  It is sufficient to prove that there is an excess of prime factors in the
product of repunits with odd prime bases defined by the sum of divisors of the
integer $N=(4k+1)^{4m+1}\prod_{i=1}^\ell ~ q_i^{2\alpha_i}$ to establish that
there do not exist any odd integers with equality between $\sigma(N)$ and 2N.
The existence of distinct prime divisors in the repunits in $\sigma(N)$ follows
from a theorem on the primitive divisors of the Lucas sequences
$U_{2\alpha_i+1}(q_i+1,q_i)$ and $U_{2\alpha_j+1}(q_j+1,q_j)$ with
$q_i,q_j,2\alpha_i+1,2\alpha_j+1$ being odd primes. The occurrence of new prime
divisors in each quotient ${{(4k+1)^{4m+2}-1}\over {4k}}$,
${{q_i^{2\alpha_i+1}-1}\over {q_i-1}}, i=1,...,\ell$ also implies that the
square root of the product of $2(4k+1)$ and the sequence of repunits will not
be rational unless the primes are matched. Although a finite set of solutions
to the rationality condition for the existence of odd perfect numbers is
obtained, it is verified that they all satisfy ${{\sigma(N)}\over N}\ne 2$
because the repunits in the product representing $\sigma(N)$ introduce new
prime divisors. Minimization of the number of prime divisors in $\sigma(N)$
leads to an infinite set of repunits of increasing mangitude or prime equations
with no integer solutions. It is proven then that there exist no odd perfect
numbers.
