if (a,b)=1 and a b=c^n, show that a and b are nth powers
To prove that a and b are nth powers, we need to show that there exist positive integers x and y such that a = x^n and b = y^n.
Given that (a, b) = 1 and a * b = c^n, we can start by expressing a and b as products of their prime factorizations.
Let's assume the prime factorization of a is given by: a = p_1^k_1 * p_2^k_2 * ... * p_m^k_m, where p_i denotes a prime number and k_i represents the power to which it is raised.
Similarly, the prime factorization of b is represented by: b = q_1^l_1 * q_2^l_2 * ... * q_n^l_n.
Since (a, b) = 1, none of the prime factors of a and b overlap. In other words, for any p_i and q_j, p_i and q_j are distinct primes.
Since a * b = c^n, we can write the prime factorization of c as: c = r_1^t_1 * r_2^t_2 * ... * r_o^t_o.
Taking r_i as an example, if p_i does not equal r_i for all i, then neither a nor b have r_i as a factor, and consequently, c does not have r_i as a factor. Therefore, there exist some r_i which is a factor of a or b.
Now, let's consider the case where r_i is a factor of a. Since p_i divides a, it follows that p_i^k_i also divides a. So let's say k_i = k_i' * n, where k_i' is a positive integer.
Then a = p_1^(k_1' * n) * p_2^k_2 * ... * p_m^k_m.
=> a = (p_1^(k_i') * p_2^k_2 * ... * p_m^k_m)^n.
=> a = (p_1^(k_i') * p_2^k_2 * ... * p_m^k_m)^n.
Here, x = (p_1^(k_i') * p_2^k_2 * ... * p_m^k_m) is a positive integer, and thus a is an nth power.
By following a similar argument for b, we can establish that b is also an nth power.
Therefore, given that (a, b) = 1 and a * b = c^n, we have shown that a and b are nth powers.