In the $k=5$ residual the Euler prime is $5 \pmod{48}$

Suppose the $k=5$ two-prime residual has already forced $p+1 = 6u^2$ for some $u$, as the proved theorem five_euler_index_is_six_times_square establishes from $B = 3^{2k+1}v^2$. Then together with the Euler-prime hypothesis $p \equiv 1 \pmod 4$ one obtains the much sharper restriction

Indeed $p \equiv 1 \pmod 4$ forces $u^2$ odd, so $u$ is odd, and an odd square is $1$ modulo $8$. Hence $p+1 = 6u^2 \equiv 6 \pmod{48}$, i.e. $p \equiv 5 \pmod{48}$.

Consequence. This upgrades the earlier parity-only conclusion $p \equiv 5 \pmod{12}$ to a congruence modulo $48$, and it is the first point at which the square structure of the linear cyclotomic factor $(p+1)/2$ feeds back into the Euler prime itself. Two further consequences follow immediately and are recorded as separate targets: $p \equiv 5 \pmod 8$ gives $p^2+p+1 \equiv 7 \pmod 8$ and $(p^2-p+1)/3 \equiv 7 \pmod 8$, which through the proved allocation $C = q u_1^2$ and $D_0 = r u_2^2$ transfer to $q \equiv r \equiv 7 \pmod 8$ and hence, with the mod-$3$ transfer, $q \equiv r \equiv 7 \pmod{24}$. Verified exhaustively for every prime $p < 200000$ with $p+1 = 6u^2$ and $p \equiv 1 \pmod 4$: 25 cases, no counterexample.