

A186699


Numbers n such that there are n numbers in arithmetic progression whose squares sum to a perfect square.


3



1, 2, 4, 9, 11, 16, 23, 24, 25, 26, 33, 36, 47, 49, 50, 52, 59, 64, 73, 74, 81, 88, 96, 97, 100, 107, 121, 122, 144, 146, 148, 169, 177, 184, 191, 193, 194, 196, 218, 225, 239, 241, 242, 244, 249, 256, 276, 289, 292, 297, 299, 311, 312, 313, 324, 337, 338
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OFFSET

1,2


COMMENTS

A positive integer n is in this sequence if and only if there is a solution to the Pelllike equation x^2ny^2=d^2(n1)n(n+1)/3 for some x,y,d integers.
A positive integer n is in this sequence if and only if it can be written in the form: (u^23w^2)/(v^2+3w^2), with u,v,w integers and gcd(v,w)=1. This can also be written as a n(v^2) + 3(n+1)(w^2) = z^2.
If n is in this sequence, then we can find an arithmetic progression of *positive* integers which satisfy this equation. (The description above does not require the sequence to be positive.)
By using the method of Legendre to find whether there exists rational numbers r,s on the curve nr^2 + 3(n+1)s^2 = 1, we get the following necessary and sufficient conditions on n:
A. Factor n=a^2b, with b squarefree, then
1. If 3 does not divide b(n+1), then b ≅ 1 (mod 3)
2. If b is divisible by 3, then b ≅ 6 (mod 9)
3. 3 is a square (mod b.)
B.
1. If n+1 is divisible by 3, then (n+1)/3 is the sum of two perfect squares
2. If n+1 is not divisible by 3, then n+1 is the sum of two perfect squares
When n is a perfect square, we can use the arithmetic sequence starting at m=(3n+2)(sqrt(n)1)/2 + 6 and common difference 6.


LINKS



EXAMPLE

For n=4, (13,19,25,31) is an arithmetic progression of length 4, and 13^2+19^2+25^2+31^2 = 46^2, so 4 is in the sequence.


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



