Use a proof by cases to show that every perfect square is either a multiple of four or one more than a multiple of four. Hint you only need two cases! 2. Use a proof by cases to show that if two integers are both odd or both even, then their sum is always even 3. Use a proof by contraposition to show that if the sum of two numbers is odd then exactly one of the two numbers must be odd.
Recall that every integer can be written as [tex]n=2k[/tex] or [tex]n=2k+1[/tex], i.e., every integer must be even or odd. Let us analyze the first case:
First case: Assume that the integer [tex]n[/tex] is even, so [tex]n=2k[/tex] for some integer [tex]k[/tex]. Then, squaring the equality we have [tex]n^2=4k^2[/tex]. Therefore, if [tex]n[/tex] is even, its square is a multiple of 4.
Second case: Assume that the integer [tex]n[/tex] is odd, so [tex]n=2k+1[/tex] for some integer [tex]k[/tex]. Then, squaring this equality we get [tex]n^2=(2k+1)^2=4k^2+4k+1 = 4(k^2+k) +1[/tex]. As the number [tex]k^2+k[/tex] is an integer, we deduce that [tex]n^2[/tex] is the multiple of 4 plus 1.
As there are no other possibilities, the statement is proven.
