Question

Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m add 1for sime integer m.

Answer 1

Answer:

Step-by-step explanation:

If a and b are two positive integers, then,

a = bq + r, 0  r  b Let b = 3

Therefore, r = 0, 1, 2

Therefore, a = 3q or a = 3q + 1 or a = 3q + 2

If a = 3q a2 = 9q2 = 3(3q2) = 3m where m = 3q^2

If a = 3q + 1  a2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1 where m = 3q^2 + 2q

If a = 3q + 2 a2 = 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1= 3m + 1, where m = 3q^2 + 4q + 1

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

Answer 2

Answer:

It is the correct answer.

Step-by-step explanation:

Hope this attachment helps you.