Finding primes so that xp+yp=zpx^p+y^p=z^p is unsolvable in the p-adic units

On my number theory exam yesterday, we had the following interesting problem related to Fermat’s last theorem:

Suppose p>2 is a prime. Show that xp+yp=zp has a solution in Z×p if and only if there exists an integer a such that p and (a+1)^p=a^p+1\pmod{p^2}.
(I wil not show this here, it’s a good exercise in Hensel’s lemma)

Now I’m interested for which primes this last property holds true.
This is what I found so far:

Proposition: for every prime p\equiv 1\pmod{3} there exists an a\in\mathbb{Z} such that p\not\mid a(a+1) and p^2\mid(a+1)^p-a^p-1.
Sketch of proof: We will often use the fact that if p\mid a-b then p^2\mid a^p-b^p for a,b,p\in\mathbb{Z}. There exists some n\in\mathbb{Z} such that p \mid n^2+n+1. We will show that we can take a=n (it is clear that p\not \mid n(n+1)). If n\equiv 1\pmod{p}, we are done because (n+1)^p\equiv 2\equiv n^p+1\pmod{p^2}. So we may assume that the order of n mod p is 3.

We see that n^{3p}\equiv 1\pmod{p^2}, hence p^2\mid \frac{n^{3p}-1}{n^p-1}=n^{2p}+n^p+1 and therefore (n+1)^p\equiv -n^{2p}\equiv n^p+1\pmod{p^2}. \blacksquare

What can we say about primes p\equiv 2\pmod{3}? Say a prime is bad if it satisfies the above property. It is easy to see that for p to be good we only have to check that for all 1\le a \le (p-1)/2 we have p^2\not\mid (a+1)^p-a^p-1.
Numerical data suggests that infinitely many primes p\equiv 2\pmod{3} are good and infinitely many are bad. I’d like to find a proof of this.
My numerical data furthermore suggests that there is no easy divisibility criterion to judge badness for primes, and that that there are a lot more good primes than bad ones, in terms of asymptotics.

I welcome all comments/ideas/references.


Edit: Here are some observations. I’ve listed the first bad primes, together with the set ofa\in\mathbb{Z} such that 1\le a \le (p-1)/2 and (a+1)^p\equiv a^p+1\pmod{p^2}:

59\qquad [3, 4, 11, 14, 15, 20] \\
83\qquad [8, 30, 36] \\
179\qquad [2, 59, 88] \\
227\qquad [36, 82, 92] \\
419\qquad[80, 110, 150] \\
443\qquad [108, 125, 201] \\
701\qquad [23, 61, 132, 146, 153, 252] \\
857\qquad [6, 143, 244] \\
887\qquad [20, 132, 168] \\
911\qquad [14, 64, 242] \\
929 \qquad[234, 253, 266] \\
971 \qquad[9, 97, 108] \\
977 \qquad[102, 182, 331] \\
1091\qquad [64, 234, 358] \\
1109 \qquad[176, 400, 523] \\
1193\qquad [224, 227, 473] \\
1217\qquad [186, 228, 410] \\
1223\qquad [304, 349, 409] \\
1259\qquad [19, 62, 264] \\
1283\qquad [19, 134, 449] \\
1289\qquad [412, 535, 593] \\
1439\qquad [93, 199, 294] \\
1487\qquad [167, 416, 649] \\
1493 \qquad[141, 179, 367] \\
1613 \qquad[227, 473, 739] \\
1637 \qquad[76, 279, 574] \\
1811 \qquad[39, 123, 498, 628, 691, 743] \\
1847\qquad [172, 362, 698] \\
1901\qquad [3, 475, 634] \\
1997\qquad [125, 671, 840] \\
2003 \qquad[31, 312, 840] \\
2087 \qquad[31, 202, 586] \\
2243 \qquad[252, 593, 1077] \\
2423\qquad [353, 704, 857] \\
2477\qquad [17, 688, 1020] \\
2579 \qquad[294, 576, 693] \\
2591 \qquad[322, 345, 368] \\
2729\qquad [1024, 1049, 1089] \\
2777\qquad [488, 1078, 1269] \\
2969\qquad [341, 789, 1158] \\
3089\qquad [677, 1015, 1053]\\
3137\qquad [1227, 1308, 1427]\\
3191\qquad [776, 916, 1383]\\
3203\qquad [1214, 1305, 1587]\\
3251\qquad [164, 337, 1260]

I also found this sequence on OEIS, along with more numerical data, where it is claimed (but not proved) that the density of bad primes is roughly \frac 16.

My numerical data also suggests that the number of a\in\mathbb{Z} satisfying 1\le a \le (p-1)/2 and p^2\mid (a+1)^p-a^p-1 is always either 3 or 6, but I think that’s rather a bold statement. (edit: indeed, 20123 is a counterexample with 9 solutions, found by Michael Stocker. The weaker statement that the number of such solutions is divisible by 3 is true, however: see here.)


Edit: Someone showed me these slides, where the above criterion for solvability of x^p+y^p=z^p in the p-adic units is stated. (There is a remark that it can be proven with eisenstein reciprocity that for all non-Wieferich primes the first case of FLT holds true – maybe there is some way to use this for some (partial) result on our problem? I’m really curious.)

Answer

Attribution
Source : Link , Question Author : ArtW , Answer Author : Community

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