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 Lucas squares (Posted on 2013-12-15)

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!

 See The Solution Submitted by broll No Rating

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 Long solution. | Comment 1 of 6
I made use of the page
http://mathworld.wolfram.com/LucasNumber.html
specifically formula [5]
<img src="http://mathworld.wolfram.com/images/equations/LucasNumber/NumberedEquation5.gif" class="numberedequation" alt=" L_n^2-L_(n-1)L_(n+1)=5(-1)^n, " border="0" height="19" width="144">
and formula [21]
<img src="http://mathworld.wolfram.com/images/equations/LucasNumber/NumberedEquation17.gif" class="numberedequation" alt=" L_n^2=L_(2n)+2(-1)^n, " border="0" height="19" width="112">

((Ln)2+(Ln+1)2)2 - 5((L2n+2)*(L2n)-1) = 0
(L2n+2(-1)n + L2n+2+2(-1)n+1)2 - 5(L2n+2)*(L2n) + 5 = 0
(L2n + L2n+2)2 - 5(L2n+2)*(L2n) + 5 = 0
(L2n)2+2L2n*L2n+2+(L2n+2)2- 5(L2n+2)*(L2n) + 5 = 0
(L2n)2-3L2n*L2n+2+(L2n+2)2 + 5 = 0
L2n(L2n-L2n+2)+L2n+2(L2n+2-L2n) - L2n*L2n+2 + 5=0
L2n(-L2n+1)+L2n+2(L2n+1)- L2n*L2n+2 + 5=0
(L2n+1)(L2n+2-L2n) - L2n*L2n+2 + 5=0
(L2n+1)(L2n+1) - L2n*L2n+2 + 5=0
(L2n+1)2 - L2n*L2n+2 + 5=0
[5(-1)2n+1+L2n*L2n+2 ]- L2n*L2n+2 + 5=0
-5 + 5 = 0
0=0 QED

 Posted by Jer on 2013-12-16 09:33:17

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