Korean Mathematical Olympiad

For any positive integer $n$, define $a_n:=2^{2^n}+1$. Since $$a_1{a}_2\cdots a_{n-1}=a_n-2,$$ $\gcd (a_n,a_m)=1$ for any $n\ne m$. Now, for any positive integer $n$ such that $n\ge a$, let $p_n$ be any prime dividing $a_n$. Note that $p_n$ is different from $p_m$ for any $m\ne n$.

Suppose that $2^{2^n}b-1$ is divisible by $p_n$ for some odd integer $b$. Since $$2^{2^{2n}}\equiv 1(\mathrm{mod}{p}_n),2^{2^{2n+1}}\equiv 1(\mathrm{mod}{p}_n),$$ $2^{2^n}\equiv 1(\mathrm{mod}{p}_n)$. This is a contradiction to the fact that $2^{2^n}\equiv -1(\mathrm{mod}{p}_n)$. Therefore, $p_n\notin S_a$ for any integer $n$ such that $n\ge a$. $\square$