(define (expmod base exp m)
  (remainder (fast-expt base exp) m))

(define (fast-expt b n)
  (cond ((= n 0) 1)
	((even? n) (square (fast-expt b (/ n 2))))
	(else (* b (fast-expt b (- n 1))))))

(define (even? n)
  (= (remainder n 2) 0))

; This appears to be equivalent to calculating the exponent modulo m with the
; exception that for large numbers, fast-expt will get very large before 
; calculating the modulus, so it is probably not suitable for a fast prime tester.