A334566 Number of solutions of the Diophantine equation z^2 - y^2 - x^2 = n > 0 when the positive integers, x, y and z, are consecutive terms of an arithmetic progression.

0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 3, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 1, 2, 0, 0, 3, 2, 0, 0, 2, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 2, 1, 0, 0, 3, 0, 0, 0, 1, 3, 0, 0, 4, 2, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 3, 1, 0, 0, 1, 1, 0, 0, 1, 3, 0, 0, 2, 0

Offset: 1

Author

Bernard Schott, May 06 2020

Keywords

• nonn

Comments

Inspired by the 135th and 136th problems of Project Euler (see links).

If d is the common difference of the arithmetic progression (x, y, z), then the Diophantine equation becomes (y+d)^2 - y^2 - (y-d)^2 = n <==> y^2 - 4dy + n = 0 <==> n = y * (4d-y).

If y is the average term, then y divides n.

Offset is 1 because for n = 0, every (x, y, z)= (3d, 4d, 5d) with d>0 would be solution.

Examples

			a(3) = 1 because 4^2 - 3^2 - 2^2 = 3.
a(15) = 3 because 5^2 - 3^2 - 1^2 = 7^2 - 5^2 - 3^2 = 19^2 - 15^2 - 11^2 = 15.
If n = 4q+3, q >= 0 then (3q+2, 4q+3, 5q+4) is a solution.
If n = 16q, q >= 1 then (3q-1, 4q, 5q+1) is a solution.
If n = 16q+4, q >= 0 then (6q+1, 8q+2, 10q+3) is a solution.
If n = 16q+12, q >= 0 then (6q+4, 8q+6, 10q+8) is a solution.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Project Euler, Problem 135: Same differences
• Project Euler, Problem 136: Singleton difference

Crossrefs

Cf. A001248, A002145, A004767, A008598, A016813, A016825, A051062.

Cf. A334567 (least value of n such that a(n) = k>0).

Programs

Maple

f:= proc(n) local r; r:= floor(sqrt(n/3));
nops(select(t -> n/t + t mod 4 = 0 and t > r, numtheory:-divisors(n)))
end proc:
map(f, [$1..100]); # Robert Israel, Jul 31 2020

Mathematica

a[n_] := Length@ Solve[(4 d - x) x == n  && x>0 && x-d>0 && x+d>0, {d, x}, Integers]; Array[a, 90] (* Giovanni Resta, May 06 2020 *)

Formula

a(n) = 0 iff n = 4q+1 (A016813), n = 4q+2 (A016825), n = 16q+8 (A051062), q>= 0.

a(n) >= 1 iff n = 4q+3, q >=0 (A004767), n = 16q, q>=1 (A008598), n = 16q+4, q>=0 (A119413), n = 16q+12, q>=0 (A098502).

a(4*q^2) >= 1, for q >= 1, since (q, 2q, 3q) is a solution.

a(p) = 1 for p = 4q+3 prime (A002145).

a(p^2) = 0 for p an odd prime (A065091).

Extensions

More terms from Giovanni Resta, May 06 2020