Question

A number can be written in the form 3m + 2, for some natural numbers m. Can this number be a perfect square?​

Answer 1

$\bf\purple{QuestioN:-}$

A number can be written in the form 3m + 2, for some natural numbers m. Can this number be a perfect square?​

$\bf\green{AnsweR:-}$

$\Large{\mathfrak{\pink{No!!}}}$

Let a be any positive integer.

Then by Euclid’s division lemma,

We have a = bq + r, where 0 ≤ r < b

For b = 3, we have

a = 3q + r, where 0 ≤ r < 3 ---------( eq i)

So, The numbers are of the form 3q, 3q + 1 and 3q + 2.

So,

$\implies\bf{(3q)^2=9q^2=3(3q^2)}$

3m, where m is a integer

$\implies\bf{(3q+1)^2=9q^2+6q+1}$

$\implies\bf{3(3q^2+2q)+1}$

$\implies\bf{3m+1}$ , where m is a integer.

$\implies\bf{(3q+2)^2=9q^2+12q+4}$

Which cannot be expressed in the form 3m + 2.

•°• Square of any positive integer cannot be expressed in the form 3m + 2.

                                                                                                                                                                       

                                                                                               

                                       

Happy Learning!♥