Team Selection Test for EGMO 2019

Answer: 2. It is easy to verify that for any odd prime number $p$ we have $s(p)=\sigma (p)=2$ and hence $2s(n)-\sigma (n)=2$. We will show that $2s(n)-\sigma (n)\le 2$ for every positive integer $n$. Let $1=d_1<d_2<\cdots <d_k=n$ be positive divisors of $n$. It is well known that $d_i{d}_{k+1-i}=n$ for $1\le i\le k$. If $d_i+1|n+1$, then $d_i+1|d_i{d}_{k+1-i}+1-(d_i+1)=d_i(d_{k+1-i}-1)$. $\Rightarrow d_i|d_{k+1-i}-1$ since $(d_i+1,d_i)=1$. $\Rightarrow i=k$ or $i<k+1-i$. $\Rightarrow i=k$ or $i\le k-i$. $\Rightarrow s(n)\le 1+\frac{k}{2}\Rightarrow 2s(n)\le k+2$ and the result follows.