Which are the Carmichael numbers?

The first Quasi–Carmichael numbers are: 35, 77, 143, 165, 187, 209, 221, 231, 247, 273, 299, 323, 357, 391, 399, 437, 493, 527, 561, 589, 598, 713, 715, 899, 935, 943, 989, 1015, 1073, 1105, 1147, 1189, 1247, 1271, 1295, 1333, 1517, 1537, 1547, 1591, 1595, 1705, 1729, (sequence A257750 in the OEIS)

How do I find my Carmichael number?

A composite integer n is a Carmichael number if and only if an ≡ a mod n for all a ∈ Z. Proof. If an ≡ a mod n for all a ∈ Z, then when (a, n) = 1 we can cancel a from both sides and get an-1 ≡ 1 mod n, so n is a Carmichael number since it is composite.

Are there infinitely many Carmichael numbers?

In (Ca2], Carmichael exhibited an algorithm to con- struct such numbers and stated, perhaps somewhat wishfully, that “this list (of Carmichael numbers) might be indefinitely extended.” Indeed, until now, no one has been able to prove that there are infinitely many Carmichael numbers, though it has long seemed highly …

Is Carmichael number 561?

3. Hence, 561 is a Carmichael number, because it is composite and b560 ≡ (b80)7 ≡ 1 mod 561 for all b relatively prime to 561. for all b relatively prime to 1105. Hence, 1105 is also a Carmichael number.

How many prime factors does a Carmichael number have?

Every Carmichael number is squarefreeand has at least three different prime factors. Korselt’s criterion (1899). For every Carmichael number n it holds that n −  1 is divisible by pi −  1, 1   ≤   i  ≤   ω (n)

How to prove the Carmichael proof of N?

Assume that n = pq, with p < q two distinct primes, is a Carmichael number. Then we have q ≡ 1 (mod q − 1) ⟹ n = pq ≡ p modq − 1 ⟹ n − 1 ≡ p − 1 (modq − 1). Here 0 < p − 1 < q − 1, so n − 1 is not divisible by q − 1. A contradiction. The same argument can be extended to prove that any prime factor of a Carmichael number n is less than √n.

Is the Carmichael number an even or odd number?

It follows from this theorem that all Carmichael numbers are odd, since any even composite number that is square-free (and hence has only one prime factor of two) will have at least one odd prime factor, and thus p − 1 ∣ n − 1 {displaystyle p-1mid n-1} results in an even dividing an odd, a contradiction.

How to find Carmichael numbers by Fermat’s theorem?

There is a pleasantly simple description of Carmichael numbers, due to Korselt: THEOREM 1. A number n is a Carmichael number if and only if n = p 1p 2…p k, a product of (at least two) distinct primes, and p j−1 divides n−1 for each j. Proof. Let n be as stated, and let gcd(a,n) = 1. By Fermat’s theorem, for each j, we have ap j−1≡ 1 mod p j.