Question

Show that the square of any positive integer is of the form 3m or 3m+1, for some integer m.

Answer 1

Let 'a' be nay positive integer and b=3
We know, $a=bq+r,0\le r<b$.
Now, $a=3q+r,0\le r<3$.
The possibilities of remainder = 0, 1 or 2
Case 1 $-a=3q{a}^2=9q^2=3\times \left(3q^2\right)=3m(wherem=3q^2)$
Case II $-a=3q+1a^2=(3q+1{)}^2=9q^2+6q+1=3(3q^2+2q)+1=3m+1(wherem=3q^2+2q)$
Case III $-a=3q+2a^2=(3q+2{)}^2=9q^2+12q+4=9q^2+12q+3+1=3(3q^2+4q+1)+1=3m+1wherem=3q^2+4q+1)$
From all the above cases it is clear that square of any positive integer (as in this case $a^2$) is either of the form 3m or 3m +1.