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Today we prove a clean Fibonacci identity that looks like pure algebra: F_{2n} = F_{n+1}^2 - F_{n-1}^2. The proof uses two main ideas. First, we reuse the standard addition formula from the previous video (the heavy lifting): F_{m+n} = F_m F_{n+1} + F_{m-1}F_n. Then we set 2n = n+n and factor to rewrite the left side in a simple product form.

Second, we attack the right side with the classic difference of squares factorization:
a^2 - b^2 = (a+b)(a-b). Using the Fibonacci recurrence F_{n+1} = F_n + F_{n-1}, we turn (F_{n+1} - F_{n-1}) into F_n, and everything matches perfectly. If the addition formula step feels mysterious, check out the earlier Fibonacci Cubed video where I prove that property (by induction).

Thumbnail Guest: Leonardo Bonacci (aka Fibonacci: 1170-1250) https://en.wikipedia.org/wiki/Fibonacci

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PROPERTIES AND CONCEPTS USED
Fibonacci Recurrence F_{n+1} = F_n + F_{n-1}
Standard Addition Formula F_{m+n} = F_m F_{n+1} + F_{m-1}F_n
Difference Of Squares a^2 - b^2 = (a+b)(a-b)
Factoring Out A Common Term
Index Manipulation 2n = n + n
Substitution Into Algebraic Factors
Commutativity Of Integers
Induction Referenced For Prior Result

CHAPTERS:
00:00 Introduction
00:45 Fibonacci Sequence Refresher
01:30 The Identity To Prove
02:15 Standard Addition Formula
03:05 Rewrite 2n As n+n
03:50 Difference Of Squares
04:35 Use The Recurrence
05:20 Final Equality
06:20 Thanks For Watching

#Fibonacci #FibonacciNumbers #FibonacciIdentity #NumberTheory #DiscreteMath #MathProof #RecurrenceRelation #DifferenceOfSquares #Induction #dogmathic