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Factorial Difference = Perfect Square (Posted on 2011-03-28) Difficulty: 3 of 5
Determine all possible pair(s) (x, y) of nonnegative integers such that (x!*y! - x! - y!) is a perfect square.

No Solution Yet Submitted by K Sengupta    
Rating: 2.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: possible pairs | Comment 2 of 3 |
(In reply to possible pairs by Dej Mar)

For totals of x + y through 1003, the only sets are (2,2), (2,3) and (3,2):

list
   10   for T=0 to 999999
   20     for X=0 to int(T/2)
   30        Y=T-X
   40        Fx=!(X):Fy=!(Y)
   50        Tst=Fx*Fy-Fx-Fy
   55        if Tst>=0 then
   60          :Sr=int(sqrt(Tst)+0.5)
   70          :if Sr*Sr=Tst then print X;Y,Tst,Sr
   80     next
   90   next
OK
run
 2  2    0       0
 2  3    4       2
Overflow in 60
?t
 1004
OK

(The program looks only for y>x, but the puzzle doesn't so require; therefore the solution includes a reversal of a found pair.)


  Posted by Charlie on 2011-03-28 14:29:09
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