square root of p2 is an integer, which of the following must

guerrero wrote:If the square root of p2 is an integer, which of the following must be true?

I. p2 has an odd number of factors

II. p2 can be expressed as the product of an even number of prime factors

III. p has an even number of factors

I
II
III
I and II
II and III

I am having trouble understanding the question . Kindly assist .

thanks !

If square root of P^2 is an integer => P is an integer

So P^2 can be represented by numbers such as 1^2, 2^2, 3^2, 4^2, 6^2, 10^2, 10^4, etc.

Always remember that a perfect square always has odd number of positive or negative factors. Why?

Because if you express a perfect suare, N as P1^(2a) * P1^(2b) * P1^(2c)... the number of positive factors of N are (2a+1)(2b+1)(2c+1)..., which is basically a product of all odds and hence the number of positive factors is odd.

The question is wrongly constructed in that statement 1 should mention number negative/positive factors and not just factors, which by default means all the factors and these happen to be even in number always.

This is a common mistake and absolutely tailor made trap for DS problems. It becomes imperative that you use the right sources to solve problems.

If statement 1 mentioned odd number of positive (or negative) factors, it would have been correct.

Statement 2 is not necessarily always true. If p^2 =1, we may not say so.

Statement 3 will not hold always. For example if P^2=10^4, p=10^2=2^2 * 5^2=> number of positive (or negative) factors 9, which is odd in number. Again the issue with construction of choice lingers around the fact that the number of factors is by default taken as total number of factors and for all natural numbers that happens to be even because basically it is the sum of total positive and total negative factors. So use the right sources for solving your problems.

An integer squared will always give you an odd number of factors as The Iceman said. You can calculate the number of factors by looking at the exponents of the prime factorization of a number.

3^4 * 5^2 has 15 factors. Just add 1 to each exponent and multiply the result. (4+1)*(2+1) = 15

So, with any number squared you will have all even exponents. When you add the one they become odd and the product of all odd numbers is odd.

Just another way. Hope it helps.