On vanishing of $p$-adic logarithms

Question

Might be related to Wieferich primes.

Let $p$ be odd prime and define the Fermat quotient

$$F(n)=\frac{(2^{n-1} -1)}{n} \mod n=\frac{(2^{n-1} \bmod n^2 )-1}{n}$$

For integer $b$ let $L_p(b)$ be the $p$-adic logarithm with precision two and let $a(p)=\frac{L_p(2)}{p}$.

$a(p)$ is integer in $[0,p-1]$ and in pari/gp the logarithm is defined as log(b+O(p^2)) and in sage Qp(p,2)(b).log()

Q1 Are there infinitely many primes for which $L_p(2)$ and $a(p)$ are zero?

Q2 Are there infinitely many primes for which $L_p(2)$ and $a(p)$ are non-zero?

Q3 If $p$ is Wieferich prime is $L_p(2)=0$?

Q4 If $p$ is non-Wieferich do we have $p=F(p)+a(p)$? this holds for up to 10^7

/*
pari/gp code for p-adic logarithm and Fermat quotient
*/
{L(p)=lift(log(2+O(p^2)))}
{a(n)=L(n)/n;}
{F(n)=lift(Mod(2,n^2)^(n-1)-1)/n}

{f(L=10^2)=
forprime(p=3,L,
b=F(p)+a(p);
if(b!=p,print([p,b,F(p),a(p)]));
)};

f(4000)
```

Julian Rosen · Accepted Answer · 2024-09-09 20:12:17Z

The $p$-adic logarithm can be computed $$ \begin{align*} \log_p(2) &= \frac{1}{p-1}\log_p(2^{p-1})\\ &=\frac{1}{p-1}\sum_{n\geq 1}(-1)^{n-1} \frac{(2^{p-1}-1)^n}{n} \\ &\equiv -pF(p)\mod p^2. \end{align*} $$ This means $a(p)+F(p)\equiv 0\mod p$, which answers part (4) in the affirmative: since $a(p)$ and $F(p)$ are in $[0,p-1]$ and their sum is divisible by $p$, the sum equals $p$ exactly when $F(p)\neq 0$. The answer to (3) is also yes, since $L_p(2)=0\Leftrightarrow a(p)=0\Leftrightarrow F(p) = 0$.

Questions 1 and 2 equivalent to: are there infinitely-many Wieferich primes to the base 2, and are there infinitely-many non-Wieferich primes to the base 2? These questions are both open, though the ABC conjecture would imply the answer to (2) is "yes". I have heard the heuristic argument that a random prime $p$ is Wieferich to the base 2 with probability $1/p$, so since $\sum_p 1/p$ diverges, we should expect there are infinitely-many Wieferich primes.

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