Show that for odd positive integer to be a perfect square ,it should be of the form 8k+1
Since, any odd positive integer n is of the form 4m\xa0+ 1 or 4m + 3.if n = 4m + 1n2 = (4m + 1)2= 16m2 + 8m + 1= 8(2m2 + m) + 1So 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) + 1So 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.
Since, any odd positive integer n is of the form 4m\xa0+ 1 or 4m + 3.if n = 4m + 1n2 = (4m + 1)2= 16m2 + 8m + 1= 8(2m2 + m) + 1So 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) + 1So 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.