# HELP! Proofs from the Book: Binomial coefficients are almost never powers

this is Paul Erdos's proof provided by the book:

Theorem: let $l$ $\geq$ $2$ and $4$ $\leq$ $k$ $\leq$ $n$ $-$ $4$, then the equation ${n\choose k}$ = $m^{l}$ does not have an integer solution.

(Part 1)

"...we may assume $n$ $\geq$ $2k$ because of ${n\choose k}$ = ${n\choose n - k}$. Assume the theorem is false. By Sylvester's theorem, there is a prime factor $p$ of ${n\choose k}$ greater than $k$, hence $p^{l}$ divides $n$$(n-1)$$(n-2)$ ... $(n-k+1)$. Clearly, only one of the factors $n-i$ can be a multiple of any such $p>k$."

(Part 2)

"Consider any factor $n-j$ of the numerator and write it in the form $a_j$$m_j^{l}, where a_j is not divisible by any non-trivial lth power. We note by part 1 that a_j only has prime divisors less or equal to k." It says that there is one of the factors a_j that can be multiple of p>k. But why not any a_j from any factor n$$(n-1)$ ... $(n-k+1)$ can have a prime divisor $p>k$?

Note that $n - (n-k+1) = k-1$. Thus, if there exists a natural number $m$ and some factor $a_j$ such that $a_j = mp$ for $p > k$, then we have $(m+1)p > a_j + k > n > (n-k+1) > a_j - k > (m-1)p$.
• $a_j$ is an integer between $n$ and $n-k+1$. Therefore $a_j \geq n-k+1$ and thus $a_j + k \geq n-k+1+k = n+1 > n$. Also, please upvote and mark my answer if this has indeed answered your question. – Berni Waterman Sep 30 '17 at 9:55