Few more properties

$\textbf{Hint for product of 2 odd primes}$

Let $p$ and $q$ be odd primes,

$(p-1)(q-1) > 2$

$pq > p+q+1$

Proof that a prime power can't be a perfect number:

Let $p$ be a prime number. If $p^n$ were a perfect number [ $n$ is an integer greater than $0$ ], then $1+p^2+p^3+\cdots p^{n-1}$ would be equal to $p$ .

In other words, $\dfrac{p^n-1}{p-1}=p^n$

Or, $p^n(2-p)=1$ .

But this is impossible since the absolute value of the left hand side is clearly greater than $1$ .

Proof of a Perfect Number cannot be a Perfect Square

Since we know that $\displaystyle \sum_{i=1}^k \frac{1}{d_i} =2$ for any perfect number where $d_i$ is a divisor of the number [ $k$ is the number of divisors].

We know that we can pair them in such a way that for some $m,n$ $\frac{1}{d_m} + \frac{1}{d_n} = \frac{d_m + d_n}{n}$ .

A property of perfect squares is that they always have odd number of divisors.

So, for any perfect square, we can write that $\displaystyle \sum_{i=1}^k \frac{1}{d_i} =\frac{\displaystyle \sum_{i=1}^k d_i - \sqrt{n}}{n} + \frac{1}{\sqrt{n}}$

So, if that number were a perfect number then we could say that $\displaystyle \sum_{i=1}^k \frac{1}{d_i} =\frac{2n - \sqrt{n}}{n} + \frac{1}{\sqrt{n}}$

Or, $2 = \frac{2n - \sqrt{n}}{n} + \frac{1}{\sqrt{n}}$

Or, $2n -1= 2n\sqrt{n}$

But, the L.S of the equation is odd when the R.S is even.

So, it is a contradiction and thus a perfect number cannot be a perfect square.