Determine all possible quadruplet(s) (A, B, C, D) of nonnegative integer(s), with A < B < C, that satisfy this equation:

A! + B! + C! = 3^D

**edited due to initial misreading of problem, I read non-negative as positive for some reason and thus left out the a=0 possibilities**

let b=a+x and c=a+y with y>x then we have
a!+(a+x)!+(a+y)!=3^d  we can then factor out a! on the left
a!(1+(a+x)!/a!+(a+y)!/a!)=3^d
then a!=3^k for some k>=0
this is only true for k=0 thus we have
a!=1 and a=0 or a=1
if a=0 then we have
b!+c!+1=3^d
we then have b<3 because otherwise both b! and c! would be multiples of 3 and thus b!+c!+1 could not be a power of 3.
so we have 0<b<3 so that leaves b=1 b=2
if b=1 we have
c!+2=3^d
again we have c<3 otherwise c!+2 won't be multiple of 3
since c>b then we are left with c=2
a!+b!+c! then is 1+1+2=4 and is not a power of 3.
if on the other hand b=2 then we have
c!+3=3^d
c!=3^d-3
c!=3(3^(d-1)-1)
and c! must have only 1 factor of 3 thus c=3,4 or 5
by inspection only c=3 and c=4 work and thus we have
(0,2,3,2) and (0,2,4,3)
for a=1 we can use similar arguments to arrive at further solutions
(1,2,3,2) and (1,2,4,3)

Edited on August 28, 2009, 9:45 pm

Posted by Daniel on 2009-08-27 11:33:12