3
$\begingroup$

Integer is powerful if all the exponents in its factorization are at least $2$.

Every powerful integer can be written in the form $a^2 b^3$.

For odd $k$, define $F(k)=(2^{2k}-1)=(2^k-1)(2^k+1)$.

This paper asks are there only finitely integers for which $F(k)$ is powerful.

It is natural to ask what properties the smallest solution to $F(K)=\textit{powerful}$ has.

$2^k+1$ is always divisible by $3$ for odd $n$ and it is divisible by $3^2$ iff $k$ is divisible by $3$. This shows $3 \mid K$, so there are no solutions with $K$ prime.

Observation 1: if odd $n \mid k$, then $2^n-1 \mid 2^k -1$ and $2^n+1 \mid 2^k+1$.

Observation 2: if odd prime $q^2 \mid 2^{2k}-1$ and $q$ is not Wieferich prime, then $q \mid k$ since the multiplicative order of $2$ modulo $q^2$ is divisible by $q$.

Conjecture 1: Assume $q$ is Wieferich prime and $q^2 \mid 2^k-1$. Then $d \mid k$ where $d$ is divisor of $q-1, d > \log_2{q}$.

We believe the observations and the conjecture imply infinitely many non-Wieferich primes.

So starting from $3 \mid K$ we are removing primes with exponents one and continue $2^3-1 \mid K, 2^7-1 \mid K, 2^7+1 \mid K, 2^{127}-1 \mid K, 2^{127}+1\mid K$.

This constraints are very strict since $2^{2^{127}-1}-1$ is extremely large.

Q1 Are these constraints true?

Q2 Can we get more constraints, hopefully showing no solutions?

Improve this question
$\endgroup$
2
  • $\begingroup$ You can combine primes also. For example, since $3|k$, and $7|k$, one must have $337|f(k)$, so $337|k$, and you can continue from there. Similarly, one has $5419|k$. $\endgroup$
    – JoshuaZ
    Jun 12, 2021 at 14:25
  • $\begingroup$ I like more to look at $\ 4^k\ $ than at $\ 2^{2k}.$ $\endgroup$
    – Wlod AA
    Jun 13, 2021 at 5:46

1 Answer 1

Reset to default
1
$\begingroup$

Note that if there are only finitely many non-Weiferich primes, then that would imply that there is such a $k$. If there were only finitely many non-Weiferich primes, one could make a sequence based on essentially your procedure and one would be forced to eventually have all the primes which divide the number be accounted for, either from the sequence, or from being Weiferich. Thus, showing that there are no numbers of your form would be imply that there are infinitely many non-Weiferich primes. Since that is a difficult open problem, your problem is likely very difficult also.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks. I just added new conjecture, do you have opinion about it? $\endgroup$
    – joro
    Jun 12, 2021 at 14:35
  • $\begingroup$ I think your claim is rather strong. The paper allows finitely many powerful cases without Wieferich consequences, unless I am mistaken. $\endgroup$
    – joro
    Jun 12, 2021 at 14:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.