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.​

Answer 1

Step-by-step explanation:

Let 'a' be any +ve integer  

And 'b' = 3

By Using Euclid's Division Lemma:

a = bq +r

a = 3q + r, where r is 0 ≤ r < b i.e. 3

Case - 1

Let r = 0

So, a = 3q

a^2 = 9q^2       ( Squaring both sides )

a^2 = 3(3q^2)

a^2 = 3m, where m = 3q^2

Case - 2

Let r = 1

So, a =  3q + 1

a^2 = (3q + 1 )^2        ( Squaring both sides )

a^2  = 9q^2 + 6q + 1

a^2 = 3(3q^2 + 2q) + 1

a^2 = 3m + 1, where m = (3q^2 + 2q)

Case - 3

Let r = 2

So, a = 3q + 2

a^2 = (3q+2)^2         ( Squaring both sides )

a^2 = 9q^2 + 12q + 4

a^2 = 9q^2 + 12q + 3 + 1

a^2 = 3( 3q^2 + 4q + 1 ) + 1

a^2 = 3m +1, where m = (3q^2 + 4q + 1 )

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