A124132 Positive integers n such that Fibonacci(2*n) is the sum of two squares.

1, 3, 6, 7, 13, 19, 31, 37, 43, 49, 61, 67, 73, 79, 91, 111, 127, 163, 169, 183, 199, 223, 307, 313, 349, 361, 397, 433, 511, 523, 541, 613, 619, 709, 823, 907, 1087, 1123, 1129, 1147, 1213, 1279

Offset: 1

Author

Melvin J. Knight (melknightdr(AT)verizon.net), Nov 30 2006

Keywords

• nonn
• more

Comments

This sequence excludes Fibonacci numbers with odd indices, all of which are sums of two squares.

Fibonacci numbers with even indices factor as F_2n = F_n L_n, L_n being n-th Lucas number. Thus F_2n factors further as F_2n = F_m L_m L_2m L_4m...L_n, with m odd. Since F_m is a sum of two squares and the pairs of Lucas numbers all have GCD dividing 2, the conclusion for F_2n depends on each Lucas number being a sum of two squares. Joint work with Kevin O'Bryant and Dennis Eichhorn.

To write Fibonacci(n) as a^2+b^2: find the a^2+b^2 representation for the individual prime factors, by using Cornacchia's algorithm, and then merge them by using the formulas (a^2+b^2)(c^2+d^2) = |ac+bd|^2 + |ad-bc|^2 = |ac-bd|^2 + |ad+bc|^2. - V. Raman, Aug 29 2012

All corresponding Fibonacci(2*n) values are the sum of two nonzero distinct squares except n = 1, 3, 6. - Altug Alkan, May 04 2016

1501 <= a(43) <= 1651. 1651, 1849, 2449 are terms. - Chai Wah Wu, Jul 22 2020

Examples

			a(4) = 7 because the first four Fibonacci numbers with even indices that are the sum of two squares are F_2, F_6, F_12 and F_14, 14 being 2*a(4) and F_14 = 377 = 11^2+16^2.
		

Links

• Christian Ballot and Florian Luca, On the equation x^2+dy^2-F_n, Acta Arithmetica, 2 (2007), 145-155.
• Blair Kelly, Fibonacci and Lucas factorizations
• Wikipedia, Cornacchia's algorithm

Crossrefs

Cf. A001906, A124130 (for Lucas numbers), A001481.

Programs

Mathematica

Select[Range[100], Length[FindInstance[x^2 + y^2 == Fibonacci[2 #], {x, y}, Integers]] > 0 &] (* T. D. Noe, Aug 27 2012 *)

PARI

for(i=2, 500, a=factorint(fibonacci(i))~; has=0; for(j=1, #a, if(a[1, j]%4==3&&a[2, j]%2==1, has=1; break)); if(has==0&&i%2==0, print((i/2)", "))) \\ V. Raman, Aug 27 2012

Python

from itertools import count, islice
from sympy import factorint, fibonacci
def A124132_gen(): # generator of terms
    return filter(lambda n:all(p & 3 != 3 or e & 1 == 0 for p, e in factorint(fibonacci(2*n)).items()),count(1))
A124132_list = list(islice(A124132_gen(),10)) # Chai Wah Wu, Jun 27 2022

Extensions

a(22)-a(38) from V. Raman, Aug 27 2012

a(39)-a(42) from Chai Wah Wu, Jul 22 2020