As usual $\sigma\_k(n)$ denotes the sum of the $k$-th powers of the positive divisors of an integer $n.$
Note that $k$ is also an integer so that it may be negative.


There are no odd perfect numbers known but there are many numbers $N$ such that
$$
\sigma(N) \equiv 2 \pmod{4}
$$
i.e.; many numbers
of the form:


$$
N = p^{4k+1}m^2
$$
with $\gcd(p,m)=1$ and $p$ a prime number with $p \equiv 1 \pmod{4}.$


Since
$$
2\sqrt{2}
$$
equals the minimum of the $2z+1/z$ when $0 < z \leq 1,$
attained for
$$
z=z\_0 = \frac{\sqrt{2}}{2},
$$


by putting


$$
z =\frac{p^{4k+1}}{\sigma(p^{4k+1})}
$$


It may have some interest the 


Question: For which numbers $k$ and prime numbers $p \equiv 1 \pmod{4}$
the $z$ above is `close` to $z\_0$, say appears as a convergent in the continued fraction of $z\_0.$


question inspired by some MO posts of Arnie Dris:


a) In his notation I am asking here when the sum below is `minimal`


$$
I(p^{4k+1}) + I(m^2) = \sigma\_{-1} (p^{4k+1}) + \sigma\_{-1}(m^2)
$$


b) Arnie wanted $2N=\sigma(N)$ so that in his case


$$
I(p^{4k+1}) = 1/z
$$


and


$$
I(m^2)= 2z
$$


so that


$$
I(p^{4k+1})+I(m^2)=2z+1/z
$$



Solutions of a exponential diophantine equation involving the $\sigma$ function
Note that
$$z = \frac{p^{4k+1}}{\sigma(p^{4k+1})} = \frac{1-p^{-1}}{1-p^{-(4k+2)}} > 1-p^{-1}.$$
Clearly, for a fixed p, the larger is $k$ the closer is $z$ to this lower bound.


Since the lower bound is greater than $z\_0$ (for any prime $p\geq 5$) and grows with $p$, it is beneficial to take $p$ as small as possible. The smallest such prime $p\equiv 1\pmod{4}$ is $p=5$.


Therefore, for any prime $p\equiv 1\pmod{4}$ and $k\geq 0$, we have
$$z - z\_0 = \frac{p^{4k+1}}{\sigma(p^{4k+1})} - \frac{\sqrt{2}}{2} > \frac{4}{5} - \frac{\sqrt{2}}{2} > 0.09$$
where
$$\frac{4}{5} = \lim\_{k\to\infty} \frac{5^{4k+1}}{\sigma(5^{4k+1})}.$$


