Re: Number theory

From: Helmut Richter <hhr-m@xxxxxx>
Date: Fri, 28 Dec 2007 22:37:07 +0100

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 <= n²

instead of

n <= p <= 2n

would suffice. Would that be easier to prove?

--
Helmut Richter
.

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Re: Number theory
... On Fri, 28 Dec 2007, Helmut Richter wrote: ... is a perfect square, so p^2 divides n!. ... Then q < n, so q divides n!, contradicting p being the largest prime dividing ... would suffice. ...
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Re: Number theory
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Re: Number theory
... is a perfect square, so p^2 divides n!. ... Then q < n, so q divides n!, contradicting p being the largest prime dividing ... For this problem a much weaker form of Bertrand's postulate, ...
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