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.
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 $${\sigma(l)}^2 - \sigma(l) = 2^{b+1}l.$$