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

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Let us start with taking a, where a is a +ve odd integer.

We apply the division algorithm with ‘a’ and ‘b’ = 4. 

Since 0 ≤ r < 4, the possible remainders are 0, 1, 2, 3. 

That is, a can be 4q, or 4q + 1, or 4q + 2 or 4q + 3, where 1 is the quotient. However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2). 

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

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