Question

use eculid divison lemma to show that the square of any positive integers is either of 3m or3m+1 for same integer m

Answer 1

Let a be any positive integer. Then it is of the form 3q,3q+1 or 3q+2

CASE I ---- When a = 3q
a^2 = (3q)^2 = 9q^2 = 3q(3q) = 3m,
where m= 3q

CASE II --- When a = 3q+1
a^2 = (3q+1)^2 = 9q^2 + 6q +1 = 3q( 3q+2) + 1
= 3m+1
where m = q(3q +2)

CASE III --- When a = 3q+2
a^2 = (3q + 2)^2 = 9q^2 + 12q + 4
= 9q^2 + 12q + 3 + 1
= 3(3q^2 + 4q + 1) = 3m +1
where m = 3q^2 + 4q + 1

Therefore, a is of the form 3m or 3m +1

Answer 2

Answer:

It is the correct answer.

Step-by-step explanation:

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