Question

Question 4
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.
[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

Answer 1

Let 'a' be any positive integer and b = 3.
Then, a = 3q + r for some integer q $\ge$ 0 and r = 0, 1, 2 because 0 $\le$ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
$a^2=\left(3q{)}^{2}or\right(3q+1{)}^{2}or(3q+2{)}^2{a}^2=(9q{)}^{2}or9q^2+6q+1or9q^2+12q+4=3\times \left(3q^2\right)or3(3q^2+2q)+1or3(3q^2+4q+1)+1=3k_{1}or3k_2+1or3k_3+1$
Where $k_1,k_2,and{k}_3$ are some positive integers.
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.