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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

50.0. "n!=x^2" by HARE::STAN () Thu Mar 15 1984 02:10

From:	ROLL::USENET       "USENET Newsgroup Distributor" 14-MAR-1984 22:06
To:	HARE::STAN
Subj:	USENET net.math newsgroup articles

Newsgroups: net.math
Path: decwrl!decvax!genrad!grkermit!masscomp!clyde!floyd!cmcl2!philabs!aecom!schwadro
Subject: prove n! isnt a perfect square for n>1
Posted: Sun Mar 11 20:50:56 1984



                  2
prove that n! <> k

where n,k are integers and n>1.

solution will be posted next week.


           schwady

p.s.

try to give an elegant proof!!!!!
T.RTitleUserPersonal
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50.1HARE::STANThu Mar 15 1984 02:127
It is a classic unsolved problem to find all the solutions to
the diophantine equation

		  2
	n! + 1 = x

Several (fallacious) proofs have appeared in the literature.
50.2RANI::LEICHTERJThu Mar 15 1984 03:578
Solution to original problem:
Consider the largest prime < k.  It divides k!, but it cannot possibly divide
it more than once (i.e. its square can't divide k!).  Hence that prime
occurs with an odd power (1) in the factorization of k! ==> k! is not a
square.

Was this supposed to be tricky in some way?
							-- Jerry
50.3ULTRA::HERBISONMon Mar 19 1984 13:317
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.
50.4HARE::STANTue Mar 27 1984 00:1635
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 ~~
    %%%%%%%%%%%%%%%%%%
50.5HARE::STANThu Mar 29 1984 00:0924
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