show that a power of 3 can never be a perfect number
Attempt:
First I asked myself what does a number to a power of 3 look like: so i said let $n^{3^k}$
Next, what would a perfect number of this form have to look like. Well if this was to be a perfect number it should look like: $$n^{3^k} = 1 + n^{0^1} + n^{2^1} + n^{3^1} +…+ n^{3^2} + n^{3^3} +…+ n^{3^{k-1}}$$ In this form all of the possible divisors of $n^{3^k}$ are captured.
When i wrote it I thought it was a geometric sequence so with some fiddling I obtained $$\frac{(n^{3})^k – 1}{n^3 – 1}$$ Looking at the expression you see that $n^3 – 1$ actually is a divisor of $(n^{3})^k – 1$. So it can be expressed as $$ (n^{3})^k – 1 = q(n^3 – 1)$$ Which then would mean $$n^{3k} = q(n^3 – 1) + 1$$ But this would mean that $n^{3^k}$ always has a remainder, so cannot be a perfect number.
notes: 1) i am concerned that the $n^{3^k}$ in my original expression and the $n^{3k}$ obtained at the end cannot be manipulated in the way I did it eventhough I used the properties of exponents, so my solution is off.
2) the solution i have seen in the book states through a string of inequalities : $$ \sigma(p^k) = \frac{p^{k+1} – 1}{p – 1} < \frac{p^{k+1}}{p – 1} = (\frac{p}{p – 1})p^k$$ where the quantity $(\frac{p}{p – 1})$ is less than 2 for any number $p > 2$, where $\sigma(n)$ = {sum of all divisors of n}
My issue with the second note is that they assumed that the quantity $\sigma(p^k)$ can never be as large as $2p^k$. but to be able to assume that you must be assuming that every perfect number is even, which although the existence of odd perfect numbers has not been established, it hasn't been disproven either. So how can this be used as an assumption?
Best Answer
In case you meant "a power of $3$":
The sum of the factors of $3^n$ is $\frac{3^{n+1}-1}{3-1}$. For $3^n$ to be perfect, we would need $$ \frac{3^{n+1}-1}{3-1}=2\cdot3^n\implies-3^n-1=0 $$