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

Let a be any odd positive integer and b = 4. By division lemma there exist integer q and r such that

a = 4 q + r, where 0 ≤ r ≤ 4


so a = 4q or, a = 4q + 1 or, a = 4q + 2 or, a = 4q + 3


4q + 1 4 is divisible by 2 but 1 is not divisible by 2, so it is an odd number


4q + 2 4 is divisible by 2 and 2 is also divisible by 2, so it is an even number


4q + 3 4 is divisible by 2 but 3 is not divisible by 2, so it is an odd number


4q + 4 4 is divisible by 2 and 4 is also divisible by 2, so it is an even number


any odd integer is of the form 4q + 1 or, 4q + 3.


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