Each of A, B and C is an integer satisfying:

(A-B)(B-C)(C-A) = A+B+C

Does 27 always divide A+B+C?
If so, prove it. If not, provide a counterexample.

Actually 54 divides A+B+C

First to show A+B+C is even:
(A-B)+(B-C)= A-C = -(C-A)
Since these terms can't all be odd, the product is even.

WLOG let A≤B≤C
Let (A-B)=-x, (B-C)=-y, (C-A)=x+y
so
B=C-y and A=C-x-y
the original equation then becomes
(-x)(-y)(x+y)=(C-x-y)+(C-y)+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 2014-03-29 15:27:39