Multiply the Fibonacci numbers on either side of F(n) and subtract its square: you always get ±1. Exactly. F(n−1)F(n+1) − F(n)² = (−1)ⁿ — Cassini's identity. However large the Fibonacci numbers grow, this combination stays pinned to a unit, flipping sign each step. It is the arithmetic behind the famous “missing square” dissection puzzle (an 8×8 board rearranged into a 5×13 rectangle, gaining one unit of area — because 5·13 − 8² = 1). A conserved quantity riding on top of exponential growth.
The demo computes the Cassini combination across several n — and it is always ±1, whatever the size of the numbers: live demo
“Products of growing numbers grow.” — this one does not: however big the Fibonaccis, the Cassini combination is always exactly ±1. The famous chessboard-into-rectangle ‘paradox’ is this identity in disguise. cited
A unit conserved on exponential growth; a paradox that is only an identity. Cassini 1680
On the canonical compiler, for n=5: F(4)·F(6) − F(5)² = 3·8 − 25 = −1 = (−1)⁵: