The two k=5 cyclotomic blocks U and V are coprime for odd p

For an odd natural $p$, the two blocks $U = p^2+p+1$ and $V = \frac{p+1}{2}\,(p^2-p+1)$ appearing in the $k=5$ factorisation $\sigma(p^5) = 2UV$ are coprime. Suppose a prime $\ell$ divides both. If $\ell$ divides $\frac{p+1}{2}$ then $\ell$ divides $p+1$, so $p \equiv -1 \pmod \ell$ and $U \equiv 1-1+1 = 1 \pmod\ell$, a contradiction. Otherwise $\ell$ divides $p^2-p+1$, and together with $\ell \mid p^2+p+1$ this gives $2p \equiv 0 \pmod\ell$. Since $U = factorisation of$\sigma(p^5)$is$\sigma(p^5) = 2UV$with$U = p^2+p+1$and$V = \frac{p+1}{2}(p^2-p+1)$. The accepted Odd p^2+p+1$ is odd when $p$ is odd, $\ell$ is odd, hence $p \equiv 0 \pmod \ell$ and again $U \equiv 1 \pmod\ell$, a contradiction. This is the copPerfectNumber.Kernel.five_cyclotomic_pair_coprime (dbed225d-85c0-4877-9ead-8c073246bf99) only shows $\gcd(p^2+p+1,rimality that the one-prime-times-square exclusion needs: the accepted prime_mul_coprime_nonsquares_not_square applies to$a = U$and$b = V$exactly p^2-p+1) = 1$, which is not the coprimality required: the prime-toggle lemma prime_mul_coprime_nonsquares_not when this gcd is $1$.