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 Factorials to Squares (Posted on 2009-02-25)
Find three different positive integers whose factorials are each one less than a perfect square, such that the total of all three factorials is also a perfect square.

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 Integer sequence for n! +1 = p^2 (help) Comment 3 of 3 |
http://www.research.att.com/~njas/sequences/A146968

4,5 and 7 appear to be the only three numbers whose factorials are one less than a perfect square.  I'm trying to see if I can prove this.

Say n! = p^2 - 1 for some natural number p
n! = (p-1)(p+1)

4! = 24 = (4)(6) = (4)(2*3)
5! = 120 = (10)(12) = (2*5)(3*4)
7! = 5040 = (70)(72) = (2*5*7)(3*4*6)

Clearly the two sides (p-1) and (p+1) must both be even and only one of them can have 4 as a factor.  In addition only one can have 3 as a factor.  Only only one can have 5 as a factor, etc...

Eventually one side gets way overbalanced.
For example 12! = 12*11*10*9*8*7*6*5*4*3*2
The 3, 6, 9, and 12 must go on the same side which gives it a factor of 4 so the 2, 4, and 8 must also go on this side.
This leaves only 5, 7, 10 and 11 for the other side.
This gives 124416 compared to 3850 as the closest these number can be.

Presumably one could find a bound for which one side will _always_ well exceed the other.  (12 may well be this bound but I'm not sure how to proceed)

 Posted by Jer on 2009-02-26 14:55:52

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