Re: Number theory

Kenshin wrote:
Does there exist a natural number n >=2 such that n! = n*(n-1)*...*2*1
= k^2 for some integer k???

There is no such n.

Suppose there were. Then, let p be the largest prime dividing n!. By hypothesis, n! is a perfect square, so p^2 divides n!. Therefore, we must have 2p <= n. But, by Bertrand's postulate, there is a prime q with p < q < 2p. Then q < n, so q divides n!, contradicting p being the largest prime dividing n!.

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Eric Schmidt

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