Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Prove it Ibero-American Mathematical Olympiad

Problem:

Find all functions $f$ on the positive integers with positive integer values such that

(1) if $x < y$ , then $f(x) < f(y)$ , and

(2) $f(y f(x)) = x^{2} f(x y)$ .

Solution

Solution:

Put $y = 1$ . Then $f(f(x)) = x^{2} f(x)$ .

Put $y = f(z)$ , then $f(f(z) f(x)) = x^{2} f(x f(z)) = x^{2} z^{2} f(x z) = f(f(x z))$ .

But $f$ is $(1,1)$ so $f(x z) = f(x) f(z)$ .

Now suppose $f(m) > m^{2}$ for some $m$ . Then by (1), $f(f(m)) > f(m^{2}) = f(m \cdot m) = f(m)^{2}$ .

But $f(f(m)) = m^{2} f(m)$ , so $m^{2} > f(m)$ . Contradiction.

Similarly, suppose $f(m) < m^{2}$ . Then $m^{2} f(m) = f(f(m)) < f(m^{2}) = f(m)^{2}$ , so $m^{2} < f(m)$ .

Contradiction. So we must have $f(m) = m^{2}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.