Show that the square of odd integer is of the form 8q + 1, for some in

1.	Show that the square of odd integer is of the form 8q + 1, for some integer q.
Answer» From Euclid’s division lemma, a = bq+r ; where 0 < r < b Putting b=4 for the question, ⇒ a = 4q + r, 0 < r < 4 For r = 0, we get a = 4q, which is an even number. For r = 1, we get a = 4q + 1, which is an odd number. On squaring, ⇒ a² = (4q + 1)² = 16q² + 1 + 8q = 8(2q² + q) + 1 = 8m + 1, where m = 2q² + q For r = 2, we get a = 4q + 2 = 2(2q + 1), which is an even number. For r = 3, we get a = 4q + 3, which is an odd number. On squaring, ⇒ a² = (4q + 3)² = 16q² + 9 + 24q = 8(2q² + 3q + 1) + 1 = 8m + 1, where m = 2q² + 3q + 1 Thus, the square of an odd integer is of the form 8q + 1, for some integer q.

Show that the square of odd integer is of the form 8q + 1, for some integer q.

Answer»

From Euclid’s division lemma,

a = bq+r ; where 0 < r < b

Putting b=4 for the question,

⇒ a = 4q + r, 0 < r < 4

For r = 0, we get a = 4q, which is an even number.

For r = 1, we get a = 4q + 1, which is an odd number.

On squaring,

⇒ a² = (4q + 1)²

= 16q² + 1 + 8q

= 8(2q² + q) + 1

= 8m + 1, where m = 2q² + q

For r = 2, we get a = 4q + 2 = 2(2q + 1), which is an even number.

For r = 3, we get a = 4q + 3, which is an odd number.

On squaring,

⇒ a² = (4q + 3)²

= 16q² + 9 + 24q

= 8(2q² + 3q + 1) + 1

= 8m + 1, where m = 2q² + 3q + 1

Thus, the square of an odd integer is of the form 8q + 1, for some integer q.