How prove this number equation $\sigma{(n)}=2n+1$ has no solution let $\sigma{(n)}$ is sums of the divisor function,
show that
$$\sigma{(n)}=2n+1$$ has no solution? Thank you 
My try:I know this: http://en.wikipedia.org/wiki/Divisor_function
 A: Not a complete solution here, but at least we can reduce the number of possibilities.
If $n$ were a power of $2$, the result is obviously false as $\sigma(n) \leq n+n/2+n/4+\ldots = 2n$. So assume $n$ has at least one odd prime $q$ such that $q|n$.
Note that we have 
$$ \sigma(n) = \prod_{p|n} {\frac{p^{v_p(n)+1}-1}{p-1} } $$
where $v_p(n)$ is the greatest power of the prime $p$ that divides $n$.
We will look at the parity. When it comes to parity, the factor where $p=2$ (if it appears) is not important. We then have $\sigma(n)\equiv \prod_{p|n ; p\neq 2} (v_p(n)+1) (\bmod. 2)$. Simply because $\frac{p^{v_p(n)+1}-1}{p-1} = 1 + \ldots + p^{v_p(n)}$ which has $v_p(n)+1$ odd terms when the prime $p$ is odd.
So if $\sigma(n)$ were even, we know there is already no solution, as $2n+1$ is odd. Hence $v_p(n)$ has to be even for all odd primes $p|n$.
Fact 1 Hence $ n = 2^k \cdot m^2$ where $m$ is an odd number. 
Thus we have $$2^{k+1} m^2 + 1 = 2n+1 = \sigma(n) = \sigma(2^k)\sigma(m^2) = (2^{k+1}-1) \sigma(m^2)$$
this is nonsense if $k$ were odd, because, by looking at the equation in module $3$ we get:  $m^2 + 1 \equiv 0(\bmod. 3)$, and $-1$ is not a quadratic residue module 3. Thus $k$ is even, and $n$ is a square.
Fact 2 $n$ is a perfect square.
