  Let $\sigma(x)$ be the sum of the divisors of $x$. If $N$ is odd and
$\sigma(N) = 2N$, then the odd perfect number $N$ is said to be given in
Eulerian form if $N = {q^k}{n^2}$ where $q$ is prime with $q \equiv k \equiv 1
\pmod 4$ and $\gcd(q,n) = 1$. In this note, we show that $q < n$ implies that
Descartes's conjecture (previously Sorli's conjecture), $k = \nu_{q}(N) = 1$,
is not true. This then implies an unconditional proof for the biconditional $$k
= \nu_{q}(N) = 1 \Longleftrightarrow n < q.$$ Lastly, following a recent result
of Cohen and Sorli, we show that if $q < n$, then either $q > 5$ or $k > 5$ is
true.
  (Note: This is withdrawn for now because this paper is currently a work in
progress.)
