A068209 -id:A068209 - cp's OEIS frontend

A165284 Primes p in A068209 whose squares never divide (x+1)^p-x^p-1 and x^x+(x+1)^(x+1) for the same x.

37493, 51941, 58073, 58901, 83813, 252341, 278321, 366521, 369821, 375101, 405689, 461861, 611801, 647837, 739061, 832721, 902201, 1001081, 1102301, 1180961, 1220801, 1269041, 1283297, 1426361, 1448081, 1483637, 1486577

Offset: 1

David Broadhurst, Sep 13 2009

nonn

A prime p belongs to A068209 if and only if p = 5 mod 6 and there are integers x with (x+1)^p - x^p - 1 = 0 mod p^2 and gcd(x^2+x,p) = 1.
This sequence is the subsequence of A068209 of primes p for which no such x solves x^x + (x+1)^(x+1) = 0 mod p^2.
For all other primes p < 1486577 in A068209, simultaneous solutions have been found by computing discrete logarithms.

			To prove that a(3) = 58073, we first show that (x+1)^p - x^p - 1 mod p^2, with gcd(x^2+x,p) = 1, has solutions when p = 58073 only for the residues x = r, -r/(1+r), 1/r, -(1+r), -1/(1+r), -(1+1/r) mod p, with r = 1281. By examining the orders of 1+1/r, 1+r, -r mod p, we prove that no x in this equivalence class can satisfy x^x + (x+1)^(x+1) = 0 mod p^2.
Similarly, we prove the absence of simultaneous roots for p = 37493, with r = 3730, and for p = 51941, with r = 15579.
By computing discrete logarithms, we provide simultaneous solutions for all other primes in A068209 with p < 58073.

David Broadhurst, On roots of n^n + (n+1)^(n+1) = 0 mod p^2
Kevin Brown, On the Density of Some Exceptional Primes

Cf. A068209

A320535 Number of solutions to (x+1)^p - x^p == 1 (mod p^2) in the range 1 <= x <= p - 2, p = prime(n).

Original entry on oeis.org

0, 0, 0, 2, 0, 2, 0, 2, 0, 0, 2, 2, 0, 2, 0, 0, 12, 2, 2, 0, 2, 8, 6, 0, 2, 0, 2, 0, 2, 0, 2, 0, 0, 2, 0, 2, 2, 2, 0, 0, 6, 2, 0, 8, 0, 2, 2, 2, 6, 2, 0, 0, 2, 0, 0, 0, 0, 2, 2, 0, 2, 0, 2, 0, 2, 0, 2, 8, 0, 2, 0, 0, 2, 2, 2, 0, 0, 2, 0, 2, 6, 8, 0, 2, 2, 6, 0

Offset: 1

Jianing Song, Oct 15 2018

nonn

For primes p > 2, (x+1)^p - x^p == 1 (mod p^2) has trivial solutions x == 0, -1 (mod p) so they are excluded.
Equivalently, a(n) is the number of solutions to x^(p-1) == (x+1)^(p-1) == 1 (mod p^2) in the range 1 <= x <= p^2 - 2, p = prime(n), that is, number of x such that both x and x + 1 occurs in the n-th row of A143548.
All terms shown here are even. The first odd terms are a(183) = 17 and a(490) = 5, with corresponding primes prime(183) = 1093 and prime(490) = 3511. a(n) is odd iff prime(n) is in A001220.
Let g be a primitive root modulo p^2, then (x+1)^p - x^p == 1 (mod p^2) has nontrivial solutions x == g^((p-1)/3) or g^(2*(p-1)/3) (mod p), and x^(p-1) == (x+1)^(p-1) == 1 (mod p^2) has nontrivial solutions x == g^(p*(p-1)/3) or g^(2*p*(p-1)/3) (mod p^2). As a result, if prime(n) == 1 (mod 6) then a(n) > 0. Primes p == 5 (mod 6) such that the equations have nontrivial solutions are listed in A068209.
a(17) = 12 is surprisingly large comparing with its nearby terms. Among the first 1000 terms there are only 7 that are larger than 12. They are a(183) = 17 and a(385) = a(552) = a(582) = a(593) = a(739) = a(922) = 14 (the corresponding primes are 1093, 2659, 4003, 4243, 4339, 5623 and 7213).

			The nontrivial solutions to (x+1)^7 - x^7 == 1 (mod 49) are x == 2, 4 (mod 7); the solutions to x^6 == (x+1)^6 == 1 (mod 49) are x == 18, 30 (mod 49), so a(4) = 2.

Robert Israel, Table of n, a(n) for n = 1..2500

Cf. A001220, A068209, A143548.

f:= proc(n) local p; p:= ithprime(n);
  nops(select(t -> (t+1)^p - t^p  mod p^2 = 1, [$1 .. p - 2]))
end proc:
map(f, [$1..100]); # Robert Israel, Mar 18 2024

a(n) = my(p=prime(n)); sum(x=1, p-2, Mod(x+1, p^2)^p-Mod(x, p^2)^p==1);

Name corrected by Robert Israel, Mar 18 2024

A358315 Primes p == 1 (mod 3) such that there exists 1 <= x <= p-2 such that (x+1)^p - x^p == 1 (mod p^2) and that p does not divide x^2 + x + 1.

Original entry on oeis.org

79, 193, 337, 421, 457, 547, 601, 619, 691, 757, 787, 907, 1039, 1093, 1231, 1237, 1303, 1489, 1531, 1657, 1993, 2089, 2113, 2251, 2311, 2377, 2389, 2437, 2539, 2647, 2659, 2713, 2731, 2749, 3001, 3037, 3109, 3229, 3319, 3331, 3511, 4003, 4177, 4243, 4273, 4339, 4447

Offset: 1

Jianing Song, Nov 08 2022

nonn

If p == 1 (mod 3) and p divides x^2 + x + 1, then p^2 divides (x+1)^p - x^p - 1; see A068209 for a proof.
Primes p == 1 (mod 3) such that A320535(primepi(p)) > 2.
Conjecture: this density of this sequence among the primes congruent to 1 modulo 3 is the same as that of A068209 among the primes congruent to 2 modulo 3. - Jianing Song, Nov 08 2022

			For p = 79, the nontrivial solutions to (x+1)^p - x^p == 1 (mod p^2) are x == 11, 23, 32, 36, 42, 46, 55, 67 (mod 79). The equivalent classes x == 11, 32, 36, 42, 46, 67 (mod 79) satisfy x^2 + x + 1 != 0 (mod 79), so 79 is a term.

Jianing Song, Table of n, a(n) for n = 1..1312 (all terms up to 2*10^5)

Cf. A068209, A320535.

isA358315(n) = if(isprime(n) && n%3==1, for(a=1, n-2, if(Mod(a+1,n^2)^n - Mod(a,n^2)^n==1 && znorder(Mod(a,n))!=3, return(1)))); return(0)

cp's OEIS Frontend

A165284 Primes p in A068209 whose squares never divide (x+1)^p-x^p-1 and x^x+(x+1)^(x+1) for the same x.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A320535 Number of solutions to (x+1)^p - x^p == 1 (mod p^2) in the range 1 <= x <= p - 2, p = prime(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

Extensions

A358315 Primes p == 1 (mod 3) such that there exists 1 <= x <= p-2 such that (x+1)^p - x^p == 1 (mod p^2) and that p does not divide x^2 + x + 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI