logo

Coset Decision Method 📂Number Theory

Coset Decision Method

Overview

The theorem, which serves as a necessary and sufficient condition for determining if a number is a Carmichael number, can also be usefully applied.

Theorem 1

Let an odd composite number be nn.

nn is a Carmichael number satisfying     \iff pnp \mid n for all primes pp such that p2np^2 \nmid n and (p1)(n1)(p-1) \mid (n-1).

Proof

Definition of Carmichael Numbers: A natural number nn is called a Carmichael number if it satisfies ana(modn)a^{n} \equiv a \pmod{n} for all 1an1 \le a \le n.


(    )( \implies )

A Carmichael number can be expressed as n=p1pmn = p_{1} \cdots p_{m} for different primes p1,,pmp_{1} , \cdots , p_{m}, excluding 22, and therefore must satisfy pi2np_{i}^2 \nmid n.

Since nn is a Carmichael number, it follows that an11(modn)a^{n-1} \equiv 1 \pmod{n} and for any kk, an1=nk+1=pi(npik)+1 a^{n-1} = n \cdot k + 1 = p_{i} \cdot \left( {{n} \over {p_{i}}} k \right) + 1 therefore an11(modpi)a^{n-1} \equiv 1 \pmod{p_{i}}.

Properties of Order: For a prime pp, if gcd(a,p)=1\gcd (a,p)=1 and an1(modp)a^{n} \equiv 1 \pmod{p} then ordp(a)n\text{ord}_{p} (a) \mid n

According to the properties of order, ordpi(a)\text{ord}_{p_{i} } (a) divides (n1)(n-1), and by the Primitive Root Theorem, having at least 11 primitive roots, for some aia_{i}, ordpi(ai)=pi1 \text{ord}_{p_{i} } (a_{i}) = p_{i} - 1 Therefore, for all p1,,pmp_{1} , \cdots , p_{m}, (pi1)(n1)(p_{i}-1) \mid (n-1) must hold.


(    )( \impliedby )

Let’s assume n:=p1pmn := p_{1} \cdots p_{m}.

For all primes p1,,pmp_{1} , \cdots , p_{m} dividing nn, since p2np^2 \nmid n, p1,,pmp_{1} , \cdots , p_{m} are distinct.

Since (p1)(n1)(p-1) \mid (n-1), for some integer kik_{i}, (n1)=(pi1)ki (n-1 ) = ( p_{i} - 1) k_{i} Now, consider 1an1 \le a \le n and whether it divides or does not divide pip_{i}.

  • Case 1. piap_{i} \mid a
    Obviously, an0a(modpi)a^{n} \equiv 0 \equiv a \pmod{p_{i}} holds.
  • Case 2. piap_{i} \nmid a
    By Fermat’s Little Theorem, an=a(pi1)ki+1=a(pi1)kia1kia(modpi)a(modpi) \begin{align*} a^{n} =& a^{ ( p_{i} - 1) k_{i} +1 } \\ =& a^{ ( p_{i} - 1) k_{i} } \cdot a \\ \equiv& 1^{k_{i}} \cdot a \pmod{p_{i}} \\ \equiv& a \pmod{p_{i}} \end{align*}

Therefore, in any case, (ana)0(modpi) ( a^{n} - a ) \equiv 0 \pmod{ p_{i} } which means that (ana)( a^{n} - a ) is divisible by p1,,pmp_{1}, \cdots , p_{m}. In other words, (ana)( a^{n} - a ) is divisible by n=p1pmn= p_{1} \cdots p_{m}, summarizing we obtain the following. (ana)0(modn) ( a^{n} - a ) \equiv 0 \pmod{ n }


  1. Silverman. (2012). A Friendly Introduction to Number Theory (4th Edition): p135. ↩︎