| Re: .2
This assumes that the prime is > k/2. I believe that for all n > 1,
there exists a prime p such that n/2 < p <= n, in fact I vaguely
remember proving something like that. However, it has been several
years since I worked with number theory.
B.J.
|
| Newsgroups: net.math
Path: decwrl!decvax!mcnc!philabs!aecom!poppers
Subject: n! (n>1) not a perfect square - a proof? (not in rot13)
Posted: Sun Mar 25 11:36:15 1984
Summing up the discussion to date, Mr. Schwadron postulated that n! is
not a perfect square for n>1, Mr. Davis (may Qrhf forgive him), stated:
> There is some largest prime less than or equal to n.
> Let this be called p. By a cute theorem p>n/2
> (no, I forget names of theorems). Hence n! contains
> exactly one factor of p and cannot be a perfect square.
> I hope this isn't considered proof by deus ex machina.
which seems to sound right, and Mr. Trigg appeared daunted by rot13. If I
may, here are my 2 cents(melted down from Jim's quarter):
Suppose n! is a square for n>1. Let p be the largest prime factor of n! .
Since all prime factors of a square have even multiplicity, 2p is also a
-
factor. But, by Bertrand's Postulate, there is a prime q with p < q < 2p ,
so that q is a factor of n! - a contradiction.
P.S. giving credit where it is due (but without interest), the 2 cents were
invested by Mr. DS of BHCA - David, your Swiss bank account is ready.
%%%%%%%%%%%%%%%%%%
$ PERITUS CLAVIS $ Michael Poppers
$ MACHINAE VIVIT $ ~~ poppers @ AECOM ~~
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|
| From: ROLL::USENET "USENET Newsgroup Distributor" 28-MAR-1984 21:06
To: HARE::STAN
Subj: USENET net.math newsgroup articles
Newsgroups: net.math
Path: decwrl!decvax!harpo!ulysses!unc!mcnc!ncsu!uvacs!gmf
Subject: n! = k**2 problem
Posted: Sat Mar 24 12:15:22 1984
From
Elementary Theory of Numbers
by W. Sierpinski (Warsaw, 1964):
p. 138:
"Corollary 4. For natural numbers > 1 number n! is not a k-th power
with k > 1 being a natural number."
Sierpinski gets this from a theorem of Tchebycheff, p. 137: "If n is
a natural number > 3, then between n and 2n-2 there is at least one
prime number." This in turn is obtained from: "If n is a natural
number > 5, then between n and 2n there are at least two different
prime numbers" (p. 137).
G. Fisher
|