Question

4. Use Euclid's division lemma to show that the square of any positive integer s either of
the form 3m or 3m + 1 for some integer m.
[Hint : Let x be any positive integer then it is of the form 3q, 39 + 1 or 39+2. Now square
each of these and show that they can be rewritten in the form 3m or 3m +1.]​

Answer 1

Answer:

Step-by-step explanation:

By lemma

a=bq+r,b=3

The remainders are 0,1,2

Then a=3q

a^2=3q^2

a^2=9q^2

a^2=3(3q^2)

a^2=3m here m=3q^2

Next

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

=9q^2+1+6q

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

=3m+1 here m=3q^2+2q

Answer 2

$\bf\huge{\underline{\underline{Correct\: Question}}}$

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

[Hint : Let x be 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.]

$\bf\huge{\underline{\underline{Solution}}}$

Let us consider an arbitrary positive integer as 'x' such that it is of the form.

⠀⠀3q, (3q + 1) or (3q + 2)

For x = 3q, we have x² = (3q)²

=> x² = 9q² = 3(3q²) = 3m ⠀⠀⠀⠀⠀ ⠀ ...(1)

Putting 3q² = m, where m is an integer.

For x = 3q + 1

⠀⠀x² = (3q + 1)² = 9q² + 6q + 1

⠀ ⠀ ⠀= 3(3q² + 2q) + 1 = 3m + 1 ⠀⠀...(2)

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

For x = 3q + 2,

⠀⠀x² = (3q + 2)²

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

⠀ ⠀⠀ = 3(3q² + 4q + 1) + 1 = 3m+1 ⠀...(3)

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

From (1), (2) and (3),

⠀⠀⠀x² = 3m or 3m + 1

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