If the second cyclotomic block is a prime times a square, its first factor is a square or three times a square

Let $p$ be a prime with $p \equiv 1 \pmod 4$ and put $A = (p+1)/2$, $B = p^2-p+1$. If the product $A B$ is a prime $c$ times a perfect square, then $A$ is either a perfect square or three times a perfect square.

Indeed the accepted child OddPerfectNumber.Kernel.second_block_gcd_dvd_three (0e45b51d) gives $\gcd A B \mid 3$, so the only prime that can divide both blocks is $3$. The square relation $A B = c y^2$ says that exactly one prime, namely $c$, occurs to odd multiplicity in $A B$. Because the odd multiplicities of $A$ and $B$ multiply to $c$, precisely one of the two blocks is a square. The block $B$ cannot be a square, since $(p-1)^2 < p^2-p+1 < p^2$ for $p > 1$. Hence $A$ carries all the odd multiplicity, and since the only prime that may be shared is $3$, the odd-multiplicity prime of $A$ is $c$ and dividing $A$ by the possible single factor $3$ leaves a square. This is the elementary shape forced on the first factor of the second cyclotomic block whenever the block is a prime times a square; it is the arithmetic content of the $k=5$ two-prime residual.