Question

Show that any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer.

Answer 1

Let n be any arbitrary positive odd integer.
On dividing n by 4, let m be the quotient and r be the remainder. So, by Euclid's division lemma, we have

n = 4m + r, where 0 ≤ r < 4.

As 0 ≤ r < 4 and r is an integer, r can take values 0, 1, 2, 3.
⇒ n = 4m or n = 4m + 1 or n = 4m + 2 or n = 4m + 3

But n ≠ 4m or n ≠ 4m + 2 (∵ 4m, 4m + 2 are multiples of 2, so an even interger whereas n is an odd integer)

⇒ n = 4m + 1 or n = 4m + 3

Thus, any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer.