Show that the square of any positive integer is either of the form 4q

1.	Show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.
Answer» According to Euclid’s division lemma, a=bq+r According to the question, When b = 4. a = 4k + r, 0 < r < 4 When r = 0, we get, a = 4k a² = 16k² = 4(4k²) = 4q, where q = 4k² When r = 1, we get, a = 4k + 1 a² = (4k + 1)² = 16k² + 1 + 8k = 4(4k + 2) + 1 = 4q + 1, where q = k(4k + 2) When r = 2, we get, a = 4k + 2 a² = (4k + 2)² = 16k² + 4 + 16k = 4(4k² + 4k + 1) = 4q, where q = 4k² + 4k + 1 When r = 3, we get, a = 4k + 3 a² = (4k + 3)² = 16k² + 9 + 24k = 4(4k² + 6k + 2) + 1 = 4q + 1, where q = 4k² + 6k + 2 Therefore, the square of any positive integer is either of the form 4q or 4q + 1 for some integer q. Hence Proved.

Show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.

Answer»

According to Euclid’s division lemma,

a=bq+r

According to the question,

When b = 4.

a = 4k + r, 0 < r < 4

When r = 0, we get, a = 4k

a² = 16k² = 4(4k²) = 4q, where q = 4k²

When r = 1, we get, a = 4k + 1

a² = (4k + 1)² = 16k² + 1 + 8k = 4(4k + 2) + 1 = 4q + 1, where q = k(4k + 2)

When r = 2, we get, a = 4k + 2

a² = (4k + 2)² = 16k² + 4 + 16k = 4(4k² + 4k + 1) = 4q, where q = 4k² + 4k + 1

When r = 3, we get, a = 4k + 3

a² = (4k + 3)² = 16k² + 9 + 24k = 4(4k² + 6k + 2) + 1

= 4q + 1, where q = 4k² + 6k + 2

Therefore, the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.

Hence Proved.