A lot of composite numbers can be separated from primes by Fermat's Little Theorem, but there are some that completely confound it.

The   Miller Rabin Test   uses a combination of Fermat's Little Theorem and Chinese Division Theorem to overcome this.

The purpose of this task is to investigate such numbers using a method based on   Carmichael numbers,   as suggested in   Notes by G.J.O Jameson March 2010.



Task

Find Carmichael numbers of the form:

  Prime1 × Prime2 × Prime3 
where    (Prime1 < Prime2 < Prime3)    for all    Prime1    up to   61.

(See page 7 of   Notes by G.J.O Jameson March 2010   for solutions.)



Pseudocode

For a given   



P
r
i
m

e

1




{\displaystyle Prime_{1}}



for 1 < h3 < Prime1
    for 0 < d < h3+Prime1
         if (h3+Prime1)*(Prime1-1) mod d == 0 and -Prime1 squared mod h3 == d mod h3
         then
               Prime2 = 1 + ((Prime1-1) * (h3+Prime1)/d)
               next d if Prime2 is not prime
               Prime3 = 1 + (Prime1*Prime2/h3)
               next d if Prime3 is not prime
               next d if (Prime2*Prime3) mod (Prime1-1) not equal 1
               Prime1 * Prime2 * Prime3 is a Carmichael Number



related task
Chernick's Carmichael numbers

 
#lang racket
(require math)
 
(for ([p1 (in-range 3 62)] #:when (prime? p1))
  (for ([h3 (in-range 2 p1)])
    (define g (+ p1 h3))
    (let next ([d 1])
      (when (< d g)
        (when (and (zero? (modulo (* g (- p1 1)) d))
                   (= (modulo (- (sqr p1)) h3) (modulo d h3)))
          (define p2 (+ 1 (quotient (* g (- p1 1)) d)))
          (when (prime? p2)
            (define p3 (+ 1 (quotient (* p1 p2) h3)))
            (when (and (prime? p3) (= 1 (modulo (* p2 p3) (- p1 1))))
              (displayln (list p1 p2 p3 '=> (* p1 p2 p3))))))
        (next (+ d 1))))))
 