Question

Write whether the square of any positive integer can be of the form 3m +2, where
m is a natural number. Justify your answer​

Answer 1

No, the square of any positive integer can not be written in the form 3m + 2 where n is natural number

Step-by-step explanation:

Since, a positive integer 'a' can be written in the form of bq + r

That is, a = bq + r

Where b, q and r are any integers,

For b = 3

a = 3 q + r

Where, r can be an integers,

For r = 0, 1 , 2, 3 ..........

3q + 0, 3q + 1, 3q + 2, 3q + 3....... are positive integers,

Since square of 3 q = 9q² = 3(3q²) = 3m ( where 3q² = m)

Square of (3q+1) = (3q+1)² = 9q²+1+6q = 3(3q²+2q)+1 = 3m + 1 ( Where, m = 3q²+2q)

Square of (3q+2) = (3q+2)² = 9q²+4+12q = 3(3q²+4q)+4 = 3m + 4 ( Where, m = 3q²+2q)

Square of (3q+3) = (3q+3)² = 9q²+9+18q = 3(3q²+6q)+9 = 3m + 9 ( Where, m = 3q²+2q)

.............. So on....

Thus, there is not any square of a positive integer that can be written in the form of 3m + 2

Hence, proved.

Answer 2

${\huge {\underline {\tt {\blue {Answer}}}}}$

No, the square of any positive integer can not be written in the form 3m + 2 where n is natural number

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${\huge {\underline {\tt {Explanation:-}}}}$

Since, a positive integer 'a' can be written in the form of bq + r

That is, a = bq + r

Where b, q and r are any integers,

For b = 3

a = 3 q + r

Where, r can be an integers,

For r = 0, 1 , 2, 3 ..........

3q + 0, 3q + 1, 3q + 2, 3q + 3....... are positive integers,

Since square of 3 q = 9q² = 3(3q²) = 3m ( where 3q² = m)

Square of (3q+1) = (3q+1)² = 9q²+1+6q = 3(3q²+2q)+1 = 3m + 1 ( Where, m = 3q²+2q)

Square of (3q+2) = (3q+2)² = 9q²+4+12q = 3(3q²+4q)+4 = 3m + 4 ( Where, m = 3q²+2q)

Square of (3q+3) = (3q+3)² = 9q²+9+18q = 3(3q²+6q)+9 = 3m + 9 ( Where, m = 3q²+2q)

Thus, there is not any square of a positive integer that can be written in the form of 3m + 2

Hence, proved.

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