**poppers@aecom.UUCP (Michael Poppers)** (03/25/84)

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 ~~ %%%%%%%%%%%%%%%%%%