For odd integer $n$, define the Euler quotient modulo $n$ to be $a(n)$:

$$ a(n)=\frac{(2^{\phi(n)}-1) \bmod n^2}{n}=\frac{2^{\phi(n)}-1}{n} \bmod n$$

$a(n)=0$ for OEIS sequence Wieferich numbers

Conjecture 1 If a Wieferich prime $p$ divides $n$, then $p$ divides $a(n)$ and in addition $\gcd(n,a(n))>1$.

Conjecture 2 For $n$ Mersenne number $M_m=2^m-1$ we have $a(2^m-1)= \phi(2^m-1)/m$.

This holds for $m$ up to 200.

Conjecture 3 If a Wieferich prime $p$ divides $M_m$, then $p$ divides $ \phi(2^m-1)/m$.

This holds for $p=1093$.

Which of the conjectures are true?

Question1 Does these conjectures and the answer(s) contribute new results about Wieferich primes?


1 Answer 1


Conjecture 1. Assume that $p$ is a Wiefrich prime, that is, $p^2$ divides $2^{p-1}-1$. Denote $n=p^km$ where $p$ does not divide $m$. By lifting the exponent lemma, $p^{1+k}$ divides $2^{(p-1)p^{k-1}\varphi(m)}-1=2^{\varphi(n)}-1$, thus $p$ divides $a(n)$.

Conjecture 2. Denote $2^m-1=t$. We should prove that $(2^{\varphi(t)}-1)/t\equiv \varphi(t)/m\pmod {t}$. Denote $\varphi(t)=mk$. Then $(2^{\varphi(t)}-1)/t=(2^{mk}-1)/(2^m-1)=(1+2^m+\ldots+2^{m(k-1)})\equiv k\pmod{t}$ as needed.

Conjecture 3. Assume that $p^2$ divides $2^{p-1}-1$ and $p$ divides $2^m-1$. Let $s$ denote the multiplicative order of $2$ modulo $p$. Then $p-1=sr$ for integer $r$, and $p^2$ divides $2^s-1$ (by lifting the exponent lemma, for example). Next, if $p$ divides $2^m-1$, then $s$ divides $m$, write $m=sp^AB$ where $p$ does not divide $B$. Then $2^m-1$ is divisible by $p^{A+2}$ by lifting the exponent lemma, therefore $p^{A+1}$ divides $\varphi(2^m-1)$, that yields the result.

  • $\begingroup$ You don't appear to address the division by $n$ and the denominator of $a(n)$. Is this typo? $\endgroup$
    – joro
    Jun 1, 2021 at 13:34
  • $\begingroup$ $n$ has $p$ in power $k$, numerator has $p$ in power at least $k+1$ $\endgroup$ Jun 1, 2021 at 13:40
  • $\begingroup$ Does these conjectures contribute new results about Wieferich primes? If there are are only finitely many non-Wieferich primes, for all n we have gcd(n,a(n)) very large. $\endgroup$
    – joro
    Jun 1, 2021 at 14:09

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