lsv@ut-ngp.UUCP (04/03/84)
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)