Question

 Use Euclid's Division Algorithm to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Answer 1

Let a and b are two positive integers such that a is greater than b; then:

a = bq + r; where q and r are also positive integers and 0 ≤ r < b

Taking b = 3, we get:

a = 3q + r; where 0 ≤ r < 3

⇒ The value of positive integer a will be

3q + 0, 3q + 1 or 3q + 2

i.e., 3q, 3q + 1 or 3q + 2.

Now we have to show that the squares of positive integers 3q, 3q + 1 and 3q + 2 can be expressed as 3m, or 3m + 1 for some integer m.

Square of 3q = (3q)²

= 9q² = 3(3q²) = 3m; 3 where m is some integer.

Square of 3q + 1 = (3q + 1)²

= 9q² + 6q + 1

= 3(3q² + 2q) + 1 = 3m + 1 for some integer m.

Square of 3q + 2 = (3q + 2)²

= 9q² + 12q + 4

= 9q² + 12q + 3 + 1

= 3(3q² + 4q + 1) + 1 = 3m + 1 for some integer m.

The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Hence the required result.