Know what's even faster? Exponentiating a 2x2 matrix of integer coefficients. And again, that's what a bit of math tells you, so I think the point of the author holds.
Since the recurrence is always the same we can e.g. use the identities F_{m+n-1} = F_mF_n + F_{m-1}F_{n-1} and F_{m+n} = F_mF_n + F_mF_{n-1} + F_{m-1}F_n to precompute and cache all of the terms for 2^{n-1} and 2^{n}, and then multiply the appropriate ones together to find whatever value we need based on a binary expansion of the index we want to find a fibonacci number for.
This is the same idea as fast matrix exponentiation but saves low-level arithmetic and half of the space, because two of the terms in the matrix are basically redundant.