Re: Number theory

On Dec 28, 11:04 pm, William Elliot <ma...@xxxxxxxxxxxxxxxxxx> wrote:
On Fri, 28 Dec 2007, Helmut Richter wrote:
On Fri, 28 Dec 2007, Eric Schmidt wrote:

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!.

For this problem a much weaker form of Bertrand's postulate, to wit

n <= p <= n&

Your post did not come through clearly.
What does & represent?

Hi, William:

In my newsgroup client it displays as the upper ASCII character
superscript two (ie. squared) in Helmut's post. In your post I
see an ampersand.

regards, chip
.

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