Question

Prove the following statements. These exercises are cumulative, covering all techniques addressed in Chapters \(4-7\). Given an integer \(a\), then \(a^{3}+a^{2}+a\) is even if and only if \(a\) is even.

Short answer

The statements are proven true: if \(a\) is even, then \(a^{3}+a^{2}+a\) is indeed even, and if \(a^{3}+a^{2}+a\) is even, then \(a\) is also even.

Step-by-step solution

Part 1: If a is even, then \(a^{3}+a^{2}+a\) is even.

First, you need to show that if \(a\) is even, then \(a^{3}+a^{2}+a\) is even. Any even number can be rewritten as \(2n\) for some integer \(n\). So, if \(a\) is even, we can replace \(a\) with \(2n\). Simplify the expression to prove the statement. \(a^{3}+a^{2}+a = (2n)^{3}+(2n)^{2}+2n = 8n^{3}+4n^{2}+2n = 2(4n^{3}+2n^{2}+n)\). This is your desired structure. Therefore, it's proven that if \(a\) is even, then \(a^{3}+a^{2}+a\) is even as well.

Part 2: If \(a^{3}+a^{2}+a\) is even, then a is even.

At this step, you need to prove the reverse statement that if \(a^{3}+a^{2}+a\) is even, then \(a\) is even. This is slightly trickier because you have to derive \(a\) is even from \(a^{3}+a^{2}+a\) is even. One way to approach this problem is by contradiction. Suppose \(a\) is not even, which means \(a\) is odd. An odd number can be expressed as \(2m+1\) for some integer \(m\). \(a^{3}+a^{2}+a = (2m+1)^{3}+(2m+1)^{2}+2m+1 = 8m^{3}+12m^{2}+6m+1+4m^{2}+4m+1+2m+1 = 2(4m^{3}+8m^{2}+6m+m) + 3\). The resulting number is not even because it has the form \(2n + 1\). This is a contradiction. Thus, if \(a^{3}+a^{2}+a\) is even, \(a\) must be even and not odd.