Show that the square of any …

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 ( 4m​² )
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 . ( Hence proved )

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