A230710 Values of x such that x^2 + y^2 = 5^n with x and y coprime and 0 < x < y.

1, 3, 2, 7, 38, 44, 29, 336, 718, 237, 2642, 10296, 8839, 16124, 108691, 164833, 24478, 922077, 2521451, 1476984, 6699319, 34182196, 35553398, 32125393, 306268562, 597551756, 130656229, 2465133864, 8701963882, 6890111163, 15949374758, 98248054847, 135250416961

Offset

1,2

Comments

The corresponding y-values are in A230711.

For all non-coprime solutions (x,y) to the equation x^2 + y^2 = p^n, x and y are both divisible by the prime p.

Using de Moivre's Theorem (in essence), define (c,d)*(e,f) as (ce-df,cf+de). Then a(n) = min{|u(n)|, |v(n)|}, where (u(n),v(n)) = (2,1)^n = (2,1)*(2,1)^[n-1]. Proof: It can be readily seen that u^2(n) + v^2(n) = 5^n. To show that u(n) and v(n) are relatively prime, assume that x,y are relatively prime. Then (2,1)*(x,y) = (2x-y, x+2y). If a prime p were to divide both of 2x-y and x+2y, then p would divide 5y, so p=5. Now suppose x == 2 (mod 5) and y == 1 (mod 5). It can be seen that 2x-y == -2 (mod 5) and x+2y == -1 (mod 5). The reverse also holds. Because u(1)=2 and v(1)=1, the result follows inductively. - Richard Peterson, May 21 2021

Links

Zak Seidov, Table of n, a(n) for n = 1..200

Chris Busenhart, Lorenz Halbeisen, Norbert Hungerbühler, and Oliver Riesen, On primitive solutions of the Diophantine equation x^2+ y^2= M, Eidgenössische Technische Hochschule (ETH Zürich, Switzerland, 2020).

Example

a(4)=7 because 7^2 + 24^2 = 625 = 5^4.

Mathematica

Table[Select[PowersRepresentations[5^n, 2, 2], CoprimeQ[#[[1]], #[[2]]] &][[1, 1]], {n, 33}] (* T. D. Noe, Nov 04 2013 *)

Crossrefs

Cf. A000351, A230711.

Cf. A188948, A230622, A230644.

Cf. A139011.

Sequence in context: A100985 A176802 A363400 * A265009 A358656 A021757

Adjacent sequences: A230707 A230708 A230709 * A230711 A230712 A230713

Keyword

nonn

Author

Colin Barker, Oct 28 2013

Extensions

Typo in data fixed by Colin Barker, Nov 02 2013

Status

approved