[size=150]Suppose a ∈ Z. If a[sup]2[/sup] is even, then a is even.[br][br]Proof. For the sake of contradiction, suppose a[sup]2[/sup] is even and a is not even.[br][br]Then a[sup]2[/sup] is even, and a is odd.[br][br]Since a is odd, there is an integer c for which a = 2c +1.[br][br]Then a[sup]2[/sup] = (2c +1)[sup]2[/sup][br] = 4c[sup]2[/sup] +4c +1 [br]= 2(2c[sup]2[/sup] +2c)+1, so a[sup]2[/sup] is odd.[br][br]Thus a[sup]2 [/sup]is even and a[sup]2[/sup] is not even, a contradiction.[/size]

[size=150]The given proof uses the method of proof by contradiction to show that if a[sup]2[/sup] is even, then a is even. Let's break down the proof step by step:[br][br]1. Suppose for the sake of contradiction that a[sup]2[/sup] is even and a is not even. This means we assume the opposite of what we want to prove, which is that a is odd.[br][br]2. We know that if a is odd, there exists an integer c such that a = 2c + 1. This is because odd numbers can be expressed in the form 2c + 1, where c is an integer.[br][sup]2[/sup][br]3. Substitute the value of a = 2c + 1 into a^2 to obtain a[sup]2[/sup] = (2c + 1)[sup]2[/sup].[br][br]4. Expand the expression (2c + 1)[sup]2[/sup] to get a[sup]2[/sup] = 4c[sup]2[/sup] + 4c + 1.[br][br]5. Rearrange the terms to get a[sup]2[/sup] = 2(2c[sup]2[/sup] + 2c) + 1.[br][br]6. From step 5, we can see that a[sup]2[/sup] can be written in the form 2k + 1, where k = 2c[sup]2[/sup]+ 2c is an integer.[br][br]7. Therefore, we have shown that a[sup]2[/sup] is odd because it takes the form of 2k + 1, where k is an integer.[br][br]8. However, this contradicts our initial assumption that a[sup]2[/sup] is even.[br][br]9. Since we have arrived at a contradiction, our initial assumption must be false.[br][br]10. Hence, we can conclude that if a is even, then a must be even.[br][br]By using contradiction, we have established the desired result, which is that if a[sup]2[/sup] is even, then a is even.[/size]