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

func forprimes(a, b, callback) {
    for (a = (a-1 -> next_prime); a <= b; a.next_prime!) {
        callback(a)
    }
}
 
forprimes(3, 61, func(p) {
   for h3 in (2 ..^ p) {
      var ph3 = (p + h3)
      for d in (1 ..^ ph3) {
         ((-p * p) % h3) != (d % h3) && next
         ((p-1) * ph3) % d && next
         var q = 1+((p-1) * ph3 / d)
         q.is_prime || next
         var r = 1+((p*q - 1)/h3)
         r.is_prime || next
         (q*r) % (p-1) == 1 || next
         printf("%2d x %5d x %5d = %s\n",p,q,r, p*q*r)
      }
   }
})