This might be related to an open problem.


For odd natural $n$ define the Euler quotient:


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



> 
> Q1 Are there infinitely many $n$ with unbounded smallest prime factor
> for which $a(n)$ is non-zero?
> 
> 
> 



> 
> Q2 Are there infinitely many $n$ for which $a(n)$ is non-zero?
> 
> 
> 


$a(n)=0$ is [OEIS Wieferich numbers (1)](https://oeis.org/A077816)


$a(n)=1$ for prime $n$ is [OEIS A125854](https://oeis.org/A125854)


$a(n)=3$ for prime $n$ is [OEIS A175866](https://oeis.org/A175866)


Conditional results are welcome, but we believe abc implies positive
answers.



Are there infinitely many nonzero Euler quotients $a(n)=\frac{2^{\phi(n)}-1}{n} \bmod n$?
Let $p$ be any odd prime. Suppose that $2^{p-1}-1=p^ku$ for some integer $u$ with $(p,u)=1$. Then by Lifting The Exponent Lemma for all integer $K\geq 0$ we have
$$
2^{p^K(p-1)}-1=p^{K+k}u\_K
$$
For some integer $u\_K$ such that $(p,u\_K)=1$. Therefore, for all $K>k$ the number $n=p^{K+1}$ satisfies
$$
2^{\varphi(n)}-1\not\equiv 0\pmod {n^2}
$$


