Vietnamese MO 2021

a) Notice that if $k$ is a positive integer such that $\gcd (k,n)=1$ and $k<n$ then $n-k$ and $n$ are relatively prime. It follows that $$\sum _{k\le n,\gcd (k,n)=1}k=\sum _{k\le n,\gcd (k,n)=1}(n-k)$$ Let $A=\{k\in \mathbb{N}\mid 1\le k\le n,\gcd (k,n)=1=\{k_1,k_2,\dots ,k_n\}\}$ then $$\sum _{i=1}^{\phi (n)}{k}_i=\frac{1}{2}\sum _{i=1}^{\phi (n)}{k}_i+\frac{1}{2}\sum _{i=1}^{\phi (n)}(n-k_i)=\frac{1}{2}\sum _{i=1}^{\phi (n)}(k_i+n-k_i)=\frac{n\phi (n)}{2}$$ So, we have $$s(n)=\sum _{i=1}^{n}i-\sum _{i=1}^{\phi (n)}{k}_i=\frac{n(n+1)}{2}-\frac{n\phi (n)}{2}=\frac{n}{2}(n+1-\phi (n)).$$

b) Suppose that there exists a positive integer $n$ such that $s(n)=s(n+2021)$. Thus, $$2s(n)=n(n+1-\phi (n))=(n+2021)(n+2022-\phi (n+2021))(\ast )$$ or $$2021(2n+2022-\phi (n+2021))=n(\phi (n+2021)-\phi (n))(1)$$ From (*), it follows that $n,n+2021$ are divisors of $2s(n)$. Otherwise, $$2s(n)=n(n+1)-n\phi (n)<n(n+1)<n(n+2021),$$ so $n,n+2021$ are not coprime, which means $(n,2021)\ne 1$. Let $d=\gcd (n,2021)>1$, so $\gcd (\frac{n}{d},\frac{2021}{d})=1$. From (1), we get $$\frac{2021}{d}(2n+2022-\phi (n+2021))=\frac{n}{d}(\phi (n+2021)-\phi (n))(2)$$ so there exists some positive integer $x$ such that $$\phi (n+2021)-\phi (n)=\frac{2021}{d}\cdot x(3)$$ $$2n+2022-\phi (n+2021)=\frac{n}{d}\cdot x(4)$$ Thus, $\phi (n+2021)-\phi (n)$ is divisible by $2021/d$. Otherwise, one can check that $\phi (n+2021),\phi (n)$ is divisible by $\phi (d)$ and $\gcd (\phi (d),2021/d)=1$ for all $d\in \{43,47,2021\}$, so $$\phi (n+2021)-\phi (n):\frac{2021\phi (d)}{d}$$ On the other hand, it follows from (3) and (4) $$d<x=d\cdot \frac{2n+2022-\phi (n)}{n+2021}<2d$$ It implies that, for all $d\in \{43,47,2021\}$ $$\frac{2021\phi (d)}{d}<\frac{2021\cdot d}{d}<\frac{2021}{d}\cdot x<2\cdot 2021<\frac{3\cdot 2021\phi (d)}{d}$$ Thus, $\phi (n+2021)-\phi (n)=2\cdot 2021\phi (d)/d$ and $x=2\phi (d)$, so $$\phi (n+2021)=\frac{2n(d-\phi (d))}{d}+2022(5)$$ $$\phi (n)=\frac{2n(d-\phi (d))}{d}+2022-\frac{2\cdot 2021\phi (d)}{d}(6)$$ If $n$ has at most 10 distinct prime divisors then $$\phi (n)>n\prod _{i=2}^{11}\left(1-\frac{1}{i}\right)=\frac{n}{11}>\frac{2n(d-\phi (d))}{d},d\in \{43,47,2021\},$$ which contradicts (6). We get $n$ has at least 11 distinct prime divisors, so $n>12!$ and $\phi (n)$ is divisible by $2^{10}$. Similarly, if $n+2021$ has at most 4 distinct prime divisors then $$\phi (n+2021)>(n+2021)\prod _{i=2}^5\left(1-\frac{1}{i}\right)=\frac{n+2021}{5}.$$ Otherwise, $n>12!$ and $$\frac{n+2021}{5}>\frac{2n(d-\phi (d))}{d}+2022,d\in \{43,47,2021\},$$ which contradicts (5). We get $n+2021$ has at least 5 distinct prime divisors, so $\phi (n+2021)$ is divisible by $2^4$. On the other hand $$v_2(\phi (n+2021)-\phi (n))=v_2\left(\frac{2\cdot 2021\phi (d)}{d}\right)\le 3,\forall d\in \{43,47,2021\}$$ which contradicts $\phi (n+2021):2^4$ and $\phi (n):2^{11}$. Hence, there does not exist a positive integer $n$ such that $s(n)=s(n+2021)$ □