SICP in Clojure: Chapter 1, Exercise 21

This is my Clojure solution to Chapter 1, Exercise 21:

(ns sicp.ch1.ex21
  (:require [sicp.ch1.ex16 :as sicp-ch1-ex16]))

(defn find-divisor
  [n test-divisor]
  (cond (> (sicp-ch1-ex16/square test-divisor) n) n
        (zero? (rem n test-divisor)) test-divisor
        :else (find-divisor n (inc test-divisor))))

(defn smallest-divisor
  [n]
  (find-divisor n 2))

(smallest-divisor 199) ; => 199

(smallest-divisor 1999) ; => 1999

(smallest-divisor 19999) ; => 7