|
Jer's Solution here does the trick very elegantly via the well-established equalities in Mathworld.
However, having reached (L2n)2-3L2n*L2n+2+(L2n+2)2 + 5 = 0, it may at once be observed that the recurrence sequence at A005248 in Sloane, namely x^2 - 5* y^2 = 4, (for x values) is identical to the sequence a^2-3ab+b^2 = -5 (for any qualifying a,b values), positive integer solutions in each case being of the form: -1/2 (3 (9-4 sqrt(5))^n-sqrt(5) (9-4 sqrt(5))^n+3 (9+4 sqrt(5))^n+sqrt(5) (9+4 sqrt(5))^n).
Indeed the relation a^2-3ab+b^2 = -5 might in a way be considered preferable to the one given in Sloane, for a(n) = b(n+1) in all cases, i.e. successive values of both a and b reproduce the required bisection of the Lucas numbers.
Thus there is an alternative route to the solution of the given problem, as follows. Let L2n = b; Let L2n+2 = a. Given (see (21) on the Mathworld page) that:
(Ln)2 = L2n +2(-1)n [1], and so (Ln+1)^2 = L2n+2 +2(-1)n the expression a^2+2ab+b^2 can be substituted for LHS in the problem. Now we have a^2+2ab+b^2=5(ab-1), or a^2-3ab+b^2 = -5, an equality that holds good iff a and b are the Lucas numbers required by the problem, as was to be shown.
|
