A Proof of the Odd Perfect Number Conjecture
  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. 
Simon Davis