Question

use euclid's division Lemma to show that any positive integer is of the form 3p,3p+1.

Answer 1

I am taken 3m Nd 3m+1 in place of 3p and 3p+1

I hope u will understand

Let N be a given integer and r denotes remainder when N is divided by 3.
Thus N=3a+r,r=0,1,2
N=3a,3a+1,3a+2
N^2=(3a)^2=9a^2=3(3a^2)=3m,where 3a^2=m
N^2=(3a+1)^2=9a^2+6a+1
=3(3a^2+2a)+1
=3m+1,where m=3a^2+2a
N^2=(3a+2)^2=9a^2+12a+4
=3(3a^2+4a+1)+1
=3m+1,where m=3a^2+4a+1
Hence we conclude N^2 is either of the form 3m or 3m+1,where m is an integer.
Hope this makes easy to understand..........

Answer 2

$\huge\sf{\underline{\underline{Solution:}}}$

Using euclid's division lemma

a = bq + r

Let a be any positive integer.

Here, b = 3 and r = 0, 1, 2 (0 ≤ r ≤ 3)

Now, take b = 3 and r = 0

 a = 3q + 0

a = 3q

Squaring both sides

a² = 9q²

3(3q²)

Put 3q² = p

 3p

Similarly, take b = 3 and r = 1

 a = 3q + 1

Squaring both sides

a² = (3q + 1)²

9q² + 6q + 1

3(3q² + 2q) + 1

Put 3q² + 2q = p

3p + 1

Now, take b = 3 and r = 2

 a = 3q + 2

Squaring both sides

a² = (3q + 2)²

9q² + 6q + 4

9q² + 6q + 3 + 1

3(3q² + 2q + 1) + 1

Put 3q² + 2q + 1 = p

3p + 1

$\therefore$Squaring of positive odd integer is of the form 3p, 3p+1.