I'm rereading an older text on fermat-quotients (see wikipedia) from which I have now the

Question for $$ b^{p-1} \equiv 1 \pmod{ p^m} \qquad \text{ with $p \in \mathbb P $, $1 \lt b \lt p$ and $m \gt 2$} $$ (This is a generalization of the question for Wieferich primes).

Note that I ask here for examples, where the bases $b$ are smaller than the prime $p$, so a very well known weaker case $3^{10} \equiv 1 \pmod {11^2 } $ were an example, but only if the exponent at $11$ where one more; however frequent and well known cases like $18^6 \equiv 1 \pmod {7^3} $ were not because the base is bigger than the prime.

The only example that I've found so far is $$ 68^{112} \equiv 1 \pmod {113^3 } $$ but I've scanned only the first 2000 primes $p \in (3 \ldots 17389)$ and my primitive brute force algorithm has more than quadratic time-characteristic, so checking 10 000 or 100 000 primes were no fun - the quadratic regression prognoses 1 hour for testing 10 000 primes and 101 hours for testing 100 000 primes...

I'm aware of a couple of webpages containing lists of fermat quotients up to much higher primes, but either there is no explicite mention of the cases of $b \lt p$ and quotient $m \gt 2$ or I've been too dense when scanning through the listings (Richard Fischer, Wilfrid Keller, Michael Mossinghoff)

For reference: my Pari/GP-code is

        r = lift(Mod(k,p3)^(p-1));
        if(r==1,print(p," ",k," ",r)));

P.s. I've no real good idea for tagging of this question; I just tried the most similar...

  • $\begingroup$ Your "p3=p1^3" line of code should be "p3=p^3", yes? $\endgroup$ Dec 5 '13 at 22:32
  • $\begingroup$ @matth: yepp. corrected.Typical error when adapting/formatting code for the wider audience...;-) $\endgroup$ Dec 5 '13 at 22:36
  • $\begingroup$ I ran your code to j=10000 and found no more solutions. $\endgroup$ Dec 6 '13 at 0:25
  • $\begingroup$ Similar question: math.stackexchange.com/questions/456744/… $\endgroup$ Aug 15 '14 at 5:59

In MO there was an answer indicating, that there shall be no more information than that of Richard Fischer's site, where he lists, that indeed that pair $(68,113)$ is the only pair up to about $p \le 3.6 \cdot 10^6$ and where also $b \lt p$ which gives a fermat-quotient greater than 2 , so I think I should "close the case" here.

For the casual reader I'll add a link to a more explanative description of the problem and my empirical table. See here.


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