#lang scheme

(define (pred-census p k) ;cant pass sum-of-3-squares as p because it needs 4 arguments, but only takes 3 here
  (define (pred-census-aux a b c counter)
    (cond ((> c k) counter)
          ((> a k) (pred-census-aux 1 (+ b 1) c counter))
          ((> b k) (pred-census-aux a 1 (+ c 1) counter))
          ((p a b c) (pred-census-aux (+ a 1) b c (+ counter 1)))
          (else (pred-census-aux (+ a 1) b c counter))))
  (pred-census-aux 1 1 1 0))

(define (pythag a b c)
  (= (* c c) (+ (* a a) (* b b))))

  (define (sum-of-3-squares a b c k)
  (= k (+ (* a a) (* b b) (* c c))))


;> (/ (pred-census pythag 50) 2) ;divide by two because pred-census detects triples twice, ie: 3 4 5 and 4 3 5
;20
;> (/ (pred-census pythag 20) 2)
;6