Question

show that the square of any positive integers cannot be of the form of 6m+2,6m+5 for any integer m

Answer 1

Let X be any integer= 3q , 3q+1,3q+2,3q+3 ,3q+4 ,3q+5 .

In case 1= (3q)^2 = 3(3q)
3m ,where as m = 3q
In case 2 = (3q +1)^2 = (9q+1)
3(3q)+1 = 3m+1 , where as m = 3q
In case 3 = (3q+3)^2= (9q+6 )
3(3q+2) = 3m where as m = (3q+2)
In case 4 = (3q+5)^2 = ( 9q+6)
3m where as m = (9q+6)
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