# Ruling out an extremely specific class of Wieferich-like primes

Recall that a prime $$p$$ is a Wieferich prime if $$p^2|2^{p-1}-1$$. The only known Wieferich primes are $$p=1093$$ and $$p=3511$$. A prime $$p$$ is a generalized Wieferich prime to base $$q$$ if $$p^2|q^{p-1}-1$$.

It is strongly suspected that there are infinitely many Wieferich primes to base $$q$$ for any $$q>1$$. In the other direction it is strongly suspected that for any $$q>1$$ there are infinitely many non-Wieferich primes to base $$q$$, but this is also open even for $$q=2$$.

The situation is slightly different if one replaces $$p^2$$ with $$p^3$$. Call a prime $$p$$ a super-Wieferich prime to base $$q$$ if $$p^3|q^{p-1}-1$$. In particular, it is a standard conjecture (using similar heuristic arguments) that for any $$q>1$$ there are only finitely many primes $$p$$ which are super-Wieferich primes to base $$q$$. (For interesting recent work related to that question see Ram Murty and Francois Seguin's recent paper "Prime divisors of sparse values of cyclotomic polynomials and Wieferich primes" in the Journal of Number Theory.)

This question is about a highly restricted version of Wieferich primes. In particular: define a prime $$p$$ to be a hyper-Wieferich prime to base $$q$$ if there is some $$m$$ such that $$p^m = \frac{q^{p-1}-1}{q-1}$$ and $$m>2$$. The idea here is that $$p$$ is not just super-Wieferich to base $$q$$, but that every potential factor of $$q^{p-1}-1$$ which isn't one of the obvious factors coming from $$q-1$$ is in fact a copy of $$p$$.

Question: Can we show that there are only finitely many hyper-Wieferich primes or at least substantially restrict what they can look like?

We at least have the following:

Claim: The ABC conjecture implies that there are only finitely many primes $$p$$ which are hyper-Wieferich to base $$q$$.

Proof: Assume that $$p$$ is hyper-Wieferich to base $$q$$. So we have $$p^m = \frac{q^{p-1}-1}{q-1}.$$ One easily has that $$m \geq p-2$$. We set $$A=p^m (q-1)$$, $$B=1$$ and $$C= q^{p-1}$$. Then $$\mathrm{rad}(ABC) = pq(q-1)$$. But $$q^{p-1}$$ is much larger than $$pq$$.

The above shouldn't be too surprising since it is essentially a very simplified version of an argument of Silverman that the ABC conjecture implies there are infinitely many non-Wieferich primes.

• in the question you might clarify whether $q$ is fixed.
– YCor
Oct 15 '21 at 15:55
• Hmm, since $p-1$ is even, with $p_2=(p-1)/2$ we have $p^m = (q^{p_2}-1) \cdot (q^{p_2}+1)/(q-1)$ and this looks (for $p_2>2$ -on a first glance- impossible to me. Do I overlook something? (2.nd thought: it seems, that the exponent at $q$ should not be $p-1$ but an odd value by zn-order of $p$ (Pari/GP-term)) Oct 15 '21 at 16:12
• I think $q^{p-1}-1=(q^{p-1/2}-1)(q^{p-1/2}+1)$ say that the existance of hyper-Wieferich is very unlikely.
– Hhan
Oct 15 '21 at 16:18

This is impossible from standard results about cyclotomic polynomials and their factors. Note that $$\frac{q^{p-1}-1}{p-1}$$ is just $$\prod_{d|(p-1), d>1}\Phi_d(q),$$ where $$\Phi_d(x)$$ is the $$d$$th cyclotomic polynomial. Taking $$p>3$$, there are two such divisors $$d$$; namely $$2$$ and $$p-1$$. By an old result attributed to Bang from 1886, these must have "new" prime divisors.
• By the way, if we had a good way to tell when $\Phi_d(q)$ is a prime power (for a single $d$) that would have some consequences regarding the odd perfect number problem. Oct 15 '21 at 17:58