The point to note is if n^2 is an integer it does not mean n is always an integer. It may or may not be

n = sqrt(2) n^2->INT n->not an INT

n=2 n^2->integer n->integer


For sqrt(n) to be an integer n has to be an integer

jeevan.Gk wrote:Is n an integer ?

1. n^2 is an integer.
2. Sqrt (n) is an integer

once you feel uncomfortable with the concepts try this method

use numbers and check with the question stem

1. n^2 is an integer.

you know n^2 is always positive so choose the first 5 positive numbers

1 - Not integer when you take its root
2 - Not integer when you take its root
3 - Integer when you take its root YOU CAN STOP HERE since you have a YES and NO , which means insuf


4
5

2. Sqrt (n) is an integer

so choose the first five squares

1 4 9 16 and 25 and all of them will give you an integer when you find their root.

SO SUF