SICP Exercise 3.35 squarer as a primitive constraint

Exercise 3.35.  Ben Bitdiddle tells Louis that one way to avoid the trouble in exercise 3.34 is to define a squarer as a new primitive constraint. Fill in the missing portions in Ben's outline for a procedure to implement such a constraint:

(define (squarer a b)
  (define (process-new-value)
    (if (has-value? b)
        (if (< (get-value b) 0)
            (error "square less than 0 -- SQUARER" (get-value b))
            <alternative1>)
        <alternative2>))
  (define (process-forget-value) <body1>)
  (define (me request) <body2>)
  <rest of definition>
  me)

 SOLUTION

 The code is here.

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