A330893 Numbers whose set of divisors contains a Pythagorean quadruple.

42, 72, 84, 126, 144, 156, 168, 198, 210, 216, 252, 288, 294, 312, 330, 336, 342, 360, 378, 396, 420, 432, 462, 468, 504, 546, 570, 576, 588, 594, 624, 630, 648, 660, 672, 684, 714, 720, 756, 780, 792, 798, 840, 864, 882, 900, 924, 930, 936, 966, 990, 1008, 1026

Offset: 1

Author

Michel Lagneau, May 01 2020

Keywords

• nonn

Comments

A Pythagorean quadruple (x, y, z, m) is a set of positive integers that satisfy x^2 + y^2 + z^2 = m^2.

The corresponding number of quadruples of the sequence is 1, 1, 2, 2, 2, 1, 3, 1, 3, 2, 4, 3, 2, 2, 1, 4, 1, 2, 3, 2, 7, 4, ... (see the sequence A330894).

It is interesting to note that each set of divisors of a(n) contains m primitive Pythagorean quadruples for some n, m = 1, 2,...

Examples:

- The set of divisors of a(1)= 42 contains only one primitive Pythagorean quadruple: (2, 3, 6, 7).

- The set of divisors of a(9) = 210 contains two primitive Pythagorean quadruples: (2, 3, 6, 7) and (2, 5, 14, 15).

- The set of divisors of a(21) = 420 contains three primitive Pythagorean quadruples: (2, 3, 6, 7), (2, 5, 14, 15) and (4, 5, 20, 21).

If k is in the sequence then so is m*k for m > 1.

Assumes the elements (x,y,z,m) in a quadruple are distinct divisors, as otherwise 6 would be in the sequence with 1^2+2^2+2^2=3^2. - Chai Wah Wu, Nov 16 2020

Examples

			168 is in the sequence because the set of divisors  {1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168} contains the Pythagorean quadruples {2, 3, 6, 7}, {4, 6, 12, 14} and {8, 12, 24, 28}. The first quadruple is primitive.
		

Links

• Eric Weisstein's World of Mathematics, Pythagorean Quadruples.

Crossrefs

Cf. A027750, A169580, A330894, A331365.

Programs

Maple

with(numtheory):
for n from 3 to 1200 do :
   d:=divisors(n):n0:=nops(d):it:=0:
    for i from 1 to n0-3 do:
     for j from i+1 to n0-2 do :
      for k from j+1 to n0-1 do:
      for m from k+1 to n0 do:
       if d[i]^2 + d[j]^2 + d[k]^2 = d[m]^2
        then
        it:=it+1:
        else
       fi:
      od:
     od:
    od:
    od:
    if it>0 then
    printf(`%d, `,n):
    else fi:
   od:

Mathematica

nq[n_] := If[ Mod[n,6]>0, 0, Block[{t, u, v, c = 0, d = Divisors[n], m}, m = Length@ d; Do[ t = d[[i]]^2 + d[[j]]^2; Do[u = t + d[[h]]^2; If[u > n^2, Break[]]; If[ Mod[n^2, u] == 0 && IntegerQ[v = Sqrt@ u] && Mod[n, v] == 0, c++], {h, j+1, m - 1}], {i, m-3}, {j, i+1, m - 2}]; c]]; Select[ Range@ 1026, nq[#] > 0 &] (* Giovanni Resta, May 04 2020 *)

PARI

isok(n) = {my(d=divisors(n), x); for (i=1, #d-3, for (j=i+1, #d-2, for (k=j+1, #d-1, if (issquare(d[i]^2 + d[j]^2 + d[k]^2, &x) && !(n % x), return(1)););););} \\ Michel Marcus, Nov 16 2020

Formula

a(n) == 0 (mod 6).