Special exponent $k=1$: $1098\,p < m^{2}$

Let $N = p\,m^{2}$ be an odd perfect number in Euler form with special exponent $k = 1$ (the configuration predicted by the Descartes–Frenicle–Sorli conjecture): $p$ is prime, $m$ is odd, $p \nmid m$, and the Dris relations

hold, with $s = \sigma(m^{2})/p$ the Dris index. Then the special prime is small compared with the square part:

Trivially $\sigma(p) = p+1$ divides $2m^{2}$, which alone gives only $p < 2m^{2}$. The point of the statement is the explicit constant: it comes from a lower bound $s \ge 13^{3} = 2197$ for the Dris index at $k = 1$, since $2m^{2} = (p+1)s$. That lower bound in turn is a counting statement: at $k = 1$ the divisor sum $\sigma(m^{2}) = p\,s$ has $p$-adic valuation $1$, so at most one of the local divisor sums $\sigma(q^{2v_q(m)})$, $q \mid m$, is divisible by $p$, and each of the others is a factor of $s$ of size at least $1 + q + q^{2} \ge 13$. Sylvester's bound $\omega(N) \ge 5$ leaves at least three such factors.

The result is unconditional given the two Dris relations: no congruence conditions on $p$, $m$ or $s$ are assumed.