Re: Number theory

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?

instead of

n <= p <= 2n

would suffice. Would that be easier to prove?

--
Helmut Richter

.

Relevant Pages

  • Re: Number theory
    ... On Fri, 28 Dec 2007, Eric Schmidt wrote: ... is a perfect square, so p^2 divides n!. ... Then q < n, so q divides n!, contradicting p being the largest prime dividing ... Helmut Richter ...
    (sci.math)
  • 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 ... see an ampersand. ...
    (sci.math)
  • 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, ...
    (sci.math)
  • Re: minimal square
    ... On Tue, 3 Apr 2007, Helmut Richter wrote: ... rest was okay. ... wit augmenting d to a perfect square. ...
    (sci.math)