A260911 Least positive integer k < prime(n) such that there are 0 < i < j < k for which i^2 + j^2 = k^2 and i,j,k are all quadratic residues modulo prime(n), or 0 if no such k exists.

0, 0, 0, 0, 5, 0, 0, 0, 0, 25, 0, 34, 0, 41, 25, 25, 5, 5, 26, 5, 37, 0, 41, 0, 0, 65, 17, 34, 5, 61, 17, 5, 17, 25, 25, 29, 37, 26, 25, 41, 5, 5, 5, 25, 25, 53, 34, 17, 34, 5, 109, 5, 5, 5, 17, 37, 34, 41, 34, 53

Offset: 1

Author

Zhi-Wei Sun, Aug 03 2015

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 25. In other words, for any prime p > 100, we have a^2 + b^2 = c^2 for some a,b,c in the set R(p) = {0

(ii) For any prime p > 50, we have a^2 + b^2 = c^2 for some a,b,c in the set N(p) = {0

(iii) For any prime p > 32, we have a^2 + b^2 = c^2 for some a,b in the set R(p) and c in the set N(p).

(iv) For any prime p > 72, we have a^2 + b^2 = c^2 for some a,b in the set N(p) and c in the set R(p).

I have verified the conjecture for primes p < 1.5*10^7.

Examples

			a(10) = 25 since 7^2 + 24^2 = 25^2, and 7, 24, 25 are all quadratic residues modulo prime(10) = 29.
		

Links

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

Crossrefs

Cf. A000040, A000290, A257364.

Programs

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]]
Do[Do[If[JacobiSymbol[k,Prime[n]]<1,Goto[bb]];Do[If[JacobiSymbol[j,Prime[n]]<1,Goto[cc]];
If[SQ[k^2-j^2]&&JacobiSymbol[Sqrt[k^2-j^2],Prime[n]]==1,Print[n," ",k];Goto[aa]];Label[cc];Continue,{j,1,k-1}];Label[bb];Continue,{k,1,Prime[n]-1}];
Print[n," ",0];Label[aa];Continue,{n,1,50}]

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