how do I prove n is perfect square iff n has odd number of positive divisors?

Question

how do I prove n is perfect square iff n has odd number of positive divisors?

3 x 3 is can be expressed as 3^2. I should say that divisors are not broken down as much as possible, they are expressed as power of prime numbers.

Answer 1

Consider the decomposition of a number n into powers of its primes:

n = 2^a 3^b 5^c 7^d ....

This means that we have the total number of divisors as follows:

(a+1)(b+1)(c+1)(d+1).....

because "0" is a choice as well as, say, from 1 to a. Now, if this total is odd, that means all of the factors like (a+1) are also odd. Therefore, all the powers a, b, c, d... are even, and the theorem is proven.

Answer 2

A perfect square can't have an odd number of divisors if it is broken down as far as possible. Have you left in a number like 9 which could have been further broken down into 3 x 3. If it is algebraic then one of your divisors could be equal to 1. This is not counted in prime factorisation.

Answer 3

let t(n) denote the number of positive divisors of n. it will suffice toshow that the product of the positive divisors of an integer n>1 = n^(t(n)/2) = P ... when t(n) is even then p is clearly an integer but when it is odd we have a problem .... let n= m^2 and p becomes n^(t(n)/2)=m^t(n) settling the problem Q.E.D.


i have excuded the details of finding p as you only asked for a hint in the right direction but it is very easy to show with very elementary steps at the disposal of anyone with at least a middleschool algebra background ... hope this helps

Answer 4

n/kjykjb,jguoi=divisor of n
UTIUHFGJHU iyo8ighk,ghoi giugjgkjiu0i;lb