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Hong Kong Preliminary Selection Contest

Algebra Difficulty 3.5 AMC 10/12 Prove it Hong Kong

Let f(x)=ax+bf(x) = ax + b where aa and bb are integers. If f(f(0))=0f(f(0)) = 0 and f(f(f(4)))=9f(f(f(4))) = 9, find the value of f(f(f(1)))+f(f(f(2)))+⋯+f(f(f(2014)))f(f(f(1))) + f(f(f(2))) + \dots + f(f(f(2014))).

Solution

From 0=f(f(0))=f(b)=ab+b=(a+1)b0 = f(f(0)) = f(b) = ab + b = (a+1)b, we get a=−1a = -1 or b=0b = 0.
If b=0b = 0, i.e. f(x)=axf(x) = ax, then we have 9=f(f(f(4)))=f(f(4a))=f(4a2)=4a39 = f(f(f(4))) = f(f(4a)) = f(4a^2) = 4a^3, which has no solution as aa is an integer.

So we must have a=−1a = -1. Then f(x)=−x+bf(x) = -x + b, and hence f(f(x))=−(−x+b)+b=xf(f(x)) = -(-x + b) + b = x. It follows that f(f(f(x)))=f(f(x))=xf(f(f(x))) = f(f(x)) = x for all xx, and thus the answer is
1+2+⋯+2014=2014×20152=2029105. 1 + 2 + \cdots + 2014 = \frac{2014 \times 2015}{2} = 2029105.

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