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

public class Test {
 
    static int mod(int n, int m) {
        return ((n % m) + m) % m;
    }
 
    static boolean isPrime(int n) {
        if (n == 2 || n == 3)
            return true;
        else if (n < 2 || n % 2 == 0 || n % 3 == 0)
            return false;
        for (int div = 5, inc = 2; Math.pow(div, 2) <= n;
                div += inc, inc = 6 - inc)
            if (n % div == 0)
                return false;
        return true;
    }
 
    public static void main(String[] args) {
        for (int p = 2; p < 62; p++) {
            if (!isPrime(p))
                continue;
            for (int h3 = 2; h3 < p; h3++) {
                int g = h3 + p;
                for (int d = 1; d < g; d++) {
                    if ((g * (p - 1)) % d != 0 || mod(-p * p, h3) != d % h3)
                        continue;
                    int q = 1 + (p - 1) * g / d;
                    if (!isPrime(q))
                        continue;
                    int r = 1 + (p * q / h3);
                    if (!isPrime(r) || (q * r) % (p - 1) != 1)
                        continue;
                    System.out.printf("%d x %d x %d%n", p, q, r);
                }
            }
        }
    }
}