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?
How to prove that $3^i5^j$ can never be a perfect number?
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number-theory
2 Answers
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If $p$ is a prime and $p^k$ is a perfect number for some $k \in N$, then $\sigma (p^k)=2.p^k$. Also, $\sigma (p^k)=\frac {p^{k+1}-1}{p-1}$.
Therefore, for $3^i.5^j$ to be a perfect number, the following,
$\frac{3^{i+1}-1}{2}.\frac{5^{j+1}-1}{4}=2.3^i.5^j$ should be true.
$3^{i+1}.5^{j+1}-3^{i+1}-5^{j+1}+1=16.3^i.5^j$
$-3^i.5^j=3^{i+1}+5^{j+1}-1$, which is clearly false.
Hence, $3^i.5^j$ is not a perfect number.
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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.