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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 () Wed Mar 14 1984 23:10

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.R	Title	User	Date	Lines
50.1		HARE::STAN	`Wed Mar 14 1984 23:12`	7
	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.
50.2		RANI::LEICHTERJ	`Thu Mar 15 1984 00:57`	8
	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
50.3		ULTRA::HERBISON	`Mon Mar 19 1984 10:31`	7
	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.
50.4		HARE::STAN	`Mon Mar 26 1984 21:16`	35
	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 ~~ %%%%%%%%%%%%%%%%%%
50.5		HARE::STAN	`Wed Mar 28 1984 21:09`	24
	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