Is $(3^p-1)/2$ always squarefree?

Question

I have little conjecture. Maybe it's stupid i don't know.

Let $p>5$ be a prime number. Then $(3^p-1)/2$ is always squarefree?

It's true for $p<192$.(I used Mathematica.)

Answer 1

This is even unknown for the usual Mersenne numbers $M_p=2^p-1$. All such numbers, where the factorisation is known, are squarefree. However, Guy (1994) believes that there are $M_p$ which are not squarefree. As far as I know, a similar conjecture holds for $N_p=3^p-1$ and $N_p/2$.

Answer 2

If $q^2$ divides $3^p-1$ with an odd prime $p$ , then $q$ must be a Wieferich-prime to base $3$. The only known Wieferich-primes to base $3$ are $11$ and $1006003$ (https://oeis.org/A014127). There is no other example below about $10^{15}$.

The example $11^2|3^5-1$ was ruled out by your condition $p>5$. There is no prime $p>5$ with $11^2|3^p-1$ because the order of $3$ modulo $11^2$ is $5$.

The order of $3$ modulo $1006003^2$ is $1006002$. Since this is not prime, $3^p-1$ cannot be divisible by $1006003^2$.

So, $3^p-1$ is almost certain squarefree for $p>5$.

Answer 3

This is true if and only if there exist a prime $q$ such that $p \mid q-1$ and $3$ is a $q(q-1) /p$ residue modulo $q^2$. There are $p$ of them, uniformly distributed among the $q(q-1) $ coprimes with $q$. For a fixed $q$, this happens with probability

$$A(q) := \sum_{p \mid q-1} \frac{p}{q(q-1) } $$

And one can see that the probability of finding a counterexample up to $x$ is

$$ P_x := 1- \exp \left (- \sum_{q \le x } A(q) \right ) $$

If the sum of contributions $\sum_q A(q) $ diverges, then $P_x \to 1$ as $x$ goes to infinity and we are almost sure that a counterexample exists. However, it seems to me that $A(q) $ can be bounded from above on average by something like $\log(q) ^4 /q(q-1) $ which converges, and this would be very bad :( in practice, it would mean that if you substitute $3$ with say $5$, the probability is the same and we don't know in advance for which bases we hit such a counterexample.

I suggest you try to compute the sum $\sum_q A(q) $ and see if this converges or not; little is known about the sum of primes dividing a number, and it maybe the case that for numbers of the form $q-1$ this is bigger than on average.

Edit. Note also that the characterization at the beginning could be used to approach the problem in a direct way. There exist a version of quadratic reciprocity law for higher residues using the Hilbert Symbol, and since it is easy to understand if $q^2$ is a $q(q-1) /p$ residue modulo 3, one has some hopes to understand if $3 $ is a residue modulo $q^2$. Unluckily, I am not an expert, and I don't know how difficult it is to compute the other term (the generalization of the sign $(-1) ^{\frac{(p-1)(q-1)}{4}}$ to higher powers).