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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.

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LudwigGR

Answer:

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.

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