I attempt integrate another factor 2 in the definition of even perfect numbers I use the method display by Florian in [1] (in true both statments of this problem are due to Florian at 99%)  to compute from $\sigma(2n)-(\sigma(n)+\sigma(n))=2^p$ (where $\sigma$ is the sum of divisor function), really there are no reason to write the second summand in the left side display as this manner, compute I said  (solving a square equation)
$$2\sigma(2n)=1+4\sigma(n)+\sqrt{1+4\sigma(n)}$$
If $n$ is an even perfect number, it is easy to prove using Euler theorem that the previous holds. Well, I attempt as I said essay Florian's trick, I write $n=2^a\cdot m$, where $a\geq 0$ and $m$ is an odd number (caution, because in this part we don't assume that $n$ is perfect, in fact it is the goal of the method, prove that only integers that safisfies the previous equation are only even perfect numbers). It is succesfull that I can prove that, after of several computations, the equation implies $\sigma(m)=2^{a+1}$ (this is in fact the first part of the method, isolated $\sigma(m)$), with the condition that $1+4\cdot\sigma(n)$ is a square (really, I should want imply from these the structure of the integer $m$ it is, $m$ is the related Mersenne prime; it is the second part of the method, derive the structure of $m$ from $\sigma(m)$). But if failed because, with a few computations with my computer, $336$ (an integer that is a power of 2 weigthed, by Mersenne primes) find the previous structure. Then I, tired, I change to display the summand in the square rooot as this manner
$$2\sigma(2n)=1+4\sigma(n)+\sqrt{1+8n}$$
and  (caution, previous relation is a conjecture, I don't derived from the method). The computer, until 10000, only finds even perfect numbers. Perhaps it is possible find a reasonable relation that satisfying only in even perfect numbers (caution I suspect that it is not well of all, because the condition “$1+4\sigma(n)$ is a square” seems have more arithmetic information that the condition “$1+8n$ is a square”).
Thus, my question (I apologize to ask in a post of this Math Stack Exchange, because the computations and solutions could go from this community)

Question. Can you find a relation that integrate the factor 2 in the definition of even perfect numbers and satisfying only by even perfect numbers?

Remark. When, in the past I tried this with odd perfect numbers, my idea was use a (known) inequality between $\sigma$ and Euler's totient function, my idea was combine the equation derived (in the easy sense) for odd perfect numbers with this inequatity (see exercise 9 a) of chapter 3 from Apostol, Introduction to Analytic Theory). I belive that this second statment isn't possible to even perfect numbers. I call your attention, too to close this ocassion, about a statment [2] by George Purdy (I have not open access) that is really very very nice. I try use this formula, but I don't obtain nothing. This statment isn't related directly with to my question, but I want put this comment here, if in future seasons someone can use this formula with the problems related with perfect numbers. 
References (there is open access only to [1] in site www.fq.math.ca):
[1] H-661 On Odd Perfect Numbers p. 377-378, from The Fibonacci Quarterly, Vol. 46/47 N. 4 November 2008/2009.
[2] George Purdy, An Integral Equal to $\sigma(n)$, Problems and Solutions, Problem E 1850 [1966, 82] American Mathematical Monthly Vol. 74 N. 5 MAY 1967, p. 594-595.
 A: Now, I am trying made some statement/answer myself question concerning the following equations, the first equation $(A)$:
$$2\phi(n)\sigma(2n)+(1-n)\sigma(n)=\phi(n)(1-4n)+\phi(n)(1+4\sigma(n)),$$
and the second, this $(B)$:
$$2\phi(n)\sigma(2n)=2\phi(n)+\sigma(n)\cdot(2\phi(n)+n-1),$$
where $\phi$ is Euler's totient function ($(B)$ follows from $(A)$ when we use $\sigma(n)=2n$); I obtain this $(A)$ by elimination of the quantity $\sqrt{1+8n}$, by multiplication of $\frac{-3-\sqrt{1+8n}}{-3-\sqrt{1+8n}}$ in one of the identities that we've computed in two of my posts about even perfect numbers $n=2^{p-1}(2^p-1)$. 

Computational facts. Thus I share with this community this information, only in the way to claim that if $n$ is an even perfect number with its Euler's form, then $n$ satisfies previous $equation$, both equations. And when I run a program written in Pascal until $10^4$ we only find even perfect numbers of one of such $(A)$, or when you run your program only for equation $(B)$. When I've tried use Florian's trick (see the reference [1] of myself question) the computations are very tedious and I don't find nothing useful to claim. Feel free to say something about this or this other implication, and critic, of course because I don't say that we could be use neccesarly Euler's totient function $\phi(n)$ in the answer of this question. Feel free to share previous statements and questions with your colleagues.

Thus, the question remains still without a full answer in the way to obtain

Goal. The answer of my question should provide us a characterisation of even perfect numbers in the way of an equation that integrate another factor $2$ inside the term $\sigma(2n)$ (this was my genuine idea, copy the characterisation given in [1] for odd perfect number in an attempt to study similarities). We should to prove that the solutions of this equation are even perfect numbers and if and only if terms of this sequence of integers. I excuse my question/answer since I believe that is an exercise that could answer an user of this site. Thanks in advance to this community.

References of my previous post: 
[2] Mathematics Stack Exchange, The question of this post I attempt integrate another factor...
[3] Mathematics Stack Exchange, If $n$ satisfies $\left(-3+\sqrt{1+8n}\right)\sigma(n)=4\left(-1+\sqrt{1+8n}\right)\phi(n)$ then is an even perfect number?
A: Previous Answer
(This is not a complete solution, but only a partial answer to Juan's question.)
Let $n$ be a positive integer such that 
$$2\sigma(2n) = 1 + 4\sigma(n) + \sqrt{1 + 4\sigma(n)}.$$
Without loss of generality, suppose that
$$n = {2^a}m$$
where $a > 0$ and $\gcd(2,m) = 1$.  We want to show that $a = p - 1$ and $m = 2^p - 1$, where $p$ and $2^p - 1$ are both primes.
Substituting, we get
$$2\sigma(2^{a+1}m) = 1 + 4\sigma({2^a}m) + \sqrt{1 + 4\sigma({2^a}m)}.$$
Simplifying, we obtain:
$$2\sigma(2^{a+1})\sigma(m) = 1 + 4\sigma(2^a)\sigma(m) + \sqrt{1 + 4\sigma(2^a)\sigma(m)}$$
$$2\left(2^{a+2} - 1\right)\sigma(m) = 1 + 4\left(2^{a+1} - 1\right)\sigma(m) + \sqrt{1 + 4\sigma(2^a)\sigma(m)}$$
$$8\cdot{2^a}\sigma(m) - 2\sigma(m) = 1 + 8\cdot{2^a}\sigma(m) - 4\sigma(m) + \sqrt{1 + 4\sigma(2^a)\sigma(m)}$$
$$2\sigma(m) - 1 = \sqrt{1 + 4\sigma(2^a)\sigma(m)}$$
Squaring both sides of the last equation:
$$\left(2\sigma(m) - 1\right)^2 = 1 + 4\sigma(2^a)\sigma(m)$$
$$4{\sigma(m)}^2 - 4\sigma(m) + 1 = 1 + 4\sigma(2^a)\sigma(m)$$
$$4{\sigma(m)}^2 - 4\sigma(m) = 4\sigma(2^a)\sigma(m)$$
Now, since $\sigma(m) \geq 1$, we get
$$\sigma(m) - 1 = \sigma(2^a)$$
$$\sigma(m) = 1 + \left(2^{a + 1} - 1\right)$$
$$\sigma(m) = 2^{a+1}$$
Added January 1, 2016
Suppose $\omega(m) \geq 2$, where $\omega(x)$ is the number of distinct prime factors of $x$.  Then
$$\omega(\sigma(m)) \geq 2$$
which contradicts
$$\omega(2^{a+1}) = 1.$$
Therefore, $\omega(m) = 1$.  So either $m = q^r$ or $m = s$ for some primes $q$ and $s$.
Suppose that $m = q^r$.  Then
$$2^{a+1} = \sigma(m) = \frac{q^{r+1} - 1}{q - 1} = 1 + q + \ldots + q^r.$$
Consequently,
$$q\sigma(q^{r-1}) \equiv 0 \pmod{2^{a+1} - 1}.$$
Our required conclusions would follow if $r = 1$.  So assume to the contrary that $r > 1$.
Can you take it from here, Juan?  ^_^
Further Edit - January 1, 2016
From the second part of Juan's original question:
Let $n$ be a positive integer such that 
$$2\sigma(2n) = 1 + 4\sigma(n) + \sqrt{1 + 8n}.$$
Without loss of generality, suppose that
$$n = {2^b}l$$
where $b > 0$ and $\gcd(2,l) = 1$.  We want to show that $b = t - 1$ and $l = 2^t - 1$, where $t$ and $2^t - 1$ are both primes.
Substituting:
$$8\cdot{2^b}\sigma(l) - 2\sigma(l) = 1 + 8\cdot{2^b}\sigma(l) - 4\sigma(l) + \sqrt{1 + 2^{b+3}l}$$
Simplifying, we obtain:
$$\left(2\sigma(l) - 1\right)^2 = 1 + 2^{b+3}l$$
Consequently, we have
$${\sigma(l)}^2 - \sigma(l) = 2^{b+1}l.$$
If $l = 2^t - 1$ is a prime, then $\sigma(l) = 2^t$, so that
$$2^{2t} - 2^t = 2^{b+1}(2^t - 1)$$
which implies that
$$b + 1 = t.$$
It remains to show that prime $l = 2^t - 1$ is the only solution to
$${\sigma(l)}^2 - \sigma(l) = 2^{b+1}l.$$
