Question

Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m. [4 MARKS]

Answer 1

Concept : 1 Mark
Application : 2 Mark
Conclusion : 1 Mark

Let x be any positive integer and b = 3

According to Euclid's division lemma, we can say that
$x=3q+r,0\le r<3$

Therefore, all possible values of x are:
$x=3q,(3q+1)or(3q+2)$

Now lets square each one of them one by one

$\left(i\right)(3q{)}^2=9q^2$
Let $m=3q^2$ be some integer, we get $9q^2=3\times 3q^2=3m$


$\left(ii\right)(3q+1{)}^2=9q^2+6q+1=3(3q^2+2q)+1$
Let $m=3q^2+2q$ be some integer, we get
$(3q+1{)}^2=3m+1$


$\left(iii\right)(3q+2{)}^2=9q^2+4+12q=9q^2+12q+3+1=3(3q^2+4q+1)+1$
Let $m=(3q^2+4q+1)$ be some integer, we get

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

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