Show that the square of any …
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Sia ? 4 years, 8 months ago
Since, any odd positive integer n is of the form 4m + 1 or 4m + 3.
if n = 4m + 1
n2 = (4m + 1)2
= 16m2 + 8m + 1
= 8(2m2 + m) + 1
So n2 = 8q + 1 ........... (i)) (where q = 2m2 + m is a positive integer)
If n = (4m + 3)
n2 = (4m + 3)2
= 16m2 + 24m + 9
= 8(2m2 + 3m + 1) + 1
So n2 = 8q + 1 ...... (ii) (where q = 2m2 + 3m + 1 is a positive integer)
From (i) and (ii) we conclude that the square of an odd positive integer is of the form 8q + 1, for some integer q.
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