Question

show that the square of any positive integer is of the type 3m or 3m+1 for same positive integer m, using euclid's division lemma.​

Answer 1

Answer:

Let 'a' be any positive integer.

On dividing it by 3, let 'q' be the quotient and 'r' be the remainder.

Such that ,

a = 3q + r , where r = 0 ,1 , 2

When, r = 0

∴ a = 3q

When r = 1

∴ a = 3q + 1

When r = 2

∴ a = 3q + 2

When a = 3q

On squaring both the sides,

a² = (3q)²

a² = 9q²

a² = 3(3q²)

a² = 3m

where m = 3q²

When, a = 3q + 1

On squaring both the sides ,

a² = (3q + 1)²

a² = 9q² + 6q + 1

a² = 3(3q² + 2q) + 1

a² = 3m + 1

where m = 3q² + 2q

When, a = 3q + 2

On squaring both the sides,

a² = (3q + 2)

a² = 9q² + 12q + 4

a² = 9q² + 12q + 3 + 1

a² = 3(3q² + 4q + 1) + 1

a² = 3m + 1

where m = 3q² + 4q + 1

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