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Path: utzoo!watmath!clyde!akgua!mcnc!unc!ulysses!mhuxl!eagle!harpo!seismo!ut-sally!ut-ngp!lsv
From: lsv@ut-ngp.UUCP
Newsgroups: net.math
Subject: re:Factorials in an Equation -- solution
Message-ID: <464@ut-ngp.UUCP>
Date: Tue, 3-Apr-84 16:38:01 EST
Article-I.D.: ut-ngp.464
Posted: Tue Apr  3 16:38:01 1984
Date-Received: Sat, 7-Apr-84 00:34:42 EST
References: <460@aecom.UUCP>
Organization: Comp. Center, Univ. of Texas at Austin
Lines: 47


Answers: (A,B,C) = (0,0,2) 
                   (0,1,2) 
                   (1,0,2) 
                   (1,1,2) 
 
Solution: Given: 
 
          (*)    A! + B! = C!, 
          note that as A! and B! are both greater than zero, 
          C! > A! and C! > B! and, therefore C > A and C > B. 
 
          Consider the case where A >= B.  Multiply both sides of (*) 
          by 1/B!, giving 
 
          (**)   A!/B! + 1 = C!/B!. 
          Assume A = 0.  Then B = 0, A! = 1, B! = 1, C! = 2, and 
          therefore, C = 2.  Assume A = 1.  Then B = 0 or 1, A! = 1, 
          B! = 1 (for both values of B), and again C = 2.  If A > 1, 
          C >= A + 1.  Rearranging the terms of (**), we have: 
 
                 1 = C!/B! - A!/C! 
                 1 = [(C)(C-1)...(C-(C-(A+1))) - 1] * A!/B!. 
          Since A >= B, A!/B! >= 1, and therefore: 
 
                 1 >= [(C)(C-1)...(C-(C-(A+1))) - 1] 
                 2 >= (C)(C-1)...(C-(C-(A+1))) 
          But, since C >= A + 1, (C)(C-1)...(C-(C-(A+1))) >= A + 1. 
          So: 
 
                 2 >= A + 1. 
          But A > 1, or A >= 2, so: 
 
                 2 >= 3, 
          which is impossible.  Therefore, if A >= B, (*) has no 
          solutions if A > 1. 
 
          By performing the above steps again using B > A (merely 
          swapping B's for A's and A's for B's everywhere from 
          "Consider the case where A >= B" on)  the last answer, (0,1,2) 
          is generated, and it is shown that if B > 1 and B >= A, 
          no solutions exist for (*).  Therefore {(0,0,2), (1,0,2), 
          (0,1,2), and (1,1,2)} is the complete solution set to (*). 

                                          Stuart Vance
                                          (...harpo!seismo!ut-sally!ut-ngp!lsv)