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%I #34 Nov 08 2022 16:45:24
%S 59,83,179,227,419,443,701,857,887,911,929,971,977,1091,1109,1193,
%T 1217,1223,1259,1283,1289,1439,1487,1493,1613,1637,1811,1847,1901,
%U 1997,2003,2087,2243,2423,2477,2579,2591,2729,2777,2969,3089,3137,3191,3203,3251
%N Primes p of the form 3k - 1 such that there exist nontrivial solutions (x other than 0 or -1 modulo p) to the congruence (x+1)^p - x^p == 1 (mod p^2).
%C Note that nontrivial solutions always exist for primes of the form 3k + 1. - _Jianing Song_, Apr 20 2019
%C From _Jianing Song_, Nov 08 2022: (Start)
%C Proof: for prime p > 3, write f_p(x) = ((x+1)^p - x^p - 1)/p; f_p(x) is a polynomial in Z[x]. We have f_p(w) = f_p(w^2) = 0, where w is a primitive cube root of 1, so f_p(x) divides x^2 + x + 1 in Q[x]. Since x^2 + x + 1 is a primitive polynomial (having coprime coefficients), it follows from Gauss's lemma for polynomials that f_p(x) divides x^2 + x + 1 in Z[x]. As a result, if p == 1 (mod 3) and p | (x^2 + x + 1) for some x, then p^2 | ((x+1)^p - x^p - 1).
%C For prime p > 2, the equation x^p + y^p = z^p has nontrivial solutions over (Z_p)* (the p-adic integers not divisible by p) if and only if there exist nontrivial solutions to the congruence (x+1)^p - x^p == 1 (mod p^2). (End)
%H Jianing Song, <a href="/A068209/b068209.txt">Table of n, a(n) for n = 1..1397</a> (all terms up to 2*10^5; first 74 terms from Robert G. Wilson v)
%H K. S. Brown, <a href="http://www.mathpages.com/home/kmath411.htm">On the Density of Some Exceptional Primes</a>
%o (PARI) isA068209(n) = if(isprime(n) && n%3==2, for(a=1, n-2, if(Mod(a+1,n^2)^n - Mod(a,n^2)^n==1, return(1)))); return(0) \\ _Jianing Song_, Nov 08 2022
%Y Cf. A001220, A320535.
%K easy,nonn
%O 1,1
%A _Benoit Cloitre_, Mar 23 2002
%E Definition corrected by _Mike Oakes_, Feb 12 2009
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