Define p ⊗ q = p+q-p*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+c-ab-ac-bc+abc=0 [by inspection (0,0,0) is a solution.]
which can be solved for c
c=(ab-a-b)/(ab-a-b+1)
an expression of the form n/(n+1) is only an integer if n=0 or n=-2
ab-a-b=0 has solution a=2, b=2 so the triplet (2,2,0)
ab-a-b=-2
can be rewritten as
b=(a-2)/(a-1) is also of the form n/(n+1)
a-2=0 has solution a=2, b=0 so the triplet (2,0,2)
a-2=-2 has solution a=0, b=2 so the triplet (0,2,2)
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Posted by Jer
on 2015-03-26 08:17:28 |
algebraic solution
