The second cyclotomic block absorbs one of the two kernel primes

This is the mirror of the proved theorem five_two_prime_cyclotomic_split, for the second cyclotomic block instead of the first.

In the $k=5$ Dris branch one writes $N = m^2 q^\alpha$ with $q \equiv 1 \pmod 4$ and $\alpha = 5$, so $\sigma(p^5) = (p+1)(p^2+p+1)(p^2-p+1)$ factors into the two cyclotomic blocks $C = p^2+p+1$ and $D = p^2-p+1$ together with the factor $p+1$.

Suppose the square-free part of the Dris index $s$ has exactly two distinct prime factors $q < r$, so $s = d_1^2 q r$ for some $d_1$, and suppose the first Dris equation $2 m^2 = \sigma(p^5) s$ holds. Then

Because $m^2$ is a square, every prime must occur to an even multiplicity in the right-hand side. The block $C$ and the block $D$ are coprime, and each is coprime to $(p+1)/2$ except for a single shared factor $3$ between $(p+1)/2$ and $D$. That single overlap is the only place where the two supports can meet, and it can contribute at most one factor of $3$.

Consequently the odd-multiplicity prime support of $D = p^2-p+1$ is a subset of $\{q, r\}$: any prime dividing $D$ to an odd power must be cancelled by the single occurrence of $q$ or $r$, since every other factor on the right is either a square or coprime to $D$. As $D$ is not a square, that support is non-empty, so $D = q x^2$ or $D = r x^2$ for some $x$.

Consequence. This supplies the missing half of the $3$-adic argument. The proved theorem five_cyclotomic_primes_one_mod_three forces both kernel primes to be $1 \pmod 3$, so in particular neither is $3$. For $p \equiv 1 \pmod 3$ the block $C$ carries exactly one factor of $3$ (the proved child three_val_first_cyclotomic), while for $p \equiv 2 \pmod 3$ it is the block $D$ that does, with $v_3(D) = 1$ (the child three_val_second_cyclotomic). Once this split is available the same Euclid argument that yields $q = 3 \lor r = 3$ from three_kernel_prime_when_p_one_mod_three applies to the second block as well, removing every $p \equiv 2 \pmod 3$ with neither kernel prime equal to $3$ from the residual five_no_two_prime_squarefree_index.