we better use the contrapositive approach.
suppose n
3m where n,m are natural numbers.
then n = 3m+1 or 3m+2
so, n2 = 9m2+6m+1 = 3(3m2+2m)+1 = 3k+1
or n2 = 9m2+12m+4 = 3(3m2+4m+1) + 1 = 3 p+1 where k , p are natural numbers.
clearly n2 is not a multiple of 3. but leaves a remainder 1 when divided by 3.
i.e.
is true. ==>
is true.
i.e. if the square of a natural number is a multiple of 3 then the number is the multiple of 3.