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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 ?

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    What's $l$ in $p^l$? Is it the same as the one in $m^l$? If so, are you assuming in the second part that the equation $\binom{n}{k}=m^l$ *does* have a solution? If not, how is $l$ determined?2012-10-19
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    l is an integer greater or equal than 2. It's the same than the one in $m^l$...2012-10-19
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    ...Yes, the proof is based on assuming the fact that $\binom{n}{k}=m^l$ has an integer solution for l$\ge$2....2012-10-19

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