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

using Primes
 
function carmichael(pmax::Integer)
    if pmax ≤ 0 throw(DomainError("pmax must be strictly positive")) end
    car = Vector{typeof(pmax)}(0)
    for p in primes(pmax)
        for h₃ in 2:(p-1)
            m = (p - 1) * (h₃ + p)
            pmh = mod(-p ^ 2, h₃)
            for Δ in 1:(h₃+p-1)
                if m % Δ != 0 || Δ % h₃ != pmh continue end
                q = m ÷ Δ + 1
                if !isprime(q) continue end
                r = (p * q - 1) ÷ h₃ + 1
                if !isprime(r) || mod(q * r, p - 1) == 1 continue end
                append!(car, [p, q, r])
            end
        end
    end
    return reshape(car, 3, length(car) ÷ 3)
end