OFFSET
1,1
COMMENTS
For x, y > 2, the solutions start with (4,11) -> (11, 29) -> (29, 76) -> ...
The sequence is infinite (see the proof in the second reference).
Consider the pairs of the form (a(2n-1), a(2n)). Limit_{n->oo} a(2n)/a(2n-1) = phi^2 = 2.618033988749894... (A104457).
Property: a(2n-1)^2 + a(2n)^2 = 3*a(2n-1)*a(2n) + 5.
Apparently a(2*n) = a(2*n+1) = A002878(n) for n >= 1. - Georg Fischer, Dec 05 2022
LINKS
Art of Problem Solving, Problem A112
Peter Vandendriessche and Hojoo Lee, Problems in Elementary Number Theory (see problem A112, p. 15). [Via Wayback Machine]
FORMULA
Conjectures from Chai Wah Wu, Jan 28 2024: (Start)
a(n) = 3*a(n-2) - a(n-4) for n > 4.
G.f.: x*(-4*x^3 - x^2 + 11*x + 4)/(x^4 - 3*x^2 + 1). (End)
MAPLE
nn:=10^6:a:=4:
for b from a+1 to nn do:
x:=(a^2-5)/b:y:=(b^2-5)/a:
if x>0 and y>0 and gcd(a, b)=1 and x=floor(x) and y=floor(y)
then
printf(`%d, `, a): printf(`%d, `, b):a:=b:
else fi:
od:
MATHEMATICA
nn = 10^6; a = 4; Reap[For[b = a+1, b <= nn, b++, x = (a^2-5)/b; y = (b^2-5)/a; If[x>0 && y>0 && GCD[a, b] == 1 && x == Floor[x] && y == Floor[y], Print[a, " ", b]; Sow[a]; Sow[b]; a = b]]][[2, 1]] (* adapted from Maple *)
(* Second program: *)
Clear[a]; a[n_] := 2^(-n-2)*((7-3*Sqrt[5])*(1-Sqrt[5])^n-(-Sqrt[5]-1)^(n+1) - (Sqrt[5]-1)^(n+1) + (3*Sqrt[5]+7)*(Sqrt[5]+1)^n); Table[a[n] // Simplify, {n, 1, 36}] (* Jean-François Alcover, Jan 25 2017 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Jan 21 2017
STATUS
approved