perplexus.info :: Just Math : Lucas squares

Lucas squares (Posted on 2013-12-15)

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

((L_n)²+(L_n+1)²)² - 5((L_2n+2)*(L_2n)-1) = 0

Prove it!

See The Solution Submitted by broll

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re: Long solution.

| Comment 2 of 6 |

(In reply to Long solution. by Jer)

Excellent!

If you don't mind, please consider this line:

(L_2n)²-3L_2n*L_2n+2+(L_2n+2)² + 5 = 0

and the generalised substitution a^2-3ab+b^2 = -5.

Are they not fully equivalent?

If so, are the remaining lines strictly necessary? I'd be grateful to hear your views, as I by no means claim to be an expert in matters of proof.

Posted by broll on 2013-12-16 14:06:09

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