Any help with this discrete math proof would be much appreci

Any help with this discrete math proof would be much appreciated.

Solution

According to the quotient-remainder theorem, any integer “n” can be written in one of three forms, for some integer “q”:

Case 1

Case 2

Case 3

n = 3q

n = 3q+1

n = 3q+2

n² = (3q)²

n² = (3q+1)²

n² = (3q+ 2)²

n² = 9q²

n² = 9q² +6q +1

n² = 9q² + 12q + 4

n² = 3(3q² )

n² = 3(3q² +2q) +1

n² = 3(3q² +4q +1) +1

n² = 3k where k = 3q²

n² = 3k+1, k = 3q²+2q

N² = 3k+1, k = 3q² + 4q +1

Therefore, any arbitrary n² can be expressed in either the form 3k or 3k+1 for some integer. Thus if n is 3q+2, then n cannot be a perfect square.