# Fermat-quotient of "order" 3: I found $68^{112} \equiv 1 \pmod {113^3}$ - are there bigger examples known?

(I've taken this from MSE, it seems to be more appropriate here)

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

for(j=2,2000,p=prime(j);p3=p^3;
for(k=2,p-1,
r = lift(Mod(k,p3)^(p-1));
if(r==1,print(p," ",k," ",r)));
);


(One correspondent took this up to the 10 000'th prime which is $p=104729$ )

[update] For the later casual reader I've included a more involved explanation and a table of data. See here (filesize 2Mb, inconvenient for modem-transfer)

• Cases b<p can be easily identified in Fischer's table fermatquotient.com/FermatQuotienten/FermatQ3 and there no other for primes below 3.6*10^8. Dec 7, 2013 at 1:52
• The question is discussed in MR2114655 (2005i:11004) Keller, Wilfrid; Richstein, Jörg; Solutions of the congruence $a^{p-1}\equiv1\pmod{p^r}$, Math. Comp. 74 (2005), no. 250, 927–936 but I don't think any solutions are found, other than the one you already have. Still, the paper might be worth a look, to see what's been done. Dec 7, 2013 at 4:34
• Thanks to Max and Gerry for the links - I could access them both and also got the article of Keller/Richstein. Dec 7, 2013 at 5:16
• @Max : if you would like to convert your comment into an answer I could mark that as accepted which would then "close the case"... Dec 7, 2013 at 15:39

Cases with $b<p$ can be easily identified in Fischer's table http://fermatquotient.com/FermatQuotienten/FermatQ3 and there are no other such instances for primes below $3.6\cdot 10^8$.