FInd missing and repeated number

;; Answer to "Two Elisp Challenges", from http://irreal.org/blog/?p=808

;; 1. Given a list of integers from 1 to n but with one of the
;; integers missing, write an efficient algorithm for finding the
;; missing integer.


;; 2. Given a list of integers from 1 to n but with one of the
;; integers missing and another integer repeated, write an efficient
;; algorithm for finding the missing and repeated integers.


(defun sum-until-n-minus-sum-l-plus-x (l x)
  "For a list l of size n, calculates 1+...+n minus the sum of all the numbers in l, and adds x it."
  (if l (sum-until-n-minus-sum-l-plus-x (cdr l) (+ (length l) (- (car l)) x)) x))


(defun sum-squares-until-n-minus-sum-squares-l-plus-x (l x)
  "For a list l of size n, calculates 1^2+...+n^2 minus the sum of the squares of all the numbers in l, and adds x to it."
  (if l (sum-squares-until-n-minus-sum-squares-l-plus-x (cdr l) (+ (expt (length l) 2) (- (expt (car l) 2)) x)) x))

(defun find-missing (l)
  "For a list l containing all the numbers 1,...,n except one with non of them repeated, returns the missing one. It is based on the fact that if S is the sum of the elements in the list l, then 1+...+n-S is the number missing in l."
  (sum-until-n-minus-sum-l-plus-x l (+ (length l) 1)))

(defun find-missing-and-repeated (l)
  "For a list l containing all the numbers 1,...,n except one with only one of them repeated, returns a list (m r) with two elements the missing one m and the repeated one r. It is based on the fact that if S is the sum of the elements in the list l and SS is the sum of the squares, then 1+...+n -S=m-r and 1^2+...+n^2-SS=r^2-m^2=(m-r)*(m-r)."
  (let* ((diff-sq (sum-squares-until-n-minus-sum-squares-l-plus-x l 0))
     (m-minus-r (sum-until-n-minus-sum-l-plus-x l 0))
     (m-plus-r (/ diff-sq r-minus-m)))
    (list (/ (+ m-plus-r m-minus-r) 2) (/ (- m-plus-r m-minus-r) 2))))