Sylvester Theorem

Question

Bonjour,

The equation $\binom{n}{k}=m^l$ has no entire solution for l$\ge$2 and 4$\le$k$\le$n-4. Suppose that n$\ge$2k (since $\binom{n}{k}=\binom{n}{n-k}$).

According to the Sylvester theorem, the binomial coefficient (for n$\ge$2k):

\begin{equation} \binom{n}{k}=\frac{n(n-1)...(n-k+1)}{k!}, \end{equation}

has always a prime factor p greater than k.

I dont't know why this fact implies that $p^l$ divides $n(n-1)...(n-k+1)$.

And I dont't understand why only one of the factors n-i can be a multiple of p.

Can someone please help ?

Answer 1

I don't understand the structure of what you're doing; it seems as if you're trying to carry out a proof by contradiction but forgot to announce it. Anyway, based on the comment in which you say that $\binom nk=m^l$ is meant to be assumed (even though you state at the beginning of the question that it can't hold), it follows that each prime factor of $\binom nk$ occurs a multiple of $l$ times, so if by Sylvester's theorem some prime $p\gt k$ divides $\binom nk$, then it must occur at least $l$ times, so $p^l$ divides $\binom nk$, so it must divide the numerator.

Answer 2

The factors $n-i$ differ by at most $k$, and $k$ is less than $p$, so only one factor can be a multiple of $p$.