \binom ni^k$.


Proof that $k$ odd fails for $p \neq 2$: For $n = p - 1$, we clearly have \[\sum_{i=0}^n \binom ni^k \equiv \sum_{i=0}^n (-1)^{Mk} \equiv 1 \pmod p.\]
Now consider $n = p(d + 1) - 1$ where $\nu_p(n) = a$, and suppose that $n = d$ satisfies the induction hypothesis for the prime $p$. Since $p - 1$ is even, there exists one more $(-1)^M = 1$ than $(-1)^M = -1$ for each block of $p$ such that $\lfloor i/p \rfloor$ remains constant. Then we have \[\sum_{i=0}^n \binom ni^k \equiv \sum_{i=0}^n (-1)^{Mk} \binom{\lfloor n/p \rfloor}{\lfloor i/p \rfloor}^k \equiv \sum_{i=0}^d \binom di^k \pmod {p^a}.\]By the induction hypothesis, $p^{a-1}$ does not divide this sum, so $p^a$ does not divide it either.


Proof that $k$ odd works for $n + 1$ a power of $2$: Let $p = 2$. For $n = 1$, clearly odd $k$ work as $1^k + 1^k \equiv 0 \pmod 2$. Now suppose odd $k$ works for $n = 2^d - 1$. If we let $n = 2^{d+1} - 1$, then we have \[\sum_{i=0}^n \binom ni^k \equiv \sum_{i=0}^n (-1)^{Mk} \binom{\lfloor n/p \rfloor}{\lfloor i/p \rfloor}^k \pmod{2^{d+1}}.\]
Notice that the since for each block of $p = 2$ such that $\lfloor i/p \rfloor$ remains constant, there is exactly one odd $M$ and one even $M$. Thus, the sum simply vanishes $\bmod~2^{d+1}$.


Since all even $k$ work for all primes $p$, it follows by CRT that all even $k$ work. For $n + 1$ not a power of $2$, there exists an odd prime $p$ such that $p \mid n + 1$, which can be easily used to show that all odd $k$ fail.
This post has been edited 1 time. Last edited by plang2008, Mar 22, 2025, 4:53 PM
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awesomeming327.
1712 posts
#35 • 1 Y
Y by cubres
Define
\begin{align*}
f_k(n,i) &= \binom{n-1}{i}^k \\
F_k(n) &= \sum_{i=0}^{n-1}f_k(n,i)
\end{align*}We want to find all $k$ such that $n\mid F_k(n)$ for all positive integers $n\ge 2$. Clearly, letting $n=3$, we have
\[3\mid F_k(n,i)=1^k+2^k+1^k\]which forces $k$ to be even.

Now we show that when $k$ is even, $n\mid F_k(n)$. Let $\nu_p(n)=a\ge 1$. First, we prove a claim.

Claim 1: $f_k(n,ip+j)\equiv f_k(n,ip+j+1)\pmod{p^a}$ for all $0\le j\le p-2$.
Note that we have
\begin{align*}
f_k(n,ip+j+1) &= \left(\frac{(n-1)!}{(ip+j+1)!(n-ip-j-2)!}\right)^k \\
&= \left(\frac{(n-1)!}{(ip+j)!(n-ip-j-1)!}\right)^k\cdot \left(\frac{n-ip-j-1}{ip+j+1}\right)^k \\
&\equiv f_k(n,ip+j)\cdot 1\pmod {p^a}
\end{align*}
This implies that $f_k(n,i)\pmod {p^a}$ is constant given that $\lfloor i/p\rfloor$ is constant. Therefore,
\[F_k(n)\equiv p\sum_{i=0}^{n/p-1}f_k(n,ip)\pmod {p^a}\]We now continue to our second claim.

Claim 2: Then $f_k(n,ip)\equiv f_k(n/p,i)\pmod{p^{a-1}}$ for all $0\le i\le p-1$.
We proceed by induction on $i$. Note that if $i=0$ this simply says $1\equiv 1\pmod {p^{a-1}}$ which is trivially true. Now assume
\[f_k(n/p,i)\equiv f_k(n,ip)\equiv f_k(n,ip+p-1)\pmod {p^{a-1}}\]and we have
\begin{align*}
f_k(n,(i+1)p)&\equiv f_k(n,ip+p-1)\cdot \frac{(n-(i+1)p)}{(i+1)p} \\ 
&\equiv f_k(n/p,i)\cdot \left(\frac{n/p-i-1}{i+1}\right)^k \\
&\equiv f_k(n/p,i+1) \pmod {p^{a-1}}
\end{align*}Which completes the induction step.
Now we have
\[F_k(n)\equiv p\sum_{i=0}^{n/p-1}f_k(n,ip)\equiv pF_k(n/p)\pmod {p^a}\]so if $p^{a-1}\mid F_k(n/p)$ then $p^a\mid F_k(n)$. By induction we are done.
This post has been edited 1 time. Last edited by awesomeming327., Mar 22, 2025, 9:27 PM
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MathLuis
1523 posts
#36 • 2 Y
Y by KevinYang2.71, cubres
If $k$ is odd then $n=2$ fails, now if $k$ is even then from CRT all we need is to prove that $p^{\ell} \mid \sum_{i=0}^{m \cdot p^{\ell}-1} \binom{m \cdot p^{\ell}-1}{i}^k$.
For this matter notice that (for $y<p$) and some positive integer $x$ such that $m \cdot p^{\ell}-1>xp+y$ that:
\[ \binom{m \cdot p^{\ell}-1}{xp+y}=\left( \frac{(m \cdot p^{\ell}-1) \cdots (m \cdot p^{\ell}-p+1)(m \cdot p^{\ell}-p-1) \cdots )}{1 \cdots (p-1)(p+1) \cdots} \right) \cdot \frac{(mp^{\ell-1}-1) \cdots (mp^{\ell-1}-x)}{1 \cdots x} \equiv \pm \binom{m \cdot p^{\ell-1}-1}{x} \pmod{p^{\ell}} \]So now using this notice that we have $\sum_{i=0}^{m \cdot p^{\ell}-1} \binom{m \cdot p^{\ell}-1}{i}^k \equiv p \cdot \sum_{i=0}^{m \cdot p^{\ell-1}-1} \binom{m \cdot p^{\ell-1}-1}{i}^k \pmod{p^{\ell}}$ so we can induct down and throw CRT until we get a degenerate case of the divisibility prompt in which case it is a trivial result thus we are done :cool:.
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deduck
227 posts
#37 • 3 Y
Y by KevinYang2.71, LostDreams, cubres
The answer is even only.

Odds fail because of $n=2$.

Now let $n+1 = \Pi p_a^{e_a}$. To show that evens work, we will induct by taking mod $p_a^{e_a}$, then CRT finishes.

We want to prove $$\sum_{i=0}^n \binom{n}{i}^k = 0 (\text{mod } p_a^{e_a})$$.

We will proceed by induction based on $v_{p_a}(n+1)$. When $v_{p_a}(n+1) = 0$ it's obvious.

For the inductive step, let's look at each $\binom{n}{i}^k$ individually. Note that $\binom{n}{i}^k = \Pi \frac{(n+1)-x}{x}$. Therefore, if $p_a \nmid x$, then both the numerator and denominator of $\frac{(n+1)-x}{x}$ are relatively prime, therefore it's $-1$. But since $k$ is even $(-1)^k=1$ and it does nothing. So just ignore all $x$ with $p_a \nmid x$.

Therefore $$\binom{n}{i}^k = \Pi \frac{(n+1)-x}{x} = \Pi_{p_a | x} \frac{\frac{n+1}{p_a} - \frac{x}{p_a}}{\frac{x}{p_a}} = \binom{\frac{n+1}{p_a}-1}{\lfloor \frac{i}{p_a} \rfloor}^k.$$
Therefore $$\sum_{i=0}^n \binom{n}{i}^k = \sum_{i=0}^n \binom{\frac{n+1}{p_a} - 1}{\lfloor \frac{i}{p_a} \rfloor}^k (\text{mod } p_a^{e_a})$$
Finishes by inductive assumption
This post has been edited 1 time. Last edited by deduck, Mar 23, 2025, 5:46 AM
Reason: typo bruh
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deduck
227 posts
#38 • 1 Y
Y by cubres
Motivation:
We can see that to move from $\binom{n}{i}$ to $\binom{n}{i+1}$ we see that we multiply by $\frac{n-i}{i+1}$, and this fraction is $-1$ when $p$ and $i$ are relatively prime. So that eliminates all odd cases in general no matter which $n$ u picked as long as $n$ is prime (i just said n=2 because it takes less explanation). Because if it's odd then $(-1)^k = -1$ and it breaks but we need $(-1)^k=1$.

However the issue is what if $i$ and $p$ aren't relatively prime?

Obviously first take a prime $p^e$ from $n$ to make it easier because CRT duh

But then everything is divisible by whichever prime $p$ that we picked and then it's easy to see we can use induction after shrink the binomial coefficient by $p$ on the top and bottom

i fakesolved it first in like 10 min then i was like sus i did not think about $i$ and $n$ aren't relatively prime, im simple minded lmao
This post has been edited 2 times. Last edited by deduck, Mar 23, 2025, 6:37 AM
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atdaotlohbh
186 posts
#40 • 1 Y
Y by cubres
The answer is all $k \ \vdots \ 2$.
To prove odd numbers don't work, consider $n=3$. Then we need $$1^k+2^k+1^k=2+2^k \ \vdots \ 3$$which tells us $k$ should be even.

Now we will prove that all evens work. To do that, let's prove that if $n+1 \ \vdots \ p^{\alpha}$, then $\sum_{i=0}^n \binom{n}{i}^k \ \vdots \ p^{\alpha}$.
We will do it using induction on $\alpha$. Base case $\alpha=0$ is trivial.
Now suppose $\alpha=u$ works, let's prove $u+1$ works. Notice that $$\frac{n+1-r}{r} \equiv -1 \pmod {p^{u+1}} \text{ for all } r \not\vdots p$$It implies that $$\binom{n}{pi}^k \equiv \binom{n}{pi+r}^k \pmod {p^{u+1}} \text{ for all } 0 \leq r \leq p-1$$Thus $$\sum_{i=0}^n \binom{n}{i}^k \equiv p\sum_{i=0}^{\frac{n+1}{p}-1} \binom{n}{pi}^k \pmod {p^{u+1}}$$So it remains to prove that $$\sum_{i=0}^{\frac{n+1}{p}-1} \binom{n}{pi}^k \ \vdots \ p^u$$But still by the fact above $$\binom{n}{pt}^k=(\prod_{i=1}^{pt} \frac{n+1-i}{i})^k \equiv (\prod_{i=1}^{t} \frac{n+1-pi}{pi})^k \equiv (\prod_{i=1}^{t} \frac{\frac{n+1}{p}-i}{i})^k \equiv \binom{\frac{n+1}{p}-1}{t}^k \pmod {p^u}$$And so we need to prove that $$\sum_{i=0}^{\frac{n+1}{p}-1} \binom{\frac{n+1}{p}-1}{t}^k \ \vdots \ p^u$$And that is our induction hypothesis. The induction is done.

As $\sum_{i=0}^n \binom{n}{i}^k$ is divisible by every prime power of $n+1$, it is divisible by $n+1$ and hence $\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k$ is an integer.
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Mathandski
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#41 • 2 Y
Y by KevinYang2.71, cubres
If you see any issue in the following solution, please email me at westskigamer@gmail.com.
This solution was what I officially submitted. I have transcribed it per verbatim.

Submitted Solution
This post has been edited 1 time. Last edited by Mathandski, Mar 26, 2025, 4:34 AM
Reason: fix typo
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Fibonacci_11235
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#42
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While solving the problem, I discovered that
${\binom{n}{0}}^2 + {\binom{n}{1}}^2 + \dots + {\binom{n}{n}}^2 = \binom{2n}{n}$

which immediately settles the case when $k=2$, using the fact that $C_n = \frac{1}{n+1} \binom{2n}{n}$ is an integer. Motivated by this observation, I tried evaluating the sum for other values of $k$, but made no progress. Is it possible to solve the problem in this way?
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Irreplaceable
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#43
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very scary indeed
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zoinkers
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#44 • 1 Y
Y by Fibonacci_11235
Fibonacci_11235 wrote:
While solving the problem, I discovered that
${\binom{n}{0}}^2 + {\binom{n}{1}}^2 + \dots + {\binom{n}{n}}^2 = \binom{2n}{n}$

which immediately settles the case when $k=2$, using the fact that $C_n = \frac{1}{n+1} \binom{2n}{n}$ is an integer. Motivated by this observation, I tried evaluating the sum for other values of $k$, but made no progress. Is it possible to solve the problem in this way?

no, as far as i know there is no closed form for the sum when $k \ge 3$
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programjames1
3046 posts
#45
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Newton Sums
Unfortunately this only works for odd $n$.
This post has been edited 2 times. Last edited by programjames1, Apr 13, 2025, 6:24 PM
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Mathgloggers
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#46
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Isn't just this babbage's congruence theorem which states that:
$\binom{ap}{bp} \equiv \binom{a}b mod(p)$,or what we trying to show here :
$\binom{mp^e -1}{ip} \equiv (-1)^k\binom{mp^{e-1}-1}{i}$,then applying our inductive hypothesis .
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