Question

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

Answer 1

Let ‘a’ be any positive integer. 

Then, 

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.