Question

Show that the square of any positive integer cannot be of the form 3m+2, where m is a natural number. ​

Answer 1

$\large\underline{\sf{Solution-}}$

Let x be any positive integer such that x = 3q + r, where 'r' can assume the values 0, 1 and 2.

So, 3 cases arises.

When x = 3q or 3q + 1 or 3q + 2

Case :- 1

When x = 3q

So,

$\rm :\longmapsto\: {x}^{2} = {(3q)}^{2} = 9 {q}^{2} = 3( {3q}^{2}) = 3m$

$\sf \: where \: m \: = \: {3q}^{2}$

$\rm :\implies\:square \: of \: positive \: integer \: is \: of \: the \: form \: 3m.$

Case :- 2

When x = 3q + 1

So,

$\rm :\longmapsto\: {x}^{2} = {(3q + 1)}^{2}$

$\: \: \: \: = \: \: \rm \: {(3q)}^{2} + {(1)}^{2} + 2 \times 3q \times 1$

$\: \: \: \: = \: \rm \: \: {9q}^{2} + 1 + 6q$

$\: \: \: \: = \: \rm \: \: 3( {3q}^{2} + 2q) + 1$

$\: \: \: \: = \: \rm \: \: 3m + 1 \: \: where \: m = {3q}^{2} + 2q$

$\rm :\implies\:square \: of \: positive \: integer \: is \: of \: the \: form \: 3m + 1$

Case :- 3

When x = 3q + 2

So,

$\rm :\longmapsto\: {x}^{2} = {(3q + 2)}^{2}$

$\: \: \: \: = \: \rm \: \: {(3q)}^{2} + {(2)}^{2} + 2 \times 3q \times 2$

$\: \: \: \: = \: \rm \: \: {9q}^{2} + 4 + 12q$

$\: \: \: \: = \: \rm \: \: {9q}^{2} + 3 + 12q + 1$

$\: \: \: \: = \: \rm \: \: 3({3q}^{2} + 1 + 4q ) \: + \: 1$

$\: \: \: \: = \: \rm \: \: 3m + 1 \: \: where \: m \: = {3q}^{2} + 1 + 4q$

$\rm :\implies\:square \: of \: positive \: integer \: is \: of \: the \: form \: 3m + 1$

So,

these 3 cases implies,

• Square of any positive integer is of the form of either 3m or 3m + 1.

So, it implies,

• square of any positive integer is not of the form 3m + 2.

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

Fundamental Theorem of Arithmetic :-

• This theorem states that Every composite number can be factorized in the form of their prime factors and this method is unique irrespective of their places.

Euclid Division Lemma :-

• This Lemma states that if a and b are positive integers such that a > b, then there exist unique integers q and r such that a = bq + r where r can assume the values r = 0, 1, 2, _ _ _, b - 1.