<s>[INST] On the equation $\varphi(n)=\left(\frac{1+\sqrt{1+8n}}{8}\right)\cdot\left(\operatorname{rad}(n)-\frac{1+\sqrt{1+8n}}{2}\right)$ An integer is said to be an even perfect number if satisifies $\sigma(n)=2n$, where $\sigma(n)$ is the sum of the positive divisors of $n$. The first few even perfect numbers are $6,28,496$ and $8128$. 

Question. I wondered if we can prove or refute the following statement:
Let $\varphi(n)$ the Euler's totient function and we denote the product of the distinct primes dividing a natural $n>1$ as 
  $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p$$ 
  (it is this arithmetic function from the Wikipedia Radical of an integer). If $n$ satisfies the equation $$\varphi(n)=\left(\frac{1+\sqrt{1+8n}}{8}\right)\cdot\left(\operatorname{rad}(n)-\frac{1+\sqrt{1+8n}}{2}\right)$$
  then $n$ is an even perfect number. 
Many thanks.

 [/INST] The converse seems to hold: If $n$ is an even perfect number, then the equation is true. 
At the end of Elements IX, Euclid shows that if $p$ and $2^p-1$ are prime, then$$n=\frac{(2^p-1)2^p}{2}$$is a perfect number.
Granting that consequently$$\phi(n)=\frac{(2^{p-1}-1)2^p}{2}=(2^{p-1}-1)2^{p-1}$$and that$$rad(n)=(2^p-1)2$$then by substitution in the given 
equation$$(2^{p-1}-1)2^{p-1}=\frac{1+\sqrt{1+8\frac{(2^p-1)2^p}{2}}}{8}\cdot [(2^p-1)2-\frac{1+\sqrt{1+8\frac{(2^p-1)2^p}{2}}}{2}]$$And since it can be shown that$$\frac{1+\sqrt{1+8\frac{(2^p-1)2^p}{2}}}{8}=2^{p-2}$$and$$\frac{1+\sqrt{1+8\frac{(2^p-1)2^p}{2}}}{2}=2^p$$then again by substitution$$(2^{p-1}-1)2^{p-1}=2^{p-2}[(2^p-1)2-2^p]$$Hence by distribution$$2^{2p-2}-2^{p-1}=2^{2p-1}-2^{p-1}-2^{2p-2}$$and eliminating and transposing$$2^{2p-2}+2^{2p-2}=2^{2p-1}$$that is$$2(2^{2p-2})=2^{2p-1}$$or$$2^{2p-1}=2^{2p-1}$$
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