Define p ⊗ q = p+qp*q, and find all possible triplets (a, b, c) of integers such that:
a ⊗ ( b ⊗ c) + b ⊗ (c ⊗ a) + c ⊗ (a ⊗ b) = 0
The equation simplifies to
a+b+cabacbc+abc=0 [by inspection (0,0,0) is a solution.]
which can be solved for c
c=(abab)/(abab+1)
an expression of the form n/(n+1) is only an integer if n=0 or n=2
abab=0 has solution a=2, b=2 so the triplet (2,2,0)
abab=2
can be rewritten as
b=(a2)/(a1) is also of the form n/(n+1)
a2=0 has solution a=2, b=0 so the triplet (2,0,2)
a2=2 has solution a=0, b=2 so the triplet (0,2,2)

Posted by Jer
on 20150326 08:17:28 