How to prove that $3^i5^j$ can never be a perfect number?

Question

I need to prove that $3^i5^j$ can never be a perfect number. How do I tackle this problem. I know that a perfect number is a positive number which is equal to the sum of its positive divisors excluding the number itself. But how do I prove that the number $3^i5^j$ is not perfect?

Answer 1

If $p$ is a prime and $p^k$ is a perfect number for some $k \in N$, then $\sigma (p^k)=2p^k$.

Also, $\sigma (p^k)=\frac {p^{k+1}-1}{p-1}$.

Therefore, for $3^i5^j$ to be a perfect number, the following,

$(\frac{3^{i+1}-1}{2})(\frac{5^{j+1}-1}{4})=2*3^i5^j$ should be true.

($3^{i+1})(5^{j+1})-3^{i+1}-5^{j+1}+1=16*3^i5^j$

$-3^i5^j=3^{i+1}+5^{j+1}-1$, which is clearly false.

Hence, $3^i5^j$ is not a perfect number.

Answer 2

Hint #1: For $p$ a prime number, you have $$\dfrac{\sigma(p^x)}{p^x} = \dfrac{p^{x+1} - 1}{{p^x}(p - 1)} < \dfrac{p^{x+1}}{{p^x}(p - 1)} = \dfrac{p}{p - 1},$$ where $\sigma(y)$ is the sum of the divisors of $y$.

Hint #2: The $\sigma$ function is multiplicative.