Ex-1.19

Applying Tpq first time
a->bq + aq + ap
b->bp + aq

Applying Tpq second time
a-> (bp + aq)q + (bq + aq + ap)q + (bq + aq + ap)p
 -> b(2pq + q2) + a(2q2 + 2pq + p2)
 -> b(2pq + q2) + a(2pq + q2) + a(p2 + q2)

b-> (bp + aq)p + (bq + aq + ap)q
 -> b(p2 + q2) + a(2pq + q2)

clearly
p' = p2 + q2
q' = q2 + 2pq

(define (fib n)
  (fib-iter 1 0 0 1 n))
(define (fib-iter a b p q count)
  (cond ((= count 0) b)
        ((even? count)
         (fib-iter a
                   b
                   (+ (square p) (square q))      ; compute p'
                   (+ (square q) (* 2 p q))      ; compute q'
                   (/ count 2)))
        (else (fib-iter (+ (* b q) (* a q) (* a p))
                        (+ (* b p) (* a q))
                        p
                        q
                        (- count 1))))) 

Note: Whoa, take a moment to notice how this example generalized the approach of fast-expt to reach logarithmic number of steps using square to fib.