0
votes
1answer
56 views

The binomial coefficients $\binom{n}{ p}$ are divisible by a prime $p$ only if $n$ is a power of $p.$

I'm looking for a "high school / undergraduate" demonstration for the: All the binomial coefficients $\binom{n}{i}=\frac{n!}{i!\cdot (n-i)!}$ for all i, $0\lt i \lt n$, are divisible by a prime $p$ ...
4
votes
1answer
257 views

Divisibility of binomial coefficient by prime power - Kummer's theorem

Let's say we have binomial coefficient $\binom{n}{m}$. And we need to find the greatest power of prime $p$ that divides it. Usually Kummer's theorem is stated in terms of the number of carries you ...
4
votes
2answers
90 views

Proofs from the Book - need quick explanation

I've been recently reading this amazing book, namely the chapter on Bertrand's postulate - that for every $n\geq1$ there is a prime $p$ such that $n<p\leq2n$. As an intermediate result, they prove ...
3
votes
1answer
89 views

On Elements of $p$th Row in n Pascal's Triangle (For Prime $p$)

If $p$ is a prime number, in Pascal's triangle all the terms in the $p$th row - except the 1s - are multiples of $p$ . It's easy to prove this property using the formula for $\binom{p}{k}$. Is there ...
1
vote
0answers
30 views

Binomial coefficient modulo composite number [closed]

For a prime number n , ${n \choose r} \mod n$ is 0 for all 0 < r < n If n is composite I want to know for which values of r the value of ${n \choose r} \mod n$ becomes 0 (for r in ...
4
votes
2answers
178 views

Proving that $n\mid(nCr)$ for all $r$ ($1 \leq r \leq n-1$), only if $n$ is prime

I'm trying to prove that $n\mid(nCr)$ for all $r$ ($1 \leq r \leq n-1$) if and only if $n$ is prime. Now proving that if $n$ is prime then $n\mid(nCr)$ is pretty easy, but how would you go about ...
3
votes
1answer
83 views

Bertrand's postulate proof

Regarding http://michaelnielsen.org/polymath1/index.php?title=Bertrand%27s_postulate I think the last inequality should be $4^{n/3}\le(2n+1)(2n)^{\sqrt{2n}}$. But even when the RHS is decreased from ...
4
votes
1answer
80 views

Binomial theorem for prime exponent

Could you explain to me why for prime $p$ we have the following? $$(x+y)^p - (x^p + y^p)= x^p + \binom{p}{1}x^{p-1}y + \binom{p}{2}x^{p-2}y^2 + \binom{p}{p-1}xy^{p-1} + y^p.$$ I found it here: ...
2
votes
0answers
195 views

Prime numbers with binomial coefficients

Let $p$ be an odd prime and $n$ a positive integer. Prove that $p+1$ divides $n$ if and only if $$\sum_{k\equiv j\pmod{p-1}}^n^{}\binom{n}{k}(-1)^{\frac{(k-j)}{p-1}}\equiv 0 \mod p$$ for every $$j\in ...
5
votes
1answer
83 views

Prime numbers with binomial coefficients

Question: Prove that for any prime $p>3$, the number $\binom{2p-1}{p-1}-1$ is divisible by $p^{3}$. Attempt: Since every integer that is relatively prime to p has a multiplicative inverse modulo ...
2
votes
1answer
77 views

Criterion for Wolstenholme Primes

Wolstenholme Theorem is a nice theorem that states that every prime $p >3$ satisfies: $$\binom{2p}{p} \equiv 2 \pmod {p^3}$$ A Wolstenholme prime is a prime $p$ such that $\binom{2p}{p} \equiv 2 ...
4
votes
2answers
364 views

Proving ${p-1 \choose k}\equiv (-1)^{k}\pmod{p}: p \in \mathbb{P}$ [duplicate]

Possible Duplicate: Prove $\binom{p-1}{k} \equiv (-1)^k\pmod p$ The question is as follows: Let $p$ be prime. Show that ${p \choose k}\bmod{p}=0$, for $0 \lt k \lt p,\space ...
10
votes
2answers
251 views

The Fermat prime 257 and binomial sum $\sum_{n=0}^\infty \frac{(-1)^n}{\binom {8n}{4n}}$?

We have, $\begin{aligned} \sum_{n=0}^\infty \frac{(-1)^n}{\binom n{n/2}} &= \frac{4}{27}(9-\pi\sqrt{3}\,)\\[2.5mm] \sum_{n=0}^\infty \frac{(-1)^n}{\binom {2n}n} &= \frac{4}{5} - ...
3
votes
0answers
315 views

Sylvester's Theorem and Schur Theorem

I'll probably end up asking more programming questions on StackExchange forums than math questions, but I'll lead off with a math question. In my Number Theory class this past semester, I worked on a ...
16
votes
7answers
3k views

Prime dividing the binomial coefficients

It is quite easy to show that for every prime $p$ and $0<i<p$ we have that $p$ divides the binomial coefficient $\large p\choose i$; one simply notes that in $\large \frac{p!}{i!(p-i)!}$ the ...
7
votes
3answers
207 views

Combinatorics question: Show divisibility

Let $a\geq2$, $b\geq2$ be two prime numbers and k be a natural number with $k\leq min(a,b)$. How can one show that $z := \binom{a+b}{k} - \binom{a}{k} - \binom{b}{k}$ is divisible by the product ...
8
votes
2answers
1k views

prime numbers in Pascal's triangle

Just wondering about this: Is it true that there are no prime numbers in Pascal's triangle, with the exception of $\binom{n}{1}$ and $\binom{n}{n-1}$? From the first 18 lines it appears that this is ...