10. We say that an integer a is nilpotent modulo m if there exists a pos- itive integer k such that a = 0 (mod m). Prove that a is nilpotent modulo m if and only if a = 0 (mod rad(m)).
To prove that an integer "a" is nilpotent modulo "m" if and only if "a ≡ 0 (mod rad(m))," we need to demonstrate both directions of this statement.
1. If "a" is nilpotent modulo "m" (i.e., there exists a positive integer "k" such that a^k ≡ 0 (mod m)):
To prove this direction, we can use the property of the radical of "m," rad(m), which is the product of all distinct prime factors of "m."
First, assume that "a" is nilpotent modulo "m," which means there exists a positive integer "k" such that a^k ≡ 0 (mod m).
Now, consider the prime factorization of "m" as m = p₁^α₁ * p₂^α₂ * ... * pk^αk, where p₁, p₂, ..., pk are distinct prime factors of "m."
Since a^k is congruent to 0 modulo "m," it is also congruent to 0 modulo each of the prime factors p₁, p₂, ..., pk, raised to their respective exponents (i.e., p₁^α₁, p₂^α₂, ..., pk^αk).
By the Chinese Remainder Theorem (CRT), we can conclude that a^k ≡ 0 (mod rad(m)) since a^k is congruent to 0 modulo each prime factor raised to its respective exponent.
Therefore, if "a" is nilpotent modulo "m," then "a" is congruent to 0 modulo the radical of "m."
2. If "a ≡ 0 (mod rad(m))":
Assume that "a" is congruent to 0 modulo the radical of "m," i.e., a ≡ 0 (mod rad(m)).
The radical of "m," rad(m), is the product of all distinct prime factors of "m."
Since "a" is congruent to 0 modulo rad(m), it means "a" is divisible by all the distinct prime factors of "m."
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