## Problem A3

Find all functions $\mathrm{f}$ on the positive integers with positive integer values such that (1) if $\mathrm{x}<\mathrm{y}$, then $f(x)<f(y)$, and $(2) f(y f(x))=x^{2} f(x y)$.


## Solution

Answer: $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$.

Note that ( 1 ) implies $f$ is $(1,1)$.

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 $f(x z)=$ $\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{z})$.

Now suppose $\mathrm{f}(\mathrm{m})>\mathrm{m}^{2}$ for some $\mathrm{m}$. Then by $(1), \mathrm{f}(\mathrm{f}(\mathrm{m}))>\mathrm{f}\left(\mathrm{m}^{2}=\mathrm{f}(\mathrm{m} \cdot \mathrm{m})=\mathrm{f}(\mathrm{m})^{2}\right.$. But $\mathrm{f}($ $\mathrm{f}(\mathrm{m}))=\mathrm{m}^{2} \mathrm{f}(\mathrm{m})$, so $\mathrm{m}^{2}>\mathrm{f}(\mathrm{m})$. Contradiction.

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

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

