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I was asked to prove for $n \in \mathbb{N}$ that $T_n=2^{2^n+1}+1$ is composite.

One solution by induction is as follows:

Base case $T_1$: $2^{2^n+1}+1=2^3+1=9=3^2$, so $T_1$ is composite with a factor of $3$.

Inductive step: $T_{n+1}=2^{2^{n+1}+1}+1 =2^{2^{n}}2^{2^{n}+1}+1$

Now, the first few terms of $T_n$ are $9, 33, 513$, suspiciously all multiples of $3$. If we reframe this inductive step as $T_n \equiv 0 \mod 3$, with a quick intermediate proof that for $n \in \mathbb{N}, 2^{2^n} \equiv 1 \mod 3$, we have:

modulo $3$: $2^{2^{n}}2^{2^{n}+1}+1 = 2^{2^{n}}(T_n-1)+1 \equiv (1)(2)+1\equiv 0$

This proves that $T_n$ is always a multiple of $3$ for $n \geq 1$. Since all $T_n > 3$, all $T_n$ are composite.

But here is my question: the proof above uses the fact that $T_n$ has a factor of $3$, more specifically than just that $T_n$ is composite. Is there a proof that doesn't use the factor of $3$ as an assumption? I've not found it. The inductive step can also be represented as $T_{n+1}=\frac{(T_n-1)^2}{2}+1$ with the assumption that $T_n$ is composite and odd.

Bill Dubuque
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RobinSparrow
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    We can even show that $2^n+1$ (with a positive integer $n$) can only be prime if $n$ is a power of $2$ ($1$ included). The proof however is again based on a concrete nontrivial factor , this time however not necessarily $3$. – Peter Jun 03 '24 at 13:52
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    If you remove the factors of 3, the rest is different ($9/3=1,33/3=11,513/27=19$), so it there’s not going to be a straightforward divisibility relationship between factors. Usually, finding a specific factor is considered better than just showing existence. – Eric Jun 03 '24 at 13:58

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Well, I'm not sure if this is quite what you want, but the key conceptual thing here (in my opinion) is that $x+1 \mid x^n+1$ as polynomials for any odd $n\ge 1$. Indeed this follows from the factor theorem, for example. The things with $2$ and $2^{n}+1$ aren't really relevant.

Sam Moore
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  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Jun 03 '24 at 14:17
  • @SamMoore: This is helpful, I wasn't thinking of it in this way. – RobinSparrow Jun 03 '24 at 16:20