Show that the square of any integer is of the form 4m + 1, for some integer m.

By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer .

On putting b = 4 we get

a = 4q + r, where 0 ≤ r < 4 (i)

If r = 0

⇒ a = 4q, 4q is divisible by 2

⇒ 4q is even.

If r = 1

⇒ a = 4q + 1, (4q + 1) is not divisible by 2.

If r = 2

⇒ a = 4q + 2,2(2q + 1) is divisible by 2

⇒ 2(2q + 1) is even.

If r = 3

⇒ a = 4q + 3, (4q + 3) is not divisible by 2.

So, for any positive integer (4q + 1) and (4q + 3) are odd integers.

Case I For (4q + 1)

Squaring both the sides using (a + b)² = a² + 2ab + b²

a² = 16q² + 1 + 8q

⇒ a² = 4(4q² + 2q) + 1 = 4m + 1

Where m = (4q² + 2q) in an integer.

Case II For (4q + 3)

Squaring both the sides using (a + b)² = a² + 2ab + b²

a² = 16q² + 9 + 24q = 16q² + 24q + 8 + 1

⇒ a² = 4 (4q² + 6q + 2) + 1 = 4m + 1

where m = (4q² + 6q + 2) is an integer.

Hence, for some integer m, the square of any odd integer is of the form 4m + 1.