Problem. Let \(\xi(s)=\tfrac12 s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)\) and let \(\Xi(z)=\xi(\tfrac12+iz)\), so \(\Xi\) is a real even entire function. Let \(\Phi\) denote the standard even Riemann theta kernel normalized by \[ \Xi(z)=2\int_0^\infty \Phi(u)\cos(zu)\,du, \qquad \Phi(u)=\sum_{n\ge1}\bigl(2\pi^2n^4e^{9u/2}-3\pi n^2e^{5u/2}\bigr)e^{-\pi n^2e^{2u}}. \] For \(z,w\) in the upper half-plane \(\mathbb H=\{z:\operatorname{Im}z>0\}\), define the Hermitian Pick/Bezout kernel \[ B_\Xi(z,w)=\frac{\Xi(z)\overline{\Xi'(w)}-\Xi'(z)\overline{\Xi(w)}}{z-\overline w}. \] Construct, directly from \(\Phi\) and explicit integral transforms, an unconditional positive-kernel representation of the form \[ B_\Xi(z,w)=\int_\Omega \Psi_z(\omega)\overline{\Psi_w(\omega)}\,d\mu(\omega),\qquad d\mu\ge0, \] valid for all \(z,w\in\mathbb H\), with convergence and Hermitian analyticity justified. The construction must not assume the Riemann Hypothesis, zero-freeness of \(\Xi\) in \(\mathbb H\), Hermite--Biehler/de Branges positivity, or the Pick positivity being proved. Prove as part of the solution that such a factorization implies that \(-\Xi'/\Xi\) is a Herglotz/Nevanlinna function on its zero-free domain and forces all zeros of \(\Xi\) to be real, hence proves the Riemann Hypothesis.