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

Answer:

https://brainly.in/question/40248511

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Sol:- Let a be a positive integer q be the quotient and r be the remainder

Dividing a by 3 using the Euclid's Division Lemma

we have

a = 3q + r,

where a = ≤ r < 3

Putting r = 0, 1 and 2, we get:

a = 3q

→ a^2 = 9q^2

= 3 × 3q^2

= 3m (Assuming m = q^2)

Then, a = 3q + 1

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

= 9q^2 + 6q + 1

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

= 3m +1 (Assuming m = 3q^2 + 2q)

Next, a = 3q + 2

→ a² = (3q + 2)^2

= 9q^2 + 12q + 4

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

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

= 3m + 1. (Assuming m = 3q^2 + 4q+1)

Therefore, the square of any positive integer (say a^2) is always of the form 3m or 3m + 1

Hence, proved