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
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i
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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

 
// Carmichael Number . Nigel Galloway: November 19th., 2017
let fN n = Seq.collect ((fun g->(Seq.map(fun e->(n,1+(n-1)*(n+g)/e,g,e))){1..(n+g-1)})){2..(n-1)}
let fG (P1,P2,h3,d) =
  let mod' n g = (n%g+g)%g
  let fN P3 = if isPrime P3 && (P2*P3)%(P1-1)=1 then Some (P1,P2,P3) else None
  if isPrime P2 && ((h3+P1)*(P1-1))%d=0 && mod' (-P1*P1) h3=d%h3 then fN (1+P1*P2/h3) else None
let carms g = primes|>Seq.takeWhile(fun n->n<=g)|>Seq.collect fN|>Seq.choose fG
carms 61 |> Seq.iter (fun (P1,P2,P3)->printfn "%2d x %4d x %5d = %10d" P1 P2 P3 ((uint64 P3)*(uint64 (P1*P2))))
 