Anyone familiar with Scheme?

[Jon Richards]

I have a super easy question... I'm working on a program to determine whether very large numbers (100 digits or more) are prime. The method I've coded here isn't 100% accurate, but it's a lot faster than something that would be more accurate so I'm ok with it. I know it's a little messy right now, but my issue is the "modulo" evaluation on line 43 doesn't actually evaluate. When I debug, it goes through the line and appears to do the calculation, but it merrily finishes out the loop, giving me a "prime number" every time. Anyone have any ideas?

(also, I realize I tagged this as Java code so it highlighted some weird stuff, but when I tagged it as "scheme" or "lisp" I lost all the cool formatting.)

#lang scheme


(define-syntax (while stx) ;; thanks to http://codeimmersion.i3ci.hampshire.edu/2009/10/15/a-scheme-while-loop/ for the code
  (syntax-case stx ()
      ((_ condition expression ...)
       #`(do ()
           ((not condition))
           expression
           ...))))

(define  (interval-list m n) ;;this comes up with every number between the 2 and the target number.
  (if (> m n)
      '()
      (cons m (interval-list (+ 1 m) n))
         )
 )

(define (refactor lis) 
  (define (remove-multiples n lis)
   (if (null? lis)
         '()
         (if  (= (modulo (first lis) n) 0)      
              (remove-multiples n (rest lis))
              (cons (first lis)
                   (remove-multiples n (rest lis))))))

 (if (null? lis)
      '()
      (cons (first lis)
             (refactor (remove-multiples (first lis) (rest lis))))))

(define (primes n)
  (refactor (interval-list 2 n)))

(define (squareRoot n)
  (primes (sqrt (sqrt (sqrt (sqrt (sqrt n)))))) 
  )

(define (divide n lis)
  (cond
    ((null? lis) "prime number")
    (eq? (modulo (first lis) n) 0 "not prime")
    (else (divide n (rest lis)))
  )
  )
      
(define (listCreator n lis)
  (define x 0)
  (while (< x 1000)
       (cons (+ x n) lis)
       (set! x (+ x 1)))
  )

  
(define (run n)
  (divide n (squareRoot n)))


;;(run 330)
;;(divide 331)