Show that the square of any positive odd integer is of the form 4m+1,for same integer n

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.

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