Prove that no perfect number of the form $3^m 5^n 7^k$ exists. For all perfect numbers $N$, $\sigma (N) = 2N$, where $\sigma$ is the divisor sigma function.
Let $s$ be a perfect number of the form $3^m 5^n 7^k$, where $m,n,k \geq 1$ are integers.
Then $\sigma (s)= \sigma (3^m 5^n 7^k)$
$ =\sigma (3^m) \sigma (5^n) \sigma (7^k)$ since $3, 5,$ and $7$ are coprime to each other.
$ =\left(\frac{3^{m+1}-1}{2}\right)\left(\frac{5^{m+1}-1}{4}\right)\left(\frac{7^{k+1}-1}{6}\right)$
$ =2(3^m 5^n 7^k)$ since $s$ is a perfect number.
$\implies 9 (3^m 5^n 7^k) = 3^{m+1} 5^{n+1}+3^{m+1} 7^{k+1} + 5^{n+1} 7^{k+1} - 3^{m+1}-5^{n+1} - 7^{k+1}-1$ after some algebra.
This is as far as I got using this method. Any and all help would be appreciated.
 A: Note that the Question stipulates the exponents considered are $m,n,k\ge 1$.  As @lulu points out, the cases where one of the exponents is zero can (if desired) be ruled out by this previous Question.
The following is a simplification of the proof that an odd perfect number cannot be divisible by $105$ found here, as previously linked under the older Question to that effect.
$N= 3^m 5^n 7^k$ is a perfect number if and only $S(N)$, the sum of all divisors of $N$ (including itself and $1$), equals $2N$.
Since $N$ is odd, it must be that $S(N)=2N$ is not divisible by $4$.  Now:
$$ \frac{S(N)}{N} = \left(1+\frac{1}{3}+\ldots+\frac{1}{3^m}\right)
\left(1+\frac{1}{5}+\ldots+\frac{1}{5^n}\right) 
\left(1+\frac{1}{7}+\ldots+\frac{1}{7^k}\right) $$
Since $m=1$ would give $\left(1+\frac{1}{3}+\ldots+\frac{1}{3^m}\right)=\frac{4}{3}$ and $k=1$ would give $\left(1+\frac{1}{7}+\ldots+\frac{1}{7^k}\right)=\frac{8}{7}$, either would imply $S(N)$ is divisible by $4$, contradicting our observation above.
Knowing thus $m,k\ge 2$, we get a contradiction:
$$ \begin{align*} 2 = \frac{S(N)}{N} 
&\ge \left(1+\frac{1}{3}+\frac{1} {3^2} \right)
\left(1+\frac{1}{5}\right) 
\left(1+\frac{1}{7}+\frac{1}{7^2}\right) \\
&= \frac{13}{9} \frac{6}{5} \frac{57}{49} = \frac{4446}{2205} \gt 2 \end{align*} $$
A: In order to get a perfect number, we need the ratio $\dfrac{\sigma(n)}{n}=2$, which also means that we need the prime powers in the denominator to appear in the sigma values on top. And we need exactly and only one power of $2$ to appear in that sigma value, because the denominator is odd.
Unrolling this into prime powers, 
$$\frac{\sigma(n)}{n}=\frac{\sigma(3^m)}{3^m} \frac{\sigma(5^n)}{5^n} \frac{\sigma(7^k)}{7^k}$$
For primes in general, $p$ does not divide $\sigma(p^j)$. So we would need to match the powers of each prime from the $\sigma$ values of the other primes.
For $\sigma(3^m)$, any odd value of $m$ makes $\sigma(3^m)$ divisible by $4$, so we need $m$ even - but in that case,  $\sigma(3^m)$ is not divisible by $5$ - we actually need $m{+}1$ divisible by $4$ to get any power of $5$, which can't happen.
Similarly, for $\sigma(7^k)$, any odd value of $k$ makes $\sigma(7^k)$ divisible by $8$, so we need $k$ even - but again in that case,  $\sigma(7^k)$ is not divisible by $5$ which once again only happens if $k{+}1$ is divisible by $4$.
So a number of that form cannot have $\dfrac{\sigma(n)}{n}=2$ and thus cannot be perfect.
A: Intuitively speaking, the left hand side is generally way smaller than the right hand side because of all the exponentials, so diving by those exponentials, $3^{m+1}, 5^{n+1}, 7^{k+1}$ is a good way to exploit that behavior.
Then you basically play around to find that no values for $m, n$, and $k$ work. (I checked, but it seems that writing it up would solve it for you.)
