2
$\begingroup$

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?

$\endgroup$
0

1 Answer 1

6
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.