Question

Question 4
Write whether the square of any positive integer can be of the form 3m + 2, where m is a natural number. Justify your answer.

Answer 1

No, we cannot express the square of any positive integer can be of the form 3m + 2.
By Euclid's lemma,
$a=bq+r,0\le r<b$, here a is any positive integer,

When we take the divisor (b), we have

$a=3q+rfor0\le r<3$

So, any positive integer is of the form 3q, 3q + 1, 3q + 2.
Now, $(3q{)}^2=9q^2=3m(wherem=3q^2)$

$(3q+1{)}^2=9q^2+6q+1=3(3q^2+2q)+1=3m+1(wherem=3q^2+2q$

$=(3q+2{)}^2=9q^2+12q+4=9q^2+12q+3+1=3(3q^2+4q+1)+1=3m+1(wherem=3q^2+4q+1)$

Thus, any square is of the form 3m or 3m + 1.

Hence, square of any positive number cannot be of the form 3m + 2.