Final answer:
To prove that the product of the three primes 6n+1, 12n+1, and 18n+1 yields a Carmichael number, we need to show that this product, denoted as P, satisfies the Carmichael property. A Carmichael number is a composite number that satisfies the equation a^(P-1) ≡ 1 (mod P) for all integers a that are relatively prime to P.
Step-by-step explanation:
To prove that the product of the three primes 6n+1, 12n+1, and 18n+1 yields a Carmichael number, we need to show that this product, denoted as P, satisfies the Carmichael property. A Carmichael number is a composite number that satisfies the equation a^(P-1) ≡ 1 (mod P) for all integers a that are relatively prime to P. Let's go step by step to prove this:
- Let's assume that 6n+1, 12n+1, and 18n+1 are prime numbers for some integer n. This means that they are all greater than 1 and have no divisors other than 1 and themselves. Since they are all greater than 1, n > 0.
- Let's create the product P = (6n+1)(12n+1)(18n+1).
- Let's select an integer a that is relatively prime to P. This means that gcd(a, P) = 1. Since P is a product of three primes, gcd(a, P) = 1 if and only if gcd(a, 6n+1) = 1, gcd(a, 12n+1) = 1, and gcd(a, 18n+1) = 1.
- Using the Chinese Remainder Theorem, we can show that if a ≡ -1 (mod 6n+1), a ≡ -1 (mod 12n+1), and a ≡ -1 (mod 18n+1), then a ≡ -1 (mod P). This means that a and P differ by a multiple of P, so a^(P-1) ≡ 1 (mod P) holds true.
Therefore, since the product of 6n+1, 12n+1, and 18n+1 satisfies the Carmichael property, it yields a Carmichael number.
To create two such Carmichael numbers, we need to choose specific values for n.
Carmichael Number 1: Let n = 1. The product becomes P = (6*1+1)(12*1+1)(18*1+1) = 7*13*19 = 1729.
Carmichael Number 2: Let n = 2. The product becomes P = (6*2+1)(12*2+1)(18*2+1) = 13*25*37 = 12025.