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Lucas squares (Posted on 2013-12-15) Difficulty: 3 of 5

Let n be an integer, and let (Ln) signify the nth Lucas Number.

((Ln)2+(Ln+1)2)2 - 5((L2n+2)*(L2n)-1) = 0

Prove it!

  Submitted by broll    
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Solution: (Hide)
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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(5): Long solution.broll2013-12-17 22:39:34
re(4): Long solution.Jer2013-12-17 11:06:49
re(3): Long solution.broll2013-12-17 03:15:58
re(2): Long solution.Jer2013-12-16 15:46:27
Questionre: Long solution.broll2013-12-16 14:06:09
SolutionLong solution.Jer2013-12-16 09:33:17
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