1) . Let there be prime $p$ dividing $3^{2^n}+1 $ then we have $3^{2^{n+1}} \equiv 1 \pmod{p}$ Let $d$ denote $order_p(3)$ . Then $d$ mist divide $2^{n+1}$ . So $d$ is of form $2^k$ . But if $k<n+1$ then , we get $3^{2^n} \equiv 1 \pmod{p}$ which is contradiction . Thus $d=2^{n+1}$ . AQlso since $d$ is order , $d|p-1$ Thus $p=2^{n+1}\cdot k+1$ . So any prime other than this form doesnot divide any term of the sequence . Thus there exist infinitely many primes , none dividing any term of sequence $\Box$.

1)$gcd(a_{n},a_{m})=2$ 2)$p\mid{3^{2^{n}}+1}\Longrightarrow{2^{n}\mid{p-1}}$ $2^{n}$ no divide $p-1$ and $p$ no divide $a_{i},i=1,2,...,n$ are infinitly

I think it even holds for sequence $(3^{2n}+1)_{n\geq 1}$. I) There is infinitely many primes of form $p=4k-1$ and none of them can divide any of this numbers (because -1 is quadratic nonresidue modulo such $p$). II) Zsigmondy theorem : http://mathworld.wolfram.com/ZsigmondyTheorem.html Q.E.D.

It even holds for $(a^n+1)_{n\geq 1}$, with arbitrary integer $a>1$ fixed; it was Problem 3 Seniors.

First part

Second part (overkill)

2)$n\ge 2$:By Zsigmondy's theorem, ∃$p$(prime) s.t. $p|3^{2^n}+1$ and $p\nmid 3^{2^m}+1(1\le m<n)$.Therefore the proof is completed.$\blacksquare$

mavropnevma wrote: It even holds for $(a^n+1)_{n\geq 1}$, with arbitrary integer $a>1$ fixed; it was Problem 3 Seniors. (2) For $n\ge 4$:By Zsigmondy's theorem, ∃$p$(prime) s.t. $p|a^n+1$ and $p\nmid a^m+1(1\le m<n)$.Therefore the proof is completed.$\blacksquare$