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%I #13 Dec 18 2019 01:14:41
%S 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,
%T 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,
%U 16,9,25,30,27,16,4,24,22,1,21,16,22,29,22,31,30,10
%N Largest positive x such that (x,x+n) is the smallest pair of quadratic residues with difference n, modulo any prime.
%C 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).
%D Richard K. Guy, Unsolved Problems in Number Theory, Springer-Verlag (1981,1994,2004), section F6 "Patterns of quadratic residues".
%H Christopher E. Thompson, <a href="/A330274/b330274.txt">Table of n, a(n) for n = 1..1000</a>
%H Emma Lehmer, <a href="https://doi.org/10.1016/0022-314X(83)90004-5">Patterns of power residues</a>, J. Number Theory 17 (1983) 37-46.
%e 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.
%K nonn
%O 1,1
%A _Christopher E. Thompson_, Dec 08 2019
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