i would go with picking numbers here.
Since N is a perfect square :
Say N=25

I - The number of distinct factors is odd
25 = 1 * 5 * 5

Disticnt factors : 1,5,25 - odd
this works for any interger thats a perfect square.
Hence,True

II - The sum of the factors is odd

1 will always be a factor,so if the other factors are all even or all odd,when u add 1 the sun will become odd.

III -
If u pick 4
4 = 2*2*1
Disntict primt factors is just 1 i.e 2.
Not always true.

Hence I and II

You have to be a bit more careful.

Both of you have assumed that N is the square of a prime number, which isn't stated in the question.

So, N could have a lot more than 5 factors. For example, there's no reason why we couldn't pick N=36, which we can write as:

(2*3)(2*3)

and which has the factors:

1, 2, 3, 2*2, 2*3, 3*3, 2*2*3, 2*3*3 and 2*2*3*3

for a total of 9 distinct factors.

Here's the important thing to recognize: a perfect square is made up of pairs of primes.

The general formula for counting the number of factors is:

Total # of factors = (a+1)(b+1)(c+1)...

in which a, b and c represent the number of each prime factor that make up the number.

For example, if our number is 25:

25 = 5^2

so a=2 and 25 has (2+1) = 3 distinct factors

36 = 2^2 * 3^2

so a=2 and b=2 and 36 has (2+1)(2+1) = 9 distinct factors

Since all perfect squares are made up of pairs of primes, the values of a, b, c, etc are always going to be even (we could certainly have more than just 2 of a prime, but we'll have pairs of pairs, giving us an even total), which means that the number in each bracket is going to be odd. So, since we're multiplying together a string of odd numbers, we'll always end up with an odd number of primes.

Of course, we don't actually need to know all of that to answer the question. Picking numbers is a great approach to take, but we need to pick more than just one number, especially if the number we're picking is "special", e.g. it's the square of a prime.