Actually 54 divides A+B+C
First to show A+B+C is even:
(AB)+(BC)= AC = (CA)
Since these terms can't all be odd, the product is even.
WLOG let A≤B≤C
Let (AB)=x, (BC)=y, (CA)=x+y
so
B=Cy and A=Cxy
the original equation then becomes
(x)(y)(x+y)=(Cxy)+(Cy)+C
x²y + xy² = 3C  x  2y
C=(x²y+xy²+x+2y)/3
Since C is an integer, (x²y+xy²+x+2y)=0mod3
if x=0mod3, y=0mod3
if x=1mod3, y²=2mod3 which is impossible
if x=2mod3, y²=2mod3 which is impossible
So both x and y are multiples of 3, and so x+y is a multiple of 3
therefore (x)(y)(x+y)=A+B+C is a multiple of 3³=27
combine this with the fact that A+B+C is even
to get A+B+C is a multiple of 54.
[There may be even stronger statements about A+B+C as I found none where A+B+C=106]

Posted by Jer
on 20140329 15:27:39 