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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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

link "factors"
 
procedure main(A)
    n := integer(!A) | 61
    every write(carmichael3(!n))
end
 
procedure carmichael3(p1)
    every (isprime(p1), (h := 1+!(p1-1)), (d := !(h+p1-1))) do
        if (mod(((h+p1)*(p1-1)),d) = 0, mod((-p1*p1),h) = mod(d,h)) then {
            p2 := 1 + (p1-1)*(h+p1)/d
            p3 := 1 + p1*p2/h
            if (isprime(p2), isprime(p3), mod((p2*p3),(p1-1)) = 1) then
                suspend format(p1,p2,p3)
            }
end
 
procedure mod(n,d)
   return (d+n%d)%d
end
 
procedure format(p1,p2,p3)
    return left(p1||" * "||p2||" * "||p3,20)||" = "||(p1*p2*p3)
end