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Question

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

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Solution

## 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 m ≠ 0 and r < 4. As m ≠ 0 and 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 ⇒⇒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. Let be any positive integer We know by Euclid's algorithm, if a and b are two positive integers, there exist unique integers q and r satisfying, where. Take Since 0 ≤ r < 4, the possible remainders are 0, 1, 2 and 3. That is, can be , where q is the quotient. Since is odd, cannot be 4q or 4q + 2 as they are both divisible by 2. Therefore, any odd integer is of the form 4q + 1 or 4q + 3.

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