The square-free part of the k=5 Dris index has at least two distinct prime factors

Let $N = n^2 q^k$ be an odd perfect number with $k = 5$, and let $s$ be its Dris index, so $2m^2 = \sigma(p^5)\, s$ for the Euler prime $p$ and the cofactor $m$. Suppose the square-free part of $s$ is written as $s = d_1^2 d_2$ with $d_2$ square-free and $s$ is not itself a square. Then the square-free part has at least two distinct prime factors: $\#\mathrm{primeFactors}(d_2) \ge 2$.

The argument is short once the two exclusions are in place. If $d_2 = 1$ then $s = d_1^2$ is a square, contradicting the hypothesis; that is the accepted child OddPerfectNumber.Kernel.sqfree_part_ne_one (b59073fb). If $d_2$ is a prime $q$ then $s = d_1^2 \cdot q$ is exactly a prime times a square, which the accepted child OddPerfectNumber.Kernel.five_index_not_prime_mul_square (2ec27119) forbids, since it uses only the first Dris equation together with the factorisation $\sigma(p^5) = 2(p^2+p+1)\frac{p+1}{2}(p^2-p+1)$ and the coprimality and non-square properties of the two cyclotomic blocks. The accepted generic lemma OddPerfectNumber.Kernel.squarefree_card_ge_two_of_not_one_or_prime (9c90129e) then converts the two exclusions into the bound of two.

This is the $\ge 2$ half of the target $\omega(\mathrm{squarefree\_part}(s)) \ge 3$ for the $k=5$ Dris branch of OddPerfectNumber.no_dris_five_s_odd_ge_five_nonsq. The remaining half is the genuine research residual: excluding $d_2 = q r$ for two distinct primes, which the two-prime case genuinely satisfies as far as the first Dris equation alone is concerned (at $p = 5$, $s = 217 = 7\cdot 31$, $m = 651$ one has exactly $2\cdot 651^2 = 3906 \cdot 217$), so the second Dris equation $\sigma(m^2) = p^5 s$ is indispensable there.