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 interger m.

Answer 1

Let the positive integer be of the form n=3q or n=3q+1
(3q)^2= 9q^2
(3q+1)^2= 9q^2+6q+1
=3(3q+2q) +1
Where m=3q+2q

Answer 2

Heyaa..

According to Euclid's Division Lemma,

$a = bq + r , \ 0 \leq r< b$ 

So when comparing to the question, we get, b = 3 and q = m...

Now applying the values of b and q in lemma, we get

$a=3m+r, \ 0 \leq r<3$  -----------------eqn 1

From this, we get ' r ' values as,

r = 0, 1, 2

So by taking square of these integers,

# $0 ^{2} = 0$

# $1 ^{2} = 1$

# $2 ^{2} = 4$

Now, applying the values of r in eqn 1,

=> $a=3m+0 \ = 3m$

=> $a=3m+1$

=> $a=3m+4$  ( But it is not possible in this case since it does not oby the rule of 0$\leq$ r < b ) ( Here b > r )

So, we can get that the square of every positive integers is of the form,

=> $a=3m$

=> $a=3m+1$

Hope it helps...

And here CutieBarbie :)