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Bonjour à tous !
Let us have some integers n, k, r satisfiying::
$n/2 \geq \ k \geq 4, r>1$
Let's suppose that there exist an integer a such that:
${n \choose k}=a^r \quad (n \geq 2k)$
From every factor in the numerator of:
${n \choose k}=\frac{n(n-1)...(n-k+1)}{k!}$
let us factor out the largest rth power that divides it:
$n-i=b_ic_i^r \quad (i=0,1,...,k-1)$
Here $b_i$ is not divisible by an rth power greater than 1. And $b_i$ cannot have a prime divisor greater than k.
Of course, if there is a prime p such that $p$ divides $a$ and $p>a$ then the denominator is not divisible by p, and in the numerator there are only k factors, and hence p can divide only one of the terms n-i. (no problem with that !)
Since the prime p doesn't divide any $b_i$, it must be included in the factor $c_i^r$. So in the canonical decomposition of the unique factor $n-i$ of the numerator that p divides, $p^r$ must appear to some power. (no problem with that)
But can someone tell me why each and every factor can be written as a multiple of an rth power ?
I know that if a prime p divides $a^r$ then $p^r$ divides $a^r$ for some $r>1$. And if we write:
$n(n-1)...(n-k+1)=k!a^r$
then we find out that $p^r$ divides the numerator $n(n-1)...(n-k+1)$
But what about factors of the numerator of $n \choose k$ taken indivually ? How do we establish that they are all of the form $b_ic_i^r \quad(i=0,1,...,k-1)$ ? What is the order of magnitude of $c_i^r$ ?
Can someone please give me an elementary arithmetical explanation ?
