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Chapter 19: Primality Testing and Carmichael Numbers
Practice HW p. 127 # 1, 2, 3a, 4
Recall Fermat’s Little Theorem says that if p is a prime number and a is an integer where
p | a , then
a p 1  1 (mod p)
If we multiply both sides by a, we obtain the equivalent expression
a  a p 1  a  1 (mod p)
or
a p  a (mod p) .
Then if a p  a (mod p) , then p is composite.
Definition 1: An integer a where a p  a (mod p) that shows p is composite is called a
witness of p.
For example, 2, 3, 7, 8, 12, and 13 are witnesses of the composite number 15 since
215  8  2 (mod 15)
315  12  3 (mod 15)
715  13  7 (mod 15)
815  2  8 (mod 15)
1215  3  12 (mod 15)
1315  7  13 (mod 15)
Note, 9 is not a witness 15 since
915  9 (mod 15)
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Carmichael Numbers
A Carmichael number is a positive integer n where a n  a (mod n) for any 1  a  n but
n is not prime.
List of Carmichael Numbers Smaller than 10000
561  3  11 17
2821  7  13  31
1105  5  13  17
6601  7  23  41
1729  7  13  19
8911  7  19  67
2465  5  17  29
How do we determine if a number is a Carmichael number?
Facts About Carmichael Numbers
1. Every Carmichael number is odd.
Proof Fact 1:
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2. Every Carmichael number is a product of distinct primes (there are no repeated prime
factors).
Proof Fact 2:
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Korselt’s Criterion for Carmichael Numbers
A composite number n is a Carmichael number if and only it is odd and for all primes p
where p | n , the following conditions are satisfied:
1. p 2 | n
2. ( p  1) | (n  1)
Proof:  Assume n is odd and for any prime p where p | n , p 2 | n and ( p  1) | (n  1) .
To show that n is Carmichael, we want to show that a n  a (mod n) for any 1  a  n  1 .
Since p 2 | n for any prime where p | n , if we compute n' s prime factorization
n  p1  p2  pr ,
then the pi ’s are distinct. Also, since ( p  1) | (n  1) for any prime p | n , for all the
primes pi in n' s factorization, we know that ( pi  1) | (n  1) or
(n  1)  ( pi  1)k
This leads to two cases.
Case 1: If pi | a , then a  0 (mod pi ) for all primes pi in n’s factorization.
Hence, a n  0 (mod pi ) or a n  a (mod pi ) .
Case 2: If pi | a , then
an
 a n 1  a (mod pi )
 a ( pi 1) k  a (mod pi )
 ( a ( pi 1) ) k  a (mod pi )
 (1) k  a (mod pi )
 a (mod pi )
Thus, a n  a (mod pi ) and pi | (a n  a ) for all of the primes in n’s factorization. Hence,
for n  p1  p2  pr , since gcd( pi , p j )  1 where i  j , it is not hard to show that
p1  p2  pr | (a n  a) or n | (a n  a ) . Thus, a n  a (mod n) for any 1  a  n and hence
n is Carmichael.
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 Assume n is a Carmichael number. We have already showed n is odd and that n is a
product of distinct primes (that is p 2 | n of all primes p that divide n). To show
( p  1) | ( n  1) , we will use the fact that for any prime p, there will always exist an
integer a for any prime p where p | a and
a p 1  1 (mod p) (Note that p  1 is the smallest exponent that makes this true)
Since n is a Carmichael number, we know that a n 1  1 (mod n) . Since p | n , it follows
that
a n 1  1 (mod p) .
Let G  gcd( p  1, n  1) . The, there exists integers x and y where
( p  1) x  ( n  1) y  G
or
( p  1) x  G  ( n  1) y
By Fermat’s Little Theorem, we know that
a ( p 1) x  ( a p 1 ) x (mod p )
 (1) x (mod p )
 1 (mod p )
Also
a ( p 1) x
 a G  ( n 1) x (mod p )
 a G  a ( n 1)(  x ) (mod p )
 a G  ( a ( n 1) ) - x (mod p )
 a G  (1) - x (mod p )
 a G (mod p )
Thus, a G  1 (mod p) . Now, we earlier stated that p  1 is the smallest exponent where
a p 1  1 (mod p) . Thus, p  1  G . Since G  gcd( p  1, n  1) , then G | ( p  1) and
G  p  1 . Thus G  ( p  1) and since G  gcd( p  1, n  1) , G | ( n  1) and hence
( p  1) | ( n  1) . This completes the proof.
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Example 1: Determine if 6601  7  23  41 is a Carmichael number.
Solution:
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Example 2: Determine if 27347  23  29  41 is a Carmichael number.
Solution:
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Note that 227347  20587  2(mod 27347) , which verifies that 27347 is not prime from
Example 2.
Verifying that Large Integers are Prime
The following property is true for all odd prime numbers.
Prime Number Property
Let p be an odd prime number and write
p  1  2k q ,
where q is an odd positive integer.
Choose an integer a where p | a . Then one of the following conditions are true (one or
the other)
1. a q  1 (mod p)
or
2. One of the numbers a q , a 2 q , a 4 q , …, a 2
the same as ( p  1) (mod p ) .
k 1
q
is congruent to  1 (mod p ) (which is
Proof:
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Taking the contrapositive of the previous property gives us another method for proving
that a number p is not prime.
Rabin-Miller Test for Composite Numbers
Let n be an odd integer and write n  1  2k q where q is an odd positive integer. Then the
number n is composite (not prime) if both the following conditions are true:
1. a q  1 (mod n)
and
2. None of the numbers a q , a 2 q , a 4 q , …, a 2
is the same as ( p  1) (mod n ) .
k 1
q
are congruent to  1 (mod n ) (which
Notes Concerning the Rabin-Miller Test
1. The base a that proves n is composite is called a Rabin-Miller witness.
2. The test is designed to show that n is not prime. However, there is no types of
numbers, like Carmichael numbers that can be used to discourage using Fermat’s
Little Theorem to show that a number is prime, to discourage using the Rabin-Miller
test to show a number is prime.
3. Although the test cannot be used directly to show a number is prime, its repeated use
can be to verify a number is prime to a certain degree of probability.
Fact: Given t integer bases a that fail to verify using the Rabin-Miller test that a number
n is not prime, then
(Probability that n is prime) = 1 (0.75)t
Example 3: Determine the approximate number of bases a that we would need to have
fail the Rabin-Miller test to show a 99.999 % chance that a given number is prime .
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To demonstrate how the Rabin-Miller test can be applied, we will use Maple.