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Question
Show that the square of any positive integer is either of the form 4q or 4q+1, for some integer q.
Show that the square of any positive integer is either of the form 4q or 4q+1, for some integer q.
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Solution
Let positive integer a = 4m + r , By division algorithm we know here 0 ≤ r < 4 , So
When r = 0
a = 4m
Squaring both side , we get
a2 = ( 4m )2
a2 = 4 ( 4m2)
a2 = 4 q , where q = 4m2
When r = 1
a = 4m + 1
squaring both side , we get
a2 = ( 4m + 1)2
a2 = 16m2 + 1 + 8m
a2 = 4 ( 4m2 + 2m ) + 1
a2 = 4q + 1 , where q = 4m2 + 2m
When r = 2
a = 4m + 2
Squaring both hand side , we get
a2 = ( 4m + 2 )2
a2 = 16m2 + 4 + 16m
a2 = 4 ( 4m2 + 4m + 1 )
a2 = 4q , Where q = 4m2 + 4m + 1
When r = 3
a = 4m + 3
Squaring both hand side , we get
a2 = ( 4m + 3)2
a2 = 16m2 + 9 + 24m
a2 = 16m2 + 24m + 8 + 1
a2 = 4 ( 4m2 + 6m + 2) + 1
a2 = 4q + 1 , where q = 4m2 + 6m + 2
Hence
Square of any positive integer is in form of 4q or 4q + 1 , where q is any integer.
When r = 0
a = 4m
Squaring both side , we get
a2 = ( 4m )2
a2 = 4 ( 4m2)
a2 = 4 q , where q = 4m2
When r = 1
a = 4m + 1
squaring both side , we get
a2 = ( 4m + 1)2
a2 = 16m2 + 1 + 8m
a2 = 4 ( 4m2 + 2m ) + 1
a2 = 4q + 1 , where q = 4m2 + 2m
When r = 2
a = 4m + 2
Squaring both hand side , we get
a2 = ( 4m + 2 )2
a2 = 16m2 + 4 + 16m
a2 = 4 ( 4m2 + 4m + 1 )
a2 = 4q , Where q = 4m2 + 4m + 1
When r = 3
a = 4m + 3
Squaring both hand side , we get
a2 = ( 4m + 3)2
a2 = 16m2 + 9 + 24m
a2 = 16m2 + 24m + 8 + 1
a2 = 4 ( 4m2 + 6m + 2) + 1
a2 = 4q + 1 , where q = 4m2 + 6m + 2
Hence
Square of any positive integer is in form of 4q or 4q + 1 , where q is any integer.
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