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COMMENT
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Conjecture: sequence is finite.
n=1 ==> k=4 or 6, n=2 ==> k=4 or 6, n=3 ==> k=3 or 9, n=4 ==> k=1 or 2 or 7 or 11, n=6 ==> k=1, 2, 6, n=7 ==> k=4, n=9 ==> k=3, n=10 ==> k=11, n=11 ==> k=10, n=12 ==> k=36, n=17 ==> k=4, n=36 ==> k=12, for other n there is no solution. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 14 2004
Joe Silverman wrote: "Here's a general finiteness result: fib(n) + c = m^2 has only finitely many solutions, regardless of the value of c. Write fib(n) = a(u^n - u^(-n)), where a=1/sqrt(5) and u=(1+sqrt(5))/2. So generalizing further, can look at the equation au^n + bu^(-n) + c = dm^2, where a,b,c,d are fixed integers (a,b,d nonzero) and u is a unit, or even just an algebraic integer, in some number field.
"Rewrite this as au^(2n) + b + cu^n = d m^2 u^n Using standard methods, it is not hard to prove that any such equation has only finitely many solutions. One way is to consider the congruence class of n modulo (say) 2. Thus write n = 2*n1+n2. We can assume that n2 is fixed (either 0 or 1). Now let x = u^n1 and y = m be the unknowns and let e=u^n2 be another fixed value.
"Then we need to solve (ae^2)x^4 + b + (ce)x^2 = dx^2y^2. This is the equation of an elliptic curve (albeit a singular one), so has only finitely many solutions in the ring of integers of any field (by a theorem of Siegel). Or one can consider n modulo a higher congruence, and then the resulting curve will have genus larger than one, in which case one can quote Faltings' theorem that there are only finitely many points with coordinates in any number field."
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