No incoming $p$-source when $(p-1)/4$ is a power of two

Let $p\equiv1\pmod4$ be prime with $p\ge5$, and suppose $(p-1)/4$ is a power of two. Then no prime $t$ can satisfy $p \mid 1+t+\dots+t^{2e}$.

Proof. $p\mid 1+t+\dots+t^{2e}$ gives $(t-1)(1+t+\dots+t^{2e}) = t^{2e+1}-1$, so $t^{2e+1}\equiv1\pmod p$ and $d=\operatorname{ord}_p(t) \mid 2e+1$; in particular $d$ is odd. By Fermat $d\mid p-1$, and since $4\mid p-1$ and $d$ is odd, $\gcd(d,4)=1$ forces $d\mid (p-1)/4$. If $(p-1)/4=2^k$ then $d$ is a power of two and also odd, so $d=1$, i.e. $t\equiv1\pmod p$. But then $1+t+\dots+t^{2e}\equiv 2e+1$, and the proved theorem sigma_square_at_one_mod_p_not_dvd_p forbids $p$ from dividing $1+t+t^2$ when $t\equiv1$. Hence $t\equiv1$, which is the only possibility.

Consequence. The second Dris equation $h_2$ requires an incoming source of $p$, since $p^5 \mid \sigma(m^2)$. For primes $p\equiv1\pmod4$ with $(p-1)/4$ a power of two this is impossible, so those $p$ are eliminated. Numerically, for $p<4000$ the eliminated primes are exactly $5$, $17$ and $257$ — in particular $p=5$ is ruled out, and $17$ survives neither. Primes with an odd factor in $(p-1)/4$ (e.g. $13,29,37,53,61,\dots$) are unaffected.