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

1,5

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<r<p: r is a quadratic residue mod p}.

(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<n<p: n is a quadratic nonresidue mod p}.

(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.

Links

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

Example

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

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}]

Crossrefs

Cf. A000040, A000290, A257364.

Sequence in context: A270030 A284104 A369732 * A228631 A101194 A196344

Adjacent sequences: A260908 A260909 A260910 * A260912 A260913 A260914

Keyword

nonn

Author

Zhi-Wei Sun, Aug 03 2015

Status

approved