Question

1. Let x ∈ Z. Prove that if 3 | 2x, then 3 | x.2. Let n ∈ Z. Prove that 3 | (2n 2 + 1) if and only if 3 - n.

Answer 1

1.Suppose 3 | 2x. Then we can write 2x = 3k for some integer k. Rearranging, we have x = (3/2)k. Since k is an integer, (3/2)k is also an integer, which means that x is divisible by 3. Hence, 3 | x.

First, suppose 3 | (2n^2 + 1). Then we can write 2n^2 + 1 = 3k for some integer k. Rearranging, we have 2n^2 = 3k - 1. Since 3k - 1 is odd, we can write it as 2m + 1 for some integer m. Substituting, we have 2n^2 = 2m + 1, which implies that n^2 = m + (1/2). But since m is an integer, (1/2)m is not an integer, which means that n^2 is not an integer. This is a contradiction, so our assumption that 3 | (2n^2 + 1) must be false.

Now suppose 3 - n. Then we can write n = 3k - 1 for some integer k. Substituting, we have 2n^2 + 1 = 18k^2 - 12k + 3. Factoring out 3, we have 2n^2 + 1 = 3(6k^2 - 4k + 1). But 6k^2 - 4k + 1 is always an integer, so if 3 - n, then 3 | (2n^2 + 1).

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