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

Author

Christopher E. Thompson, Dec 08 2019

Keywords

• nonn

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).

Examples

			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.
		

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.