A conjecture concerning the number of divisors and the sum of divisors. I stumbled upon the following conjecture:

Let $n$ be a positive integer. Let $\sigma\left(n\right)$ be the sum of all (positive) divisors of $n$, and let $\tau\left(n\right)$ be the number of these divisors.
  Then,
  $$\tau(n)+\sigma(n)\equiv 1 \mod 2 \iff n = 2m^2 \text{ for some integer } m .$$

Does anybody have an idea on how to prove this, maybe in parts? Or maybe someone has a counterexample?
 A: In fact, this conjecture holds as we will show. 
Proof: Let $n = \prod_{i=1}^n p_i^{e_i}$, then
$$\tau(n) = \prod_{i=1}^n (e_i +1)$$
and
$$\sigma(n) = \prod_{i=1}^n \Big( \sum_{j=0}^{e_i} p_i^j \Big).$$
For $p_i >2$, we see that $(e_i +1)$ is even if and only if $ \sum_{j=0}^{e_i} p_i^j$ is even (sum of $n$ odd numbers is even if and only if $n$ is even). In other words
$$(e_i+1) \equiv \sum_{j=0}^{e_i} p_i^j \quad \mathrm{mod} \ 2.$$
Thus, if only one $e_i$ of the $p_i >2$ is odd, we get already that
$$\tau(n) + \sigma(n)  \equiv 0 \quad \mathrm{mod} \, 2.$$
Assuming that 
$$\tag{1} \tau(n) + \sigma(n) \equiv 1 \quad \mathrm{mod} \ 2$$
we see that all $e_i$ are even if $p_i >2$. If $p_1=2$, then (1) reduces to
$$(e_1+1) + 2^{e_1+1} -1 \equiv 1 \quad \mathrm{mod} \ 2.$$
Thus $e_1 \equiv 1 \mod 2$. Writing $e_1 = 2 k_1 +1 $ and $e_i = 2k_i$, we get
$$n = 2 m^2,$$
where
$$m = \prod_{i=1}^n p_i^{k_i}.$$
This calulcations shows also that for $n = 2m^2$ we have $\tau(n)+\sigma(n) \equiv 1 \ \mathrm{mod} \ 2$.
A: In a more descriptive style of argument:
The parity of $\tau(n)$, the number of divisors of $n$, depends on whether the divisors are composed entirely of distinct factor pairs $(d_1,d_2)$ or whether there is a square root $n=d_s\cdot d_s$ to make the count odd.
$\tau(n)$ is odd $\iff n$ is a perfect square.
The parity of $\sigma(n)$, the sum of divisors of $n$, depends on the count of its odd factors. We can thus ignore all even factors of $n$, so focus on the odd number $\ell$ defined by $n=2^k\ell$. Then as before if $\ell$ has only distinct factor pairs $(d_1,d_2)$ the sum will be even, and only if $\ell$ has a square root $d_s$ will the sum be odd.
$\sigma(n)$ is odd $\iff \ell$, the largest odd factor of $n$, is a perfect square.
So for $\tau(n)+\sigma(n)$ to be odd, we cannot have $n$ a perfect square - that would make both $\tau(n)$ and $\sigma(n)$ odd, so the sum even. So we need $\ell$ square (for $\sigma(n)$ odd) and $2^k$ not square (for $\tau(n)$ even), which makes $2^{k-1}$ square. Thus we can set $m=\sqrt{2^{k-1}\ell\ \ }$ giving $n=2m^2$.
By the same calculations we can see that $n=2m^2$ will give $\tau(n)+\sigma(n)$ odd.
