Show that the square of an …

Let x be an odd integer.

On dividing x by 4, let q be the quotient and r be the remainder.

So, by Euclid's division lemma, we have:

                    x = 4q + r, where 0 <_ r < 4.

Therefore,  x² = ( 4q + r )² .

                        = 16q² + r² + 8qr                                   .....(i).where 0 <_ r < 4.

Case l.      When r = 0

                 Putting r = 0 in (I), we get: 

                 x² = 16q²  = 4 ( 4q² ) = 4 Q, where Q = 4q² is an integer.

Case ll.    When r = 1 

                  Putting r = 1 in (I), we get: 

                 x² = ( 16q² + 1 + 8q ) = 4 ( 4q² + 2q ) + 1 = ( 4 Q + 1),  

                                                                         where Q = ( 4q² + 2q ) is an integer. 

Clearly, 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.