Question

Prove the following statements using either direct or contrapositive proof. If \(n \in Z,\) then \(4 \nmid\left(n^{2}-3\right)\)

Short answer

The proof has shown that for any integer \(n\), \(4\) does not divide into \(n^{2} - 3\) without a remainder. This has been validated using a contrapositive proof method. Therefore, the statement 'If \(n \in Z\), then \(4 \nmid\left(n^{2}-3\right)\' is true.

Step-by-step solution

Understand the problem

We need to prove that for any integer \(n\), \(4 \nmid\left(n^{2}-3\right)\), meaning that \(n^{2}-3\) can't be divided by 4 without remainder. We'll use the contrapositive proof method, which involves proving that if \(4 | \left(n^{2}-3\right)\), then \(n\) is not an integer.

Contrapositive Statement

We assume the contrapositive: if \(4 | \left(n^{2}-3\right)\), then \(n\) is not an integer. This is the same as saying that if \(n^{2}-3\) can be divided by 4 without remainder, \(n\) is not an integer. We now prove this statement to be true.

Evaluate the squares of integers modulo 4

Observing the integers, you can see that an integer squared is always 0 or 1 modulo 4. This can be split into two cases: If \(n\) is even (\(n = 2m\)), then \(n^{2} = 4m^{2} = 0\) (mod 4). If \(n\) is odd (\(n = 2m + 1\)), then \(n^{2} = 4m^{2} + 4m + 1 = 1\) (mod 4). So, in both cases, \(n^{2} - 3 = 0 - 3 = -3\) or \(1 - 3 = -2\), neither of which is 0 mod 4.

Conclude the proof

Hence, \(n^{2} - 3\) cannot be congruent to 0 modulo 4, meaning 4 does not divide \(n^{2} - 3\). Therefore, the contrapositive is proven which means that the original statement is true: If \(n \in Z\), then \(4 \nmid\left(n^{2}-3\right)\).