A260946 Least positive integer k < prime(n) such that i^2 + j^2 = k^2 for some 0 < i < j with i*j*k a primitive root modulo prime(n), or 0 if no such k exists.

0, 0, 0, 0, 10, 0, 15, 5, 5, 5, 0, 17, 5, 15, 5, 10, 10, 26, 10, 17, 5, 5, 5, 5, 5, 13, 15, 5, 10, 15, 15, 10, 13, 13, 5, 17, 5, 10, 5, 10, 10, 10, 25, 61, 10, 13, 17, 25, 5, 50, 15, 13, 17, 10, 15, 5, 5, 10, 17, 10, 26, 10, 10, 17, 5, 10, 5, 5, 5, 13

Offset: 1

Author

Zhi-Wei Sun, Aug 05 2015

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 11. In other words, for any prime p > 31 there are a,b,c among 1,...,p-1 with a^2 + b^2 = c^2 such that a*b*c is a primitive root modulo p.

Examples

			 a(5) = 10 since 10^2 = 6^2 + 8^2, and 6*8*10 = 480 is a primitive root modulo prime(5) = 11.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A260911, A260947.

Programs

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]]
Dv[n_]:=Divisors[Prime[n]-1]
Do[Do[If[SQ[k^2-j^2]==False, Goto[aa]];Do[If[Mod[(Sqrt[k^2-j^2]j*k)^(Part[Dv[n],t]),Prime[n]]==1,Goto[aa]];Continue,{t,1,Length[Dv[n]]-1}];Print[n," ",k];Goto[bb];Label[aa];Continue,{k,1,Prime[n]-1},{j,1,k-1}];Print[n," ",0];Label[bb];Continue,{n,1,70}]