This a repost of a question I asked at Stack Exchange: 


<https://math.stackexchange.com/questions/35264/congruences-for-fermat-quotients>


I didn't get a complete answer to my question, so I'm trying again here.


If $p$ is a prime number and $a$ is relatively prime to $p$, then by Fermat's Little Theorem, the **Fermat quotient** $q\_p(a) = (a^{p-1}-1)/p$ is an integer. To summarize that other post, I stumbled across some congruences relating Fermat quotients and wonder 1) if these and/or similar congruences are known, and 2)where I can find other proofs than my own, either simple enough to post here or written up somewhere.


Here are some examples:


If $p = 2^a-1$ is a Mersenne prime, so $a$ is prime, then $q\_p(2) \equiv 2q\_a(2) \pmod{p}$. 


If $p = 2^n-3$ is prime for some $n$, then $3nq\_p(2) + 1 \equiv 3q\_p(3) \pmod{p}$


If $p = 3^n-4$ is prime for some $n$, then $4nq\_p(3) + 1 \equiv 8 q\_p(2) \pmod{p}$


I have proved a few others of this nature and seem to have a way to generate more if I wish.


Edit: These and other similar congruences follow from an elementary argument of Edmund Landau. See my answer to my own question belowl.



Congruences between Fermat quotients
Let just try a look at your first result:


Corollary 1 in Johnson, Wells paper above says:


Let $\epsilon = 1,-1$ and let $p=2^r+ \epsilon$ be a prime number.
Then


$$
q\_p(2) \equiv \frac{\epsilon}{r} \pmod{p} 
$$


Your result for a Mersenne prime (taking $\epsilon =-1$ above) follows from this:


Observe that (trivially) $2^r \equiv 1 \pmod{p}$. Thus $2^r-2 \equiv -1 \pmod{p}$


In other words:


$$
2q\_r(2) \equiv (2^r -2) \frac{1}{r} \equiv (-1) \cdot \frac{1}{r} \equiv \frac{\epsilon}{r} \equiv q\_p(2) \pmod{p}
$$


Thus 
$$
q\_p(2) \equiv 2q\_r(2) \pmod{p}.
$$


