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Let $T_n=3^{{2^n}}+1$, $n\geq 1$. I conjecture $$T_n=2^1\prod_{k=1}^m p_k^1$$ where $p_k$ are primes of the form $1+4k$, $k\in\mathbb{N}^*$. Note that the exponents on the $p_k$ are all $1$.

Numerically, it appears that this is true. I will provide the first few values as an example

(DFP stands for "prime factorization" ):

n = 1, 3^(2^n) + 1 = 10, DFP : 2^1 x 5^1
n = 2, 3^(2^n) + 1 = 82, DFP : 2^1 x 41^1
n = 3, 3^(2^n) + 1 = 6562, DFP : 2^1 x 17^1 x 193^1
n = 4, 3^(2^n) + 1 = 43046722, DFP : 2^1 x 21523361^1

It is easy to see that $2|T_n$ and the exponent $1$ on $2$ comes from the fact that $4$ does not divide $T_n$, We can prove that in the prime factorization of $T_n$, the $p_k\not\equiv 3\pmod 4$.

Addition If $p>3$ is a prime number and divides $T_n$, then $p=1+k2^{n+1}$ for some $k\in \mathbb N$. If we assume that $p^2$ divides $T_n$, then $3^{2^n}\equiv -1 \mod p^2$, and with this integer $k$ ($k=\frac{p-1}{2^{n+1}}$), we have $3^{k2^{n+1}}=(3^{2^n})^{2k}\equiv 1 \mod p^2$. Thus, $3^{p-1}\equiv 1 \mod p^2$, which implies that $p^2$ divides $3^{p-1}-1$. But i dont see how to use thisenter image description here

Pascal
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  • Instead of your program, can you please include how "one can prove that in the prime factorization of $N$, the $p_k\not\equiv 3\pmod 4$"? So, there only remains to prove that $N$ s squarefree? – Anne Bauval Apr 17 '23 at 04:58
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    If $p$ is an odd prime factor of $N$, it follows that the residue class of $-1$ modulo $p$ is a $2^n$th power. This immediately implies that $p\equiv1\pmod{2^{n+1}}$. See here if not earlier. – Jyrki Lahtonen Apr 17 '23 at 05:07
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    Mathematica verified easily the square-freeness when $n\le8$. Full factorization of $3^{256}+1$ took a while, so I am not going to wait and see if it can handle $3^{512}$ :-) – Jyrki Lahtonen Apr 17 '23 at 05:18
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    A faster partial method is to use modular exponentiation to test whether $3^{2^n}$ is congruent to $-1$ modulo $p^2$. It only takes a minute to verify that none of the first $200$ numbers $3^{2^n}+1$ are divisible by the squares of any of the first $10{,}000$ primes. Numbers with this property are 99.9999% likely to be truly squarefree (in the sense of density—that's an actual number, not hyperbole), so this is pretty strong evidence for those $200$ terms. – Greg Martin Apr 17 '23 at 06:44
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    Nevertheless, the question is basically equivalent to asking: if $2^k$ is the smallest power of $2$ such that $3^{2^k}\equiv1\pmod p$, could it accidentally be the case that $3^{2^k}\equiv1\pmod{p^2}$ as well? Such coincidences are known to happen (see Wieferich primes for an example of this phenonemon), so I would speculate that there are some rare nonsquarefree numbers in this sequence. – Greg Martin Apr 17 '23 at 06:46
  • I saw you asked a related question on MO, does that question helps answer this one? I believe it disproves your conjecture here but only if provided that one of the prime divisors of the Fermat number is also Wieferich prime so you still need to know if that could in fact happen but maybe I understood wrong and more (less?) is being proved – Dabed Apr 21 '23 at 23:46
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    @Dabed I can also prove that a prime divisor $p$ of $T_n$ is a Wieferich prime if and only if $p^2$ divides $T_n$. The difficulty is that I have no way to prove the conjecture and it's also difficult to provide a counterexample because if $p|T_n$ and $p^2|T_n$, as above, then $p$ is a Wieferich prime, but only two numbers of this type are known at the moment. In summary, we cannot prove or disprove the conjecture. – Pascal Apr 22 '23 at 00:26
  • The numbers are of the form $m^2+1$ , hence prime factors of the form $4k+3$ are impossible. factordb gives the known partial factorizations. In fact, they all seem to be squarefree. – Peter Jun 29 '23 at 07:14
  • Since the prime factors except $2$ must have the form $2^{n+1}\cdot k+1$ with positive integer $k$ , the chance for a non-squarefree case decreases drastically with growing $n$. – Peter Jun 29 '23 at 07:20
  • That the prime factors must apparently also be a Wieferich prime to base $3$ (to give a non-squarefree case) makes it almost impossible that there is a non-squarefree case. But I agree , this is still not a proof , but I do not think that anyone can offer more currently. – Peter Jun 29 '23 at 07:33
  • Among other results : There is no prime $p\le 10^{12}$ such that $p^2$ can divide any of those numbers. – Peter Jun 29 '23 at 17:32

1 Answers1

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Since $3^{2^n}+1|3^{2^{n+1}}-1$, we might gain information about the left expression by looking at the right. One theorem we can now use is Euler's theorem. Euler's theorem states that if $gcd(x,y)=1$ then $x|y^{\phi\left(x\right)}-1$. Where $\phi()$ is the euler totient function. One important thing to note is that for the smallest $q\in\Bbb{N}$ such that $x|y^q-1$ then $q|\phi(x)$. The reason for this is explained in the properties section of this page on the Carmichael function.

If we examine when $p^k|3^{\phi(p^k)}-1\rightarrow p^k|3^{(p-1)(p^{k-1})}-1$ where $p$ is prime. Notice that regardless of how big $k$ is this doesn't change the number of factors of 2 in the exponent of the power number of $3$. Since $3^{p-1}-1|3^{(p-1)(p^{k-1})}-1$ at minimum we are looking for $p^2|3^{p-1}-1$.

These are wieferich primes with base three. Only two of these primes are known (11 and 1006003).

For regular wieferich primes (not of the base three kind) it is suspected that there are infinitely many of them but very rare in comparison to the natural numbers.

Now examining even further if $p$ is of the form $2^a+1$ so that we can use the totient function to produce $3^{2^a}-1$. These primes are fermat primes, it suspected that there are only five them (3, 5, 17, 257, and 65537).

So now we need something even stronger than wieferich primes in base 3. Let $b$ the largest natural number such that $2^b|p-1$. We need $p^2|3^{2^c}-1$ where $c\in\Bbb{N}$ and $c\le b$ for any $c$.

quantus14
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