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

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)));
);


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

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