Spring 2016, problem 13

Consider the binomial coefficients $\binom{n}{k}=\frac{n!}{k!(n-k)!}$, where $k\in\{1,2,\ldots, n-1\}$. Determine all positive integers $n$ for which $\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}$ are all even numbers.

2 years ago

Hi, I'm Hubert from Paris ... Luca's theorem shows that $n=2^q$;

using the Wikipedia article notations with $p=2$ we want $N=\binom mn$ even for $1\leq n\leq m-1$,

• if $m$ is not a power of $2$, then $m_k=m_j=1$ for some $j\lt k$ and the given formula says $\binom m{2^j} \equiv \binom 11 (mod\; 2)$ with $n_j=1,\; n_k=0, \; k\neq j$, odd;

• now if $m=2^k,\; m_j=0,\; j\lt k$ and $n\lt m\Rightarrow~n_k=0$, and at least one of the $n_0,n_1,\cdots n_{k-1}$ is one, putting a $0=\binom 01$ in the given product, for each $n\in(0,m)$, making each $N$ even.

Note: this strange $\lt$ is the symbol "less than", can someone fix that ?

I need your help now ... :-(