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

Given,

For any given two integers a and b , there exists a unique integers between them q and r ,which satisfies

a=bq+r , where zero is less than or equal to r

Here b=3

Case 1:

If r=0

a^2=(3q+0)^2

=(3q)^2

=9q^2

=3(3q^2)

= 3m [where m=3q^2]

Case 2:

If r=1

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

=(9q^2 + 2×3q×1 + 1)

=(9q^2 + 6q + 1)

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

=3m +1 [where m= 3q^2 + 2q]

therefore square of any positive integer is either of

the form 3m or 3m + 1 for some integer m.

Answer 2

Step-by-step explanation:

Let n be any random 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

So, n² = 9q² + r² + 6qr .... (i), where 0 ≤ r ≤ 3 Case I

When r = 0.

Putting r = 0 in (i), we get

n² = 9q²

= 3(3q²)

= 3m, where m = 3q² is an integer.

Case II

When r = 1.

Putting r = 1 in (i), we get

n² = (9q² + 1 6q)

= 3(3q² + 2q) + 1

= 3m + 1,

where m = (3q² + 2q) is an integer.

Case III

When r = 2.

Putting r = 2 in (i), we get

n² = (9q² + 4 + 12q)

= 3(3q² + 4q + 1)

= 3m + 1,

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

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

Please mark my answer as the brainliest answer.....