Argentine National Olympiad 2015

All $n\in \mathbb{N}$ are representable except the powers of $2$. Set $f(a,b,c)=[a,b]+[b,c]+[c,a]$. Take an arbitrary $k\in \mathbb{N}$ and let $a=k$, $b=c=1$ to obtain $f(k,1,1)=2k+1$. Hence all odd $n$, $n\ge 3$, are representable. If $n$ is representable then so is $2n$

because $[2u,2v]=2[u,v]$ implies $f(2a,2b,2c)=2f(a,b,c)$. So each $n\ge 3$ is representable if it has an odd divisor greater than $1$. There remain the powers $2^k$ of $2$, with $k\ge 0$. We show that they are not representable. This is true for $k=0,1$ since clearly $f(a,b,c)\ge 3$ for all $a,b,c\in \mathbb{N}$. Suppose that $f(a,b,c)=2^k$ with $k\ge 2$. Consider the least $k$ with this property. Observe that at least two numbers among $a,b,c$ are even. Otherwise $f(a,b,c)$ is odd while $2^k$ is even. If $a,b,c$ are all even, they can be divided by $2$ to yield $f\left(\frac{a}{2},\frac{b}{2},\frac{c}{2}\right)=2^{k-1}$, which contradicts the minimality of $k$. Hence one may assume that $a,b$ are even and $c$ is odd. Here $[a,b]=2\left[\frac{a}{2},\frac{b}{2}\right]$. Note also that $$[a,c]=2\left[\frac{a}{2},c\right]\text{ holds because }a\text{ is even and }c\text{ is odd.}$$ Analogously $[b,c]=2\left[\frac{b}{2},c\right]$. It follows that $$f\left(\frac{a}{2},\frac{b}{2},c\right)=\frac{1}{2}f(a,b,c)=2^{k-1},\text{ contradicting the minimality of }k\text{ again.}$$