Question

4. 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

Let any positive integer be a then,

Case 1 : When a = 3q

→ a = 3q

→ a² = (3q)²

→ a² = 9q²

→ a² = 3(3q²)

→ a² = 3m ; where 3q² = m.

Case 2 : When a = 3q + 1

→ a² = (3q + 1)²

→ a² = 9q² + 6q + 1

→ a² = 3(3q² + 2q) + 1

→ a² = 3m + 1 ; where m = 3q² + 2q

Case 3 : When a = 3q + 2

→ a² = 9q² + 12q + 8

→ a² = 3(3q² + 4q + 2) + 2

→ a² = 3m + 2 ; where m = 3q² + 4q + 2

∴ Square of any positive integer by Euclids lemma is in form of 3m, 3m + 1 and 3m + 2.

Q.E.D

Answer 2

Solution:-

Let "a" be any positive Integer and b=3.

By Euclid Division Lemma,

The Positive Value of "r" can be 0, 1 and 2.

Hence,

The Positive Value of "a" be ( 3b ), ( 3b + 1) & (3b + 2).

Case |,

=) a = 3b

Squaring on both the sides.

=) a² = (3b)²

=) a² = 9b²

=) a² = 3( 3b²)

Let (a² = x) and (3b² = m)

=) x = 3m

Case ||,

=) a = (3b + 1)

Squaring on both the sides.

=) a² = ( 3b +1)²

=) a² = 9b² + 1 + 6b

=) a² = 3( 3b² + 2b) + 1

Taking ( a² = x) and ( 3b² + 2b = m)

=) x = 3m + 1.

Case |||,

=) a = ( 3b +2)

Squaring on both the sides.

=) a² = ( 3b + 2)²

=) a²= 9b² + 4 + 12b

=) a² = 9b² + 12b + 3 + 1

=) a² = 3( 3b² + 4b + 1) +1

Taking (a² = x) and ( 3b² + 4b +1 = m)

=) x = 3m + 1.

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

Square of any Positive Integer is of the form either 3m or 3m+1.