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Question

Suppose that n is an integer such that n^2 is divisible by 3. Prove that n is divisible by 3.

Answer 1

According to the fundamental theorem of arithmetic, we have n^2 = p_1^a1....p_n, where the p_is are pairwise distinct prime numbers and the a_i >0 are their integer exponents.

Since n^2 is a perfect square, all a_i must be even number, and since n^2 is divisible by 3, one of the p_is must be 3. So, each a_i = 2b_i for some positive integer b_i and we have

n^2 = p_1^(2b_1) *....p_n^(2b_n). Therefore, n = p_1^b1...* p_n^b_n, which shows n is multiple of each of the primes p_i. Simce one of then is 3, it follows n is divisible by 3.

Answer 2

n^2 is divisible by 3
Let n^2 = 3k where k is an integer.

Now if we factorize a perfect square, all the factors occur in pairs. So if n^2 has one factor as 3, it must also have another factor as 3 which implies that k is divisible by 3.

let k = 3p

therefore,
n^2 = 3(3p) = 9p
or n = 3 * sq. rt. p

Hence n is also divisible by 3.

Hope that helped

Answer 3

use Inductive reasoning, prove the base and then prove the rest