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