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.
[Hint: Letxbe any positive integer then it is of the form 3q ,3q + 1 or 3q + 2. Now square
each of these and show that they can be rewritten in the form 3m or 3m + 1.​

Answer 1

This is the answer for the question

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let n be the arbitrary positive integer.

On dividing n by 3,

let q be the quotient

And r be the remainder.

Then,

By Euclid's division lemma, we have

n = 3q + r, where 0 ≤ r < 3.

Therefore,

n = 3q + r² + 6qr ...... (i)

where 0 ≤ r < 3

Case 1. when r = 0

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

n² = 3q² = 3(3q)² = 3m,

where m = 3q² is an integer.

Case 2. when r = 1.

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

n² = 9q² + 1 + 6q = 3(3q² + 2q) = 3(3q² + 2q) + 1

= 3m + 1, where m = 3q² + 2q is an integer.

Case 3. when r = 2.

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

n² = (9q² + 4 + 12q) = 3(3q² + 4q + 1) + 1

= 3m + 1, where m = (3q² + 4q + 1) is an integer.

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