Prove that the square of any positive integer is of the form 5q or 5q

1.	Prove that the square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.
Answer» Let ‘a’ be any positive integer. Then, According to Euclid’s division lemma, a = bq+r According to the question, when b = 5. a = 5k + r, 0 < r < 5 When r = 0, we get, a = 5k a² = 25k² = 5(5k²) = 5q, where q = 5k² When r = 1, we get, a = 5k + 1 a² = (5k + 1)² = 25k² + 1 + 10k = 5k(5k + 2) + 1 = 5q + 1, where q = k(5k + 2) When r = 2, we get, a = 5k + 2 a² = (5k + 2)² = 25k² + 4 + 20k = 5(5k² + 4k) + 4 = 4q + 4, where q = 5k² + 4k When r = 3, we get, a = 5k + 3 a²= (5k + 3)² = 25k² + 9 + 30k = 5(5k² + 6k + 1) + 4 = 5q + 4, where q = 5k² + 6k + 1 When r = 4, we get, a = 5k + 4 a² = (5k + 4)² = 25k² + 16 + 40k = 5(5k² + 8k + 3) + 1 = 5q + 1, where q = 5k² + 8k + 3 Therefore, the square of any positive integer is of the form 5q or, 5q + 1 or 5q + 4 for some integer q.

Prove that the square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.

Answer»

Let ‘a’ be any positive integer.

Then,

According to Euclid’s division lemma,

a = bq+r

According to the question, when b = 5.

a = 5k + r, 0 < r < 5

When r = 0, we get, a = 5k

a² = 25k² = 5(5k²) = 5q, where q = 5k²

When r = 1, we get, a = 5k + 1

a² = (5k + 1)²

= 25k² + 1 + 10k

= 5k(5k + 2) + 1

= 5q + 1, where q = k(5k + 2)

When r = 2, we get, a = 5k + 2

a² = (5k + 2)²

= 25k² + 4 + 20k

= 5(5k² + 4k) + 4

= 4q + 4, where q = 5k² + 4k

When r = 3, we get, a = 5k + 3

a²= (5k + 3)²

= 25k² + 9 + 30k

= 5(5k² + 6k + 1) + 4

= 5q + 4, where q = 5k² + 6k + 1

When r = 4, we get, a = 5k + 4

a² = (5k + 4)²

= 25k² + 16 + 40k

= 5(5k² + 8k + 3) + 1

= 5q + 1, where q = 5k² + 8k + 3

Therefore, the square of any positive integer is of the form 5q or, 5q + 1 or 5q + 4 for some integer q.