The smallest solution to $2^{2k}-1=\text{powerful}$

Question

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?

Added: There is some progress in the related question Congruence obstructions for three consecutive powerful numbers

Answer 1

Let $2^k-1$ be powerful (allowing $k$ be odd). Note that $k>1$ and $k\ne 6$, meaning that there exists a primitive prime factor $p\mid 2^k-1$, for which we have $k=\mathrm{ord}_p(2)\mid p-1$ and thus $p\nmid k$. From the powerfulness we have $p^2\mid 2^k-1$, implying with necessity that $p$ is Wieferich.

Noticing that the known Wieferich primes 1093 and 3511 cannot be obtained this way, it follows that any powerful number of the form $2^k-1$ will necessarily give us a new Wieferich prime. Furthermore, there are no more powerful numbers of this form than Wieferich primes, and finiteness of the latter would imply finiteness of the former.

Answer 2

Note that if there are only finitely many non-Wieferich primes, then that would imply that there is such a $k$. If there were only finitely many non-Wieferich 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 Wieferich. Thus, showing that there are no numbers of your form would be imply that there are infinitely many non-Wieferich primes. Since that is a difficult open problem, your problem is likely very difficult also.

Answer 3

I have now made this post Community Wiki, since I have changed it several times. I decided to revert to something close to my original answer, but to highlight different consequences in a different fashion. For each odd prime $p$, let $e_{p}$ denote the order of $4$ in the multiplicative group of $\mathbb{Z}/p\mathbb{Z},$ and let $f_{p} = \nu_{p}(4^{e_{p}}-1).$

As usual, when $p$ is a prime and $a\neq 0$ is an integer, we write $p^{r}|| a$ if $p^{r}| a$ and $p^{r+1} \not | a,$ so that we also write $p^{f_{p}} || 4^{e_{p}}-1.$

Notice that $e_{p}$ is always a divisor of $\frac{p-1}{2},$ and that we have $f_{p} > 1 $ if and only if $p$ is a Wieferich prime.

As the question suggests, we consider a positive integer $k$ which is minimal subject to $4^{k}-1$ being powerful, assuming any such integer exists.

Let $p$ be a prime divisor of $k$, and set $m = \frac{k}{p}.$ Then $4^{m}-1$ is not powerful by the minimal choice of $k$, so there is a prime $q$ such that $q||(4^{m}-1).$ Then $q^{2} | 4^{k}-1$ by hypothesis, so that $q$ divides both $4^{m}-1$ and $\frac{4^{k}-1}{4^{m}-1}$ and hence $q | \frac{k}{m}= p.$ Note also that $q$ is odd, so that $p$ must be odd.

Thus we see that $4^{m}-1 =ph$ where $h$ is a powerful integer coprime to $p$, and since $p$ is odd it follows that $p^{2} || 4^{k}-1 = 4^{pm}-1 .$ It also follows that $e_{p}$ is a divisor of $m$ and that $p || 4^{e_{p}}-1,$ so that $p$ is not a Wieferich prime.

On the other hand, if $p$ is a prime divisor of $4^{k}-1$ which does not divide $k$, we know that $p^{2}|4^{k}-1$, so that $e_{p} |k$. If $f_{p} =1$, we must have both $p|4^{e_{p}}-1$ and $p | \frac{4^{k}-1}{4^{e_{p}}-1},$ so that $p | \frac{k}{e_{p}},$ contrary to $p \not | k.$ Thus $f_{p} \geq 2$ and $p$ is a Wieferich prime.

We may now conclude that $4^{k}-1 = k^{2}A,$ where $A$ is the product (over all Wieferich primes $p$ such that $e_{p} | k$) of $p^{f_{p}}.$

Now suppose that $k$ has $n$ different prime factors. Then these $n$ prime factors are exactly the the non-Wieferich primes which divide $4^{k}-1,$ and all divide $4^{k}-1$ to the second power only.

We claim that at least $2^{n+1}-n-2$ different Wieferich primes divide $4^{k}-1$ (at least to the second power, of course).

Note that a prime $p$ is a divisor of $4^{k}-1$ if and only if $p$ is a primitive prime divisor of $4^{d}-1$ for some divisor $d$ of $k$.

For each divisor $d$ of $k$ (other than $1$ and $3$), there are at least two primitive prime divisors of $4^{d}-1$- one being a primitive prime divisor of $2^{d}-1$ and another being a primitive prime divisor of $2^{2d}-1$. For different such divisors $d$ and $d^{\prime},$ we see that the respective pairs of associated primes are disjoint, because $4$ has multiplicative order $d$ in $\mathbb{Z}/p\mathbb{Z}$ for either prime $p$ in the first pair and likewise order $d^{\prime}$ for each prime in the second pair.

Now $k$ has $2^{n}$ distinct divisors, so this procedure shows that $4^{k}-1$ has at least $2^{n+1} - 2$ different prime divisors (taking account of $3$ and $7$). Each of these prime factors $p$ divides $4^{k}-1$ to at least the second power.

We have seen that exactly $n$ of these primes are divisors of $k$, and these are exactly the prime divisors of $4^{k}-1$ which are NOT Wieferich primes. Hence at least $2^{n+1}-n-2$ of the prime divisors of $4^{k}-1$ are Wieferich primes.

Answer 4

It seems that most arguments here is essentially equivalent to an identity given as Theorem 1.1 of https://arxiv.org/pdf/2112.04173. If $p$ is an odd prime and $p|(a^k-1)$ then

$(*) \;\;\; \nu_p(a^k-1)-\nu_p(k)=\nu_p(a^{p-1}-1)$

As noted above and in the comment a primitive divisor $p$ of $a_k:=a^k-1$ satisfies $p-1 \ge ord_p(a)=k$ so that $\nu_p(k)=0$, so $\nu_p(a^{p-1}-1)= \nu_p(a_k) \ge 2$. But it is also known there is at least one primitive $p$ dividing $a_k$ to an odd order so at least one $p$ is Wieferich of order $\ge 3$.

There is also a bootstrap argument. If $a_k=a^k-1$ is powerful, and $p$ is a non-Wieferich prime such that $ord_p(a)|k$ then $p|a_k$ and $p|k$.

Proof : $ord_p(a)|k $ implies $p|a_k$ which implies $\nu_p(k) \ge 1$ by (*).

This allows us to boostrap from any $t|k$ to find primes primes $p_j$ with $ord_{p_j}(a)|t$ to deduce $p_j|k$. We now look for more primes $p$ such that $ord_{p}(a) | t \prod_j p_j$ and eventually maybe prove $k$ must be divisible by infinitely many primes.

For example for $a=2$ if we know $2|k$, we can deduce $3|k$ since $ord_3(2)=2$ and then $7|k$ since $ord_7(2)=3$, we then get primes $p=43,127,337,5419$ (upto $10^6$) with $ord_p(2)|2.3.7$, now look for primes with $ord_p(2)|2.3.7.43.127.337.5419$ which gives new primes $431,1033,2287,3049,9719$ which must all divide k and also $a_k$. This was the initial observations given earlier, may be made precise .

So most likely there is no powerful $a^k-1$. One can't turn the highly unlikely existence of many Wieferich primes to a proof because it seems nothing is known about them .