Prove that square of any positive integer is of the form 4m or 4m+1 fo

1.	Prove that square of any positive integer is of the form 4m or 4m+1 for some integer
Answer» Let positive integer a = 4m + r , By division algorithm we know here 0 ≤ r < 4 , SoWhen r = 0a = 4m\xa0Squaring both side , we geta2 = ( 4m )2a2 = 4 ( 4m\u200b2)a2 = 4 q , where q = 4m2When r = 1a = 4m + 1squaring both side , we geta2 = ( 4m + 1)2a2 = 16m2 + 1 + 8m\xa0a2 = 4 ( 4m2 + 2m ) + 1\xa0a2 = 4q + 1 , where q = 4m2 + 2mWhen r = 2a = 4m + 2\xa0Squaring both hand side , we geta2 = \u200b( 4m + 2 )2a2 = 16m2 + 4 + 16m\xa0a2 = 4 ( 4m2 + 4m + 1 )a2 = 4q , Where q = \u200b 4m2 + 4m + 1When r = 3\xa0a = 4m + 3Squaring both hand side , we geta2 = \u200b( 4m + 3)2a2 = 16m2 + 9 + 24m\xa0a2 = 16m2 + 24m \u200b + 8 + 1a2 = 4 ( 4m2 + 6m + 2) + 1a2 = 4q + 1 , where q = 4m2 + 6m + 2\xa0Hence\xa0Square of any positive integer is in form of 4q or 4q + 1 , where q is any integer.

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