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

Cases[Cases[
  Cases[Table[{p1, h3, d}, {p1, Array[Prime, PrimePi@61]}, {h3, 2, 
     p1 - 1}, {d, 1, h3 + p1 - 1}], {p1_Integer, h3_, d_} /; 
     PrimeQ[1 + (p1 - 1) (h3 + p1)/d] && 
      Divisible[p1^2 + d, h3] :> {p1, 1 + (p1 - 1) (h3 + p1)/d, h3}, 
   Infinity], {p1_, p2_, h3_} /; PrimeQ[1 + Floor[p1 p2/h3]] :> {p1, 
    p2, 1 + Floor[p1 p2/h3]}], {p1_, p2_, p3_} /; 
   Mod[p2 p3, p1 - 1] == 1 :> Print[p1, "*", p2, "*", p3]]