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From: poppers@aecom.UUCP (Michael Poppers)
Newsgroups: net.math
Subject: n! (n>1) not a perfect square - a proof? (not in rot13)
Message-ID: <442@aecom.UUCP>
Date: Sun, 25-Mar-84 14:36:15 EST
Article-I.D.: aecom.442
Posted: Sun Mar 25 14:36:15 1984
Date-Received: Mon, 26-Mar-84 20:49:52 EST
Organization: Albert Einstein Coll. of Med., NY
Lines: 31


    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.


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    $ PERITUS CLAVIS $                        Michael Poppers
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