In the literature (see for example sites and paper concerning to the abc conjecture, I say this as reference and by caution to avoid mistakes) is defined the arithmetical function $rad(n)$ as $1$ if $n=1$ and $rad(n)=\prod_{p|n}p$ the product of distinct primes dividing $n$. For example $$rad(18)=rad(2\cdot 3^2)=2\cdot 3=6,$$ and yes, is multiplicative.
We have the following
Proposition. If $N$ is an even perfect number then $$\frac{2+rad(N)}{8}\left(\sum_{\substack{d|N,d<rad(N)}}d\right)=N.$$
Proof. By Euler's Theorem for even perfect numbers, $$\sum_{\substack{d|N,d<rad(N)}}d=1+\cdots+2^{p-1}+(2^p-1)=2(2^p-1),$$ and $$\frac{2+rad(N)}{8}=\frac{2+2(2^p-1)}{2}=\frac{2^{p+1}}{8}=2^{p-2}.$$ And this is the end of proof (if there aren't mistakes seems short and cool).
My question is
Question. What about the reciprocal statement? Can you find a counterexample?
I've made computations until $10^3$ and only find (with a computer) even perfect numbers. I sure that I've made a program in Lazarus Free Pascal, but my algorithm is bad, I tried run my program until $10^4$ but don't runs. I don't implement the parity when I divide by $8$. Only I write the condition $M$ such that $$\frac{2+rad(M)}{8}\left(\sum_{\substack{d|M,d<rad(M)}}d\right)=M$$ in my algorithm.
Please feel free to made feedback, suggestions, hints or critics are welcome. Thanks in advance.
Update to delete a previous code, I was using this post as draf. I'm sorry.
References:
Wikipedia or Mathworld pages about Perfect Numbers.