Question

Prove that the square of any positive integer is of the form 3m or, 3m + 1 but not of the form 3m +2.

Answer 1

By Euclid’s division algorithm
a = bq + r, where 0 ≤ r ≤ b
Put b = 3
a = 3q + r, where 0 ≤ r ≤ 3
If r = 0, then a = 3q

If r = 1, then a = 3q + 1
If r = 2, then a = 3q + 2
Now, (3q)² = 9q²
= 3 × 3q²
= 3m, where m is some integer
(3q + 1)² = (3q)² + 2(3q)(1) + (1)²
= 9q² + 6q + 1
= 3(3q² + 2q) + 1
= 3m + 1, where m is some integer
(3q + 2)² = (3q)² + 2(3q)(2) + (2)²
= 9q² + 12q + 4
= 9q² + 12q + 4
= 3(3q² + 4q + 1) + 1
= 3m + 1, where m is some integer
Hence the square of any positive integer is of the form 3m, or 3m +1
But not of the form 3m + 2