A330274 Largest positive x such that (x,x+n) is the smallest pair of quadratic residues with difference n, modulo any prime.

9, 4, 1, 10, 4, 14, 9, 1, 9, 12, 5, 4, 11, 13, 1, 9, 10, 15, 11, 10, 4, 14, 4, 1, 15, 10, 9, 26, 16, 12, 9, 4, 16, 21, 1, 21, 23, 14, 16, 9, 15, 14, 17, 16, 4, 22, 9, 1, 16, 25, 25, 29, 19, 16, 9, 25, 30, 27, 16, 4, 24, 22, 1, 21, 16, 22, 29, 22, 31, 30, 10

Offset

1,1

Comments

There is a finite limit for any n. By considering the pairs (1,n+1), (n^2,n^2+n), (n,2n), (4n,5n), (9n,10n) it can be seen that a(n) <= max(9n,n^2).

References

Richard K. Guy, Unsolved Problems in Number Theory, Springer-Verlag (1981,1994,2004), section F6 "Patterns of quadratic residues".

Links

Christopher E. Thompson, Table of n, a(n) for n = 1..1000

Emma Lehmer, Patterns of power residues, J. Number Theory 17 (1983) 37-46.

Example

If each of the pairs (1,5),(4,8),(6,10),(3,7) are not both quadratic residues, then (10,14) must be. Moreover, if 3 is a quadratic residue but 2,5,7 and 13 are not, then (10,14) is the smallest pair (x,x+4) which are both quadratic residues. Therefore, a(4)=10.

Crossrefs

Sequence in context: A021519 A199780 A292684 * A248197 A199291 A091661

Adjacent sequences: A330271 A330272 A330273 * A330275 A330276 A330277

Keyword

nonn

Author

Christopher E. Thompson, Dec 08 2019

Status

approved