Show that the square of an odd integer is of the form 4 Q + 1 for some

1.	Show that the square of an odd integer is of the form 4 Q + 1 for some integer q
Answer» Let x\xa0be an odd integer.On dividing x\xa0by 4, let\xa0q\xa0be the quotient and\xa0r\xa0be the remainder.So, by Euclid\'s division lemma, we have: x\xa0= 4q\xa0+\xa0r, where 0 <_\xa0r\xa0< 4.Therefore, x2\xa0= ( 4q + r\xa0)2\xa0. = 16q2 +\xa0r2\xa0+ 8qr.....(i).where 0 <_\xa0r\xa0<\xa04.Case l. When\xa0r\xa0= 0 Putting\xa0r\xa0= 0 in (I), we get:\xa0 x2\xa0= 16q2\xa0\xa0= 4 ( 4q2\xa0) = 4 Q, where Q = 4q2\xa0is an integer.Case ll. When\xa0r\xa0= 1\xa0 Putting\xa0r\xa0= 1 in (I), we get:\xa0 x2\xa0= ( 16q2\xa0+ 1 + 8q\xa0) = 4 ( 4q2\xa0+ 2q\xa0) + 1 = ( 4 Q + 1), where Q = ( 4q2\xa0+ 2q\xa0) is an integer.\xa0Clearly, 4Q is even and since x is odd. So x is not equal to 4Q.Hence, the square of an odd integer is of form ( 4Q + 1) for some integer Q.\xa0\xa0

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