Question 404544: the equation is written to the base 3n-1 instead of base 10 (a,b,c) range from 0 to 3n-2
i have solved it up to the point but can't find a method to find a,b,c
2[a(3n-1)^2 + b(3n-1) + c)] = c(3n-1)^2 + b(3n-1) + a
a(18n^2 - 12n + 1) + b(3n-1) = c(9n^2 -6n -1)
(18a-9c)^2 + (-12a+3b+6c)n + (a+b+c) = 0
this is the bit where am stuck how to obtain a,b and c
the answers should be a=n-1 b=3n-2 c=2n-1
thanks
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! The easiest way is to try small cases (i.e. n = 1, 2, 3, etc.), then prove your result using induction. However there's quite a bit of algebra, and the only flaw is that the solution doesn't prove uniqueness.
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