# Euler quotients modulo $n$

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

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.

• You don't appear to address the division by $n$ and the denominator of $a(n)$. Is this typo?
– joro
Jun 1 '21 at 13:34
• $n$ has $p$ in power $k$, numerator has $p$ in power at least $k+1$ Jun 1 '21 at 13:40
• 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.
– joro
Jun 1 '21 at 14:09