Question

Show that the square of any positive integer cannot be of the form 3m+2 where m is a natural number

Answer 1

Answer:

Each integer n has the form n=3a+b, for some a and b, where b = 0, 1 or -1.

So each square n² has the form:

n² = (3a+b)² = 9a² + 6ab + b² = 3(3a²+2ab) + b² = 3m + b², where m is an integer.

Since b = 0, 1 or -1, the term b² = 0 or 1.

That is, n² has either the form 3m or 3m+1.  Therefore the form 3m+2 is not possible.