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

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

If 'a' be any positive integer is b=3 then,

By Euclid division lemma,

a=bq+r (a=3q+r)

So, r = 0,1,2 because (0<r≤b)

a=3q+r

Squaring on both sides,

Now consider,

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

For r=0

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

W.K.T,

____________________

(a+b)^2 = a^2 + 2ab + b^2

____________________

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

= (9q^2) + 0 + 0

= 9q^2

= 3(3q^2)

= 3m (let,m = 3q^2)

:. a^2 = 3m.

For r=1

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

W.K.T,

____________________

(a+b)^2 = a^2 + 2ab + b^2

____________________

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

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

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

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

:. a^2 = 3m+1.

For r=2

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

W.K.T,

____________________

(a+b)^2 = a^2 + 2ab + b^2

____________________

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

= (9q^2) + 12q + 4

= (9q^2) + 12q + 3 + 1

= 3(3q^2 + 4q + 1) + 1

= 3m+1 (let m=3q^2 + 4q +1)

a^2 = 3m+1.

Here,

Any positive integer is of the form 3m or 3m+1.