2.21 A composite number \( n \) is a Carmichael number if \( a^{n-1} \equiv 1 \) modulo \( n \) for all \( a \) relatively prime to \( n \). - Prove that 1729 is a Carmichael number. - Prove that if \( n \) is a square-free integer (that is, \( n \) is not a multiple of a perfect square) such that \( p-1 \) divides \( n-1 \) for all prime factors \( p \) of \( n \), then \( n \) is a Carmichael number. (Hint: Do the second part before doing the first.) It is known that there exist infinitely many Carmichael numbers, but this was only proved in 1994.
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Let \( n = p_1 p_2 ... p_k \) be the prime factorization of \( n \). Since \( n \) is square-free, all \( p_i \) are distinct. By the Chinese Remainder Theorem, it suffices to show that \( a^{n-1} \equiv 1 \) modulo \( p_i \) for all \( i \). Since \( p_i - 1 Show more…
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Show that if $n=p_{1} p_{2} \cdots p_{k},$ where $p_{1}, p_{2}, \ldots, p_{k}$ are distinct primes that satisfy $p_{j}-1 | n-1$ for $j=1,2, \ldots, k$ then $n$ is a Carmichael number.
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Solving Congruences
Show that 1729 is a Carmichael number.
a) Use Exercise 48 to show that every integer of the form $(6 m+1)(12 m+1)(18 m+1),$ where $m$ is a positive integer and $6 m+1,12 m+1,$ and $18 m+1$ are all primes, is a Carmichael number. b) Use part (a) to show that $172,947,529$ is a Carmichael number.
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