Pinko wrote: Prove that there exist infinitely many positive integers $n$, for which at least one of the numbers $2^{2^n}+1$ and $2018^{2^n}$ is composite. Isn't $2018^{2^n}$ already composite? Also, I think it is an open problem whether there exists infinitely many integers $n$ for which $2^{2^n}+1$ is composite.

Check it again I fixed it. I forgot to add "+ 1" after $2018^{2^{n}}$.

2018 Serbia TST

Rickyminer wrote: 2018 Serbia TST Please give a link, it's very hard

If for enough large $n$, $2^{2^n}+1$ and $2018^{2^n}+1$ are all prime, we write $2^{2^n}+1=p_{n}=p$, and consider $2018^{2^{2^n}} \equiv 1 (modp)$, then $ord_{p}(2018)=2^s$ and $s$ is lagre enough. Then $2018^{2^{s-1}} \equiv -1 (modp)$, so $2018^{2^{s-1}}+1$ is not a prime, a contradiction.