Question

Prove the following statements. These exercises are cumulative, covering all techniques addressed in Chapters \(4-7\). If \(a \in \mathbb{Z},\) then \(4 \nmid\left(a^{2}-3\right)\)

Short answer

For all the possible representative values of \(a\) in integer form, the remainder is not zero when \(a^2 - 3\) is divided by \(4\). Therefore, \(4\) does not divide \(a^2 - 3\).

Step-by-step solution

Express \(a\) in terms of \(4n\)

Since \(a\) is an integer \(\mathbb{Z}\), it can be represented in the form of \(4n\), \(4n+1\), \(4n+2\), \(4n+3\) where \(n\) is an integer.

Substitute each representation of \(a\) into \(a^2 - 3\)

We substitute \(a\) with \(4n\), \(4n+1\), \(4n+2\), \(4n+3\) in the expression \(a^2 - 3\) and evaluate each expression.

Evaluate remainder for each expression

We find the remainder when \(4n^2 - 3\), \((4n+1)^2 - 3\), \((4n+2)^2 - 3\), and \((4n+3)^2 - 3\) are divided by \(4\). If all remainders are different from zero, then it is proven that \(4\) does not divide \(a^2 - 3\).

Generalization

Since all the possible forms of \(a\) result in a remainder that is not equal to zero, we can generalise that no matter the value of \(a\), \(4\) does not divide \(a^2-3\).