cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

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

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Author

David Broadhurst, Sep 13 2009

Keywords

Comments

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.

Examples

			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.
		

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