[size=100][size=150]An integer a is odd if and only if a[sup]3 [/sup]is odd.[br][br]Proof. Suppose that a is odd. Then a = (a)________________.[br][br]for some integer n[sup]3[/sup], and a[sup]3[/sup] = (2n+1)[sup]3[/sup] = 8n[sup]3[/sup]+12n[sup]2[/sup] +6n+1 = 2(4n[sup]3[/sup] +6n[sup]2[/sup] +3n)+1. [br][br]This shows that a[sup]3[/sup] is twice an integer, plus 1, so a[sup]3[/sup] is odd. Thus,[br][br]we’ve proved that if a is (b)______________ then a[sup]3[/sup] is (d)__________________.[br][br]Conversely, we need to show that if a[sup]3[/sup] is odd, then a is odd. For this we employ[br][br](e)_________________________. Suppose a is not odd. Thus, a is even, so a = 2n for some[br][br]integer n. Then a[sup]3[/sup] = (2n)[sup]3[/sup] = 8n[sup]3[/sup] = 2(4n[sup]3[/sup]) is even (not odd).[/size][/size]

[size=150]Prove the following statement. This exercise is cumulative, covering all techniques addressed in this module.[br]"There is a prime number between 90 and 100."[/size]