1. Number Theory (2)

Find all $\displaystyle n \in \mathbb{N}$ such that $\displaystyle 4 \mid \binom{2n}{n}.$

2. Spoiler:

$\displaystyle 4 \mid \binom{2n}{n}$ if and only if $\displaystyle n$ is not of the form $\displaystyle 2^k$.
Proof:

Using $\displaystyle n!!$ as the double factorial of $\displaystyle n$:

$\displaystyle \binom{2n}{n} = \frac{(2n)!}{n!^2} = \frac{(2n)!! (2n-1)!!}{n!^2} = \frac{2^n n!(2n-1)!!}{n!^2} = \frac{2^n(2n-1)!!}{n!}$

$\displaystyle 2 \nmid (2n-1)!!$ and therefore $\displaystyle 4 \mid \binom{2n}{n}$ if and only if $\displaystyle d \le n-2$, where $\displaystyle d$ satisfies $\displaystyle 2^d \mid \mid n!$ (that is, $\displaystyle 2^d \mid n!$, but $\displaystyle 2^{d+1} \nmid n!$).
We'll use the fact that $\displaystyle n! = \prod_{p \le n} p^{\sum_{k=1}^{\infty} \lfloor n/p^k \rfloor}$ ($\displaystyle p$ prime) (Legendre's formula). The exponent of $\displaystyle 2$ in $\displaystyle n!$ is:
$\displaystyle d = \sum_{k=1}^{\infty} \lfloor \frac{n}{2^k} \rfloor = \sum_{k=1}^{\lfloor \log_2 n \rfloor} \lfloor \frac{n}{2^k} \rfloor \le \sum_{k=1}^{\lfloor \log_2 n \rfloor} \frac{n}{2^k} = n (1-2^{- \lfloor \log_2 n \rfloor}) \le n(1-2^{- \log_2 n}) = n-1$
Where the equality signs of $\displaystyle \le$ are taken if and only if $\displaystyle n$ is a power of two. Thus, $\displaystyle d \le n-2$ if and only if $\displaystyle n$ is not a power of $\displaystyle 2$, and so $\displaystyle 4 \mid \binom{2n}{n}$ if and only if $\displaystyle n$ is not of the form $\displaystyle 2^k$.
Q.E.D

3. Originally Posted by Unbeatable0
Spoiler:

$\displaystyle 4 \mid \binom{2n}{n}$ if and only if $\displaystyle n$ is not of the form $\displaystyle 2^k$.
Proof:

Using $\displaystyle n!!$ as the double factorial of $\displaystyle n$:

$\displaystyle \binom{2n}{n} = \frac{(2n)!}{n!^2} = \frac{(2n)!! (2n-1)!!}{n!^2} = \frac{2^n n!(2n-1)!!}{n!^2} = \frac{2^n(2n-1)!!}{n!}$

$\displaystyle 2 \nmid (2n-1)!!$ and therefore $\displaystyle 4 \mid \binom{2n}{n}$ if and only if $\displaystyle d \le n-2$, where $\displaystyle d$ satisfies $\displaystyle 2^d \mid \mid n!$ (that is, $\displaystyle 2^d \mid n!$, but $\displaystyle 2^{d+1} \nmid n!$).
We'll use the fact that $\displaystyle n! = \prod_{p \le n} p^{\sum_{k=1}^{\infty} \lfloor n/p^k \rfloor}$ ($\displaystyle p$ prime) (Legendre's formula). The exponent of $\displaystyle 2$ in $\displaystyle n!$ is:
$\displaystyle d = \sum_{k=1}^{\infty} \lfloor \frac{n}{2^k} \rfloor = \sum_{k=1}^{\lfloor \log_2 n \rfloor} \lfloor \frac{n}{2^k} \rfloor \le \sum_{k=1}^{\lfloor \log_2 n \rfloor} \frac{n}{2^k} = n (1-2^{- \lfloor \log_2 n \rfloor}) \le n(1-2^{- \log_2 n}) = n-1$
Where the equality signs of $\displaystyle \le$ are taken if and only if $\displaystyle n$ is a power of two. Thus, $\displaystyle d \le n-2$ if and only if $\displaystyle n$ is not a power of $\displaystyle 2$, and so $\displaystyle 4 \mid \binom{2n}{n}$ if and only if $\displaystyle n$ is not of the form $\displaystyle 2^k$.
Q.E.D
that's an unbeatable solution!

4. Originally Posted by Unbeatable0
Spoiler:

$\displaystyle 4 \mid \binom{2n}{n}$ if and only if $\displaystyle n$ is not of the form $\displaystyle 2^k$.
Proof:

Using $\displaystyle n!!$ as the double factorial of $\displaystyle n$:

$\displaystyle \binom{2n}{n} = \frac{(2n)!}{n!^2} = \frac{(2n)!! (2n-1)!!}{n!^2} = \frac{2^n n!(2n-1)!!}{n!^2} = \frac{2^n(2n-1)!!}{n!}$

$\displaystyle 2 \nmid (2n-1)!!$ and therefore $\displaystyle 4 \mid \binom{2n}{n}$ if and only if $\displaystyle d \le n-2$, where $\displaystyle d$ satisfies $\displaystyle 2^d \mid \mid n!$ (that is, $\displaystyle 2^d \mid n!$, but $\displaystyle 2^{d+1} \nmid n!$).
We'll use the fact that $\displaystyle n! = \prod_{p \le n} p^{\sum_{k=1}^{\infty} \lfloor n/p^k \rfloor}$ ($\displaystyle p$ prime) (Legendre's formula). The exponent of $\displaystyle 2$ in $\displaystyle n!$ is:
$\displaystyle d = \sum_{k=1}^{\infty} \lfloor \frac{n}{2^k} \rfloor = \sum_{k=1}^{\lfloor \log_2 n \rfloor} \lfloor \frac{n}{2^k} \rfloor \le \sum_{k=1}^{\lfloor \log_2 n \rfloor} \frac{n}{2^k} = n (1-2^{- \lfloor \log_2 n \rfloor}) \le n(1-2^{- \log_2 n}) = n-1$
Where the equality signs of $\displaystyle \le$ are taken if and only if $\displaystyle n$ is a power of two. Thus, $\displaystyle d \le n-2$ if and only if $\displaystyle n$ is not a power of $\displaystyle 2$, and so $\displaystyle 4 \mid \binom{2n}{n}$ if and only if $\displaystyle n$ is not of the form $\displaystyle 2^k$.
Q.E.D
Wow! Where did you learn how to do all that?