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

 
for i = 3 to max_sieve step 2
	isprime[i] = 1
next i
 
isprime[2] = 1
for i = 3 to sqr(max_sieve) step 2
	if isprime[i] = 1 then
		for j = i * i to max_sieve step i * 2
			isprime[j] = 0
		next j
	end if
next i
 
subroutine carmichael3(p1)
	if isprime[p1] <> 0 then
		for h3 = 1 to p1 -1
			t1 = (h3 + p1) * (p1 -1)
			t2 = (-p1 * p1) % h3
			if t2 < 0 then t2 = t2 + h3
			for d = 1 to h3 + p1 -1
				if t1 % d = 0 and t2 = (d % h3) then
					p2 = 1 + (t1 \ d)
					if isprime[p2] = 0 then continue for
					p3 = 1 + (p1 * p2 \ h3)
					if isprime[p3] = 0 or ((p2 * p3) % (p1 -1)) <> 1 then continue for
					print p1; " * "; p2; " * "; p3
				end if
			next d
		next h3
	end if
end subroutine
 
for i = 2 to 61
	call carmichael3(i)
next i
end
 