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. A well-known collection of theorems beginning with the work of Wieferich shows that if the first case of Fermat's Last Theorem holds for $p$, then $q_p(a)$ is divisible by $p$ for every prime number $a \leq 89$.

I have stumbled across some congruences between Fermat quotients, and haven't turned up similar ones in a Google search. I hope someone here is an expert on these and can tell me 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 $q_p(2) \equiv 2q_a(2) \pmod{p}$. In particular, since $q_a(2) < p$, a Mersenne prime is not a Wieferich prime (i.e., $q_p(2)$ is not divisible by $p$), a well-known result but not by this proof, I think.

If $p = 2^n-3$ is prime for some $n$, then $3nq_p(2) + 1 \equiv 3q_p(3) \pmod{p}$, so in particular by the result mentioned above, the first case of FLT must hold for all such $p$.

If $p = 3^n-4$ is prime for some $n$, then $4nq_p(3) + 1 \equiv 8 q_p(2) \pmod{p}$, so again, the first case of FLT must hold for all such $p$.

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

Thanks in advance for any information that you can provide.


I absolutely am not an expert, but I did informally study Fermat quotients a few years ago, and I don't recall seeing this sort of thing in the literature, so I will make a mild guess that these are not known. Unfortunately I don't expect this approach will help much with FLT because e.g. primes of the form $2^n-3$ are probably extremely sparse (possibly there are only finitely many; I'm not sure how/whether the heuristics for Mersenne primes would carry over).

If you are interested, there is another class of quotients whose indivisibility by $p$ implies the first case of FLT for $p$; I don't recall if they have an established name, but one might call them Lucas quotients. For $m\equiv 1\bmod 4$ and a particular Lucas sequence $L$ depending on $m$, one has the congruence $$L_{p-\left(\frac{m}{p}\right)}\equiv 0\bmod p$$ (note that $L_0=0$; there is an obvious analogy with $a^{p-1}\equiv a^0\bmod p$), and the first case of FLT holds for $p$ if $p$ does not divide the quantity $$\frac{L_{p-\left(\frac{m}{p}\right)}}{p}$$ The only reference I'm aware of for this fact is this paper by Granville (p.13, equation 6.9).

In the particular instance of $m=5$, the Lucas sequence is just the Fibonacci numbers, and the above criterion is called the Wall-Sun-Sun criterion. There has been more in-depth investigation of this case; no Wall-Sun-Sun primes (i.e. $p^2\mid F_{p-\left(\frac{p}{5}\right)}$) are currently known.

  • $\begingroup$ Thank you Zev. I'll wait to see if anyone else chimes in. If not, then I'll pick your answer and I guess repost at MO. I wasn't aware of the Wall-Sun-Sun criterion and the paper by Granville, so I'll have some fun reading if I find the time. Thanks again! $\endgroup$ – Barry Smith Apr 27 '11 at 19:56
  • $\begingroup$ @Barry: No problem, I just wanted to chime in since I happened to have some (small) amount of interaction with them before. Hope you have better luck on MO! $\endgroup$ – Zev Chonoles Apr 28 '11 at 1:16

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