Answer
See explanation
Work Step by Step
Below is a standard “factorization” proof of the Fibonacci‐number identity that often appears in texts:
**Claim (a common form).** For all integers \(k \ge 2\),
\[
F_k^2 \;-\; F_{k-1}^2 \;=\; F_{k-2}\,F_{k+1}.
\]
*(Depending on the exact source, the identity may be written in a slightly different form or with different index shifts. The core idea is the same.)*
---
## Proof
Recall the Fibonacci recurrence:
\[
F_k \;=\; F_{k-1} \;+\; F_{k-2},
\quad\text{and}\quad
F_{k+1} \;=\; F_k + F_{k-1}.
\]
Consider the left‐hand side:
\[
F_k^2 - F_{k-1}^2
\;=\;
\bigl(F_k - F_{k-1}\bigr)\,\bigl(F_k + F_{k-1}\bigr).
\]
From the recurrence:
1. \(F_k - F_{k-1} = F_{k-2}.\)
2. \(F_k + F_{k-1} = F_{k+1}.\)
Hence,
\[
F_k^2 - F_{k-1}^2
\;=\;
(F_k - F_{k-1})(F_k + F_{k-1})
\;=\;
F_{k-2}\,\cdot F_{k+1}.
\]
This completes the proof that
\[
\boxed{F_k^2 \;-\; F_{k-1}^2 \;=\; F_{k-2}\,F_{k+1}
\quad \text{for all } k \ge 2.}
\]
