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India — Team Selection Test, 2013

Algebra Difficulty 5.4 AIME, harder Prove it India

Let h≥3h \ge 3 be an integer and XX the set of all positive integers that are greater than or equal to 2h2h. Let SS be a nonempty subset of XX such that the following two conditions hold:
* if a+b∈Sa + b \in S with a≥h,b≥ha \ge h, b \ge h, then ab∈Sab \in S;
* if ab∈Sab \in S with a≥h,b≥ha \ge h, b \ge h, then a+b∈Sa+b \in S.

Prove that S=XS = X.

Solution

Let f:X→{0,1}f: X \to \{0, 1\} be such that f(x)=1f(x) = 1 if and only if x∈Sx \in S. Then f(a+b)=f(ab)f(a+b) = f(ab) whenever a≥h,b≥ha \ge h, b \ge h. If a≥h+2a \ge h+2 then
f(2a−1)=f(a2−a)=f(a3−2a2)=f(a2+a−2)=f(2a+1). f(2a - 1) = f(a^2 - a) = f(a^3 - 2a^2) = f(a^2 + a - 2) = f(2a + 1).
Given n≥2hn \ge 2h, it is easy to see that there exists an a≥h+2a \ge h+2 such that f(n)=f(2a−1)f(n) = f(2a-1). This proves that ff is constant and hence S=XS = X. □\square

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