In the k=5 two-prime case both cyclotomic kernel primes divide m

Let $p \equiv 1 \pmod 4$ be an odd prime, let $q < r$ be primes, and suppose the $k=5$ first Dris relation holds with square-free kernel $d_1^2 q r$. Assume in addition that $q r (p^2+p+1) \frac{p+1}{2}(p^2-p+1)$ is a perfect square. Then both $q$ and $r$ divide $m$.

The accepted child five_two_prime_cyclotomic_split (81a70179) shows that the cyclotomic block $U = p^2+p+1$ is $q x^2$ or $r x^2$, and the same parity argument applied to $V$ gives $V = t y^2$ for the remaining kernel prime $t$. Substituting into the square relation $m^2 = d_1^2 U V q r$ gives $m^2 = d_1^2 q^2 r^2 x^2 y^2$, hence $m = d_1 q r x y$, so $q \mid m$ and $r \mid m$.

This is the first structural bridge from the parity split into the second Dris equation: it shows that each kernel prime must be supplied to $\sigma(m^2)$ by a different prime factor of $m$, since $\sigma(q^{2a}) \equiv 1 \pmod q$.