Do there exist finitely many primes $p$ such that there exists $k \in [1,p]$ with $\mathrm{ord}_p(k) = q$ such that $k^{p-1} \equiv 1 \pmod{p^2}$?

This is extended computational evidence for the Conjecture from this question, focusing on the $n=1$ case.

Conjecture A for $n=1$ states: for all sufficiently large primes $p$ with $q \mid p-1$, $$v_p\!\left(\prod_{k=1}^{p} \Phi_q(k)\right) = q-1.$$

Reduction to $v_p(k^{p-1}-1)$. By the Proposition in the linked question, the product receives $p$-adic contribution only from terms where $\mathrm{ord}_p(k) = q$, and for such $k$, $v_p(\Phi_q(k)) = v_p(k^{p-1}-1)$. There are exactly $q-1$ residue classes of order $q$ mod $p$, each appearing exactly once in $[1,p]$, so $$v_p\!\left(\prod_{k=1}^{p}\Phi_q(k)\right) = \sum_{\substack{k=1 \\ \mathrm{ord}_p(k)=q}}^{p} v_p(k^{p-1}-1).$$ Now each term is at least $1$ by Fermat, giving the lower bound at least $q-1$. The conjecture says that for all sufficiently large $p$, every term is exactly $1$ — which is to say no $k \in [1,p]$ with $\mathrm{ord}_p(k) = q$ satisfies $k^{p-1} \equiv 1 \pmod{p^2}$.

We therefore check the following equivalent conjecture:

For a fixed prime $q$ and sufficiently large primes $p$, there exists no $k \in [1,p]$ such that $\mathrm{ord}_p(k) = q$ and $k^{p-1} \equiv 1 \pmod{p^2}$.

Computational results. We searched all primes $p \equiv 1 \pmod{q}$ up to $10^{12}$ (for $q=11$) and $10^{11}$ (for $q=3$):

• $q=3$: no violations found.
• $q=11$: two violations, $p = 400263469$ with witness $k = 160169173$, and $p = 9544876463$ with witness $k = 3519317465$. No violations for $p > 9544876463$ up to $10^{12}$, consistent with these being the finitely many exceptions below the "sufficiently large" threshold.

Heuristic. The probability that a given $k$ of order $q$ satisfies $k^{p-1}\equiv 1\pmod{p^2}$ is heuristically $1/p$. Summing over all $q-1$ such $k$ and all primes $p\leq X$, $$\mathbb{E}[\#\text{ violations up to }X] \;\approx\; \ln\ln X.$$ At $X=10^{12}$: predicted $\approx 3.3$, observed $2$ for $q=11$. This is the direct analogue of classical Wieferich primes, where only $p=1093$ and $p=3511$ would be known below $6.7\times 10^{15}$.

Sharpness. Allowing $k\leq p^{3/2}$ instead of $k\leq p$, violations become dense: a search over $p\in[10^6,10^8]$ finds $1271$ violations for $q=3$ and $1100$ for $q=11$. The $O(p)$ threshold is sharp.

Question. For $q=11$, would $p = 400263469$ and $p = 9544876463$ be the only primes for which some $k \in [1,p]$ with $\mathrm{ord}_p(k) = 11$ satisfies $k^{p-1} \equiv 1 \pmod{p^2}$? More generally, for any fixed prime $q$, is the set of such exceptional primes finite?

Remark. Our conjecture is logically independent of the classical Wieferich problem. For a fixed base $a > 1$, it is an open question whether there are infinitely many primes $p$ with $a^{p-1} \equiv 1 \pmod{p^2}$; the heuristic predicts infinitely many but only finitely many are known for any base. Our conjecture makes no claim about any fixed base. Instead, for each prime $p$, we ask about the $q-1$ values of $[1,p]$ of order $q$ mod $p$ — a set of bases that changes entirely with $p$. Whether infinitely or finitely many primes $p$ are Wieferich for some fixed $k$ has no bearing on whether, for large $p$, the order-$q$ elements of $[1,p]$ happen to include a Wieferich witness. The quantifier structures are different: the Wieferich problem fixes $k$ and varies $p$; our conjecture fixes $p$ and asks about all $k \in [1,p]$ of a given order simultaneously.

Remark (sharpness of the $[1,p]$ threshold). The bound $k \leq p$ in the Conjecture is not arbitrary. The following computational data for $q=11$ shows a sharp transition precisely at $k = p$:

• For $k \in [1, p]$: searching all $p \equiv 1 \pmod{11}$ up to $10^{12}$, only two violations appear, both at $p < 10^{10}$. No violations for $p > 9544876463$.

• For $k \in [1, p^{11/10}]$: four violations appear in $p \in [10^8, 10^{10}]$, including new ones at $p = 1788375733$ and $p = 8960980403$ that do not violate the Conjecture since $k > p$.

• For $k \in [1, p^{3/2}]$: $1100$ violations in $p \in [10^6, 10^8]$.

So violations are essentially absent inside $[1,p]$ for large $p$, appear almost immediately once $k$ is allowed to exceed $p$, and become dense at $k \leq p^{3/2}$. The $O(p)$ threshold captures something structurally real, not a coincidence of the search range.