1994-04-11 - Re: (n!+1)^(1/2) Oops! I’m wrong.

Header Data

From: collins@newton.apple.com (Scott Collins)
To: Matthew J Ghio <mg5n+@andrew.cmu.edu>
Message Hash: 49dab908684fc472791872df34f4bbbf862c1435d8975753b2bd320e01cffe9a
Message ID: <9404112043.AA28093@newton.apple.com>
Reply To: N/A
UTC Datetime: 1994-04-11 21:59:54 UTC
Raw Date: Mon, 11 Apr 94 14:59:54 PDT

Raw message

From: collins@newton.apple.com (Scott Collins)
Date: Mon, 11 Apr 94 14:59:54 PDT
To: Matthew J Ghio <mg5n+@andrew.cmu.edu>
Subject: Re: (n!+1)^(1/2) Oops! I'm wrong.
Message-ID: <9404112043.AA28093@newton.apple.com>
MIME-Version: 1.0
Content-Type: text/plain


    >For any number a, 1<a<=n, n! mod a == 0; therefore, n!+1 mod a == 1.  n!+1
    >is prime.  Prime numbers don't have integral square roots.

  >For example :
  >
  >(4!+1)^(1/2)=5
  >(5!+1)^(1/2)=11
  >(7!+1)^(1/2)=71

I am completely wrong.  I replied too hastily.  Please accept my apologies.
 In fact, n!+1 is relatively prime to any a, 1<a<=n, however plainly it is
much larger than n itself and when n>3, (n!+1)>(n^2) and may have factors
(including an integral square root) larger than n.

Oops :-)


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