Question

Using Euclid's division lemma, show that the square of any positive integer is either of the form 3m or (3m+1) for some integer m.

Answer 1

this is the answer for your question

Answer 2

Answer:

Any integer p can be of three forms - (3q), (3q+1), (3q+2)

Case I(p=3q)

p=3q

$p^{2} = (3q)^{2}$

$p^{2} = 9q^{2}$

$p^{2} = 3$ × $3q^{2}$

$p^{2} = 3m$ (where m = $3q^{2}$)

Case II(p=3q+1)

p=3q+1

$p^{2} =(3q+1)^{2}$

$p^{2} = 9q^{2} + 6q +1$

$p^{2} =3(3q^{2} + 2q ) +1$

$p^{2} = 3m + 1$ (where m =$3q^{2} + 2q$)

Case III(p=3q+2)

p=3q+2

$p^{2} =(3q+2)^{2}$

$p^{2} = 9q^{2} + 12q + 4$

$p^{2} = 3 (3q^{2} + 4q + 1) + 1$

$p^{2} = 3m + 1$ (Where m =$3q^{2} + 4q +1$)