Euclid's Division Lemma question

  Question :

Using Euclid's Division Lemma, show that the square of any positive integer is either of the form 3m or (3m + 1) for some integer m.

Solution:

Let us consider any arbitrary integer be n.

On dividing n by 3, we get q as quotient and r as remainder. Then by Euclid's division lemma, we have

n = 3q + r , where 0=< r < 3.

Now, the question is asking to show the square of any positive integer. Therefore squaring both the sides.

n^2 = (3q + r)^2

n^2 = 9q^2 + r^2 + 6qr ....................(i)

Now, three cases arises as of remainder can be either 0, 1 or 3.

CASE I

When r = 0

Putting r = 0 in equation (i) , we get

n^2 = 9q^2

Now this 9q^2 can be written as the factor of 3.

n^2 = 3(3q^2)

n^2 = 3m , where m = (3q^2) and is an integer.

CASE II

When r = 1

Putting r = 1 in equation (i) , we get

n^2 = 9q^2 + 1 + 6q

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

n^2 = 3m + 1 , where m = (3q^2 + 2q) and is an integer.

CASE III

When r = 2

Putting r = 2 in equation (i) , we get

n^2 = (9q^2 + 4 + 12q)

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

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

n^2 = 3m + 1 , where m = (3q^2 + 4q + 1) and is an integer.

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

Watch the video solution here:

PS: Question is from secondary school Mathematics for class 10 by R S Aggarwal and V Aggarwal of Bharti Bhawan Publication (Publication : 2021)

Number System

Real Numbers

Euclid's Division Lemma

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