Let $n \geq 3$ and $\frac{n}{2} \leq p \leq \ n-1$. Let $a, b \in {\Bbb R}$ satisfy $b\geq \frac1n a^2$. Let the $n \times n$ symmetric matrix ${\bf C}_p$ be hollow (zero entries on the main diagonal) and binary (entries are either $0$ or $1$) have $p$ ones in either the upper or lower triangular part. Consider the following quadratically-constrained quadratic program (QCQP).
$$ \begin{array}{ll} \underset {{\bf x} \in {\Bbb R}^n} {\text{minimize}} & \frac12 {\bf x}^\top {\bf C}_p {\bf x} \\ \text{subject to} & {\bf 1}_n^\top {\bf x} = a \\ & {\bf x}^\top {\bf x} = b \end{array} \tag{QCQP} $$
The purpose of $b \geq \frac1n a^2$ is to guarantee that the hyperplane $\left\{ {\bf x} \in {\Bbb R}^n : {\bf 1}_n^\top {\bf x} = a \right\}$ and the spherical surface $\left\{ {\bf x} \in {\Bbb R}^n : {\bf x}^\top {\bf x} = b \right\}$ do intersect, thereby ensuring that the QCQP's feasible region is non-empty.
I have been stuck on this QCQP for a while now.
My work
My work is based on an answer to this question on Mathoverflow.
Consider the Lagrangian function $$ \mathcal{L}(\textbf{y},\theta_1,\theta_2) = \frac{1}{2}x^\top\textbf{C}_p{\bf y}-(\textbf{1}_n-a)\theta_1-({\bf y}^\top {\bf y}-b)\theta_2. $$ Lagrange multipliers' rule then implies that if the quadratic form ${\bf y}^\top {\bf C}_p {\bf y}$ attains its minimum at $\bf x$, then there are $\theta_1,\theta_2\in\mathbb{R}$ such that $\nabla_\textbf{x}\mathcal{L}(\textbf{x},\theta_1,\theta_2)=0$. Thus, \begin{equation} \tag{1} \textbf{C}_p\textbf{x}-\textbf{1}_n\theta_1-2\textbf{x}\theta_2 = 0. \end{equation} Multiply both sides of (1) with $\textbf{1}_n^\top$ then yields $$ \textbf{1}^\top_n\textbf{C}_p\textbf{x} - n\theta_1-2\textbf{1}^\top_n\textbf{x}\theta_2=0. $$ Now the constraint $\textbf{1}_n^\top\textbf{x}=a$ along with the fact that $\textbf{1}_n^\top\textbf{1}_n=n$ give $$ \textbf{1}_n^\top \textbf{C}_p\textbf{x}-n\theta_1-2a\theta_2=0, $$ which allows us to express $\theta_1$ in terms of other variables as \begin{equation} \tag{2} \theta_1 =\frac{1}{n}(\textbf{1}_n^\top \textbf{C}_p\textbf{x}-2a\theta_2). \end{equation} By substituting (2) into (1), we get \begin{equation} \tag{3} \Biggl[\biggl(I_n-\frac{1}{n}\textbf{1}_n\textbf{1}_n^\top\biggl)\textbf{C}_p-2\theta_2I_n\Biggl]\textbf{x}=\frac{2a\theta_2}{n}\textbf{1}_n. \end{equation} Now the matrix $B=I_n-\frac{1}{n}\textbf{1}_n\textbf{1}_n^\top$ can be explicitly computed as: $$ B = \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix} -\frac{1}{n} \begin{pmatrix} 1 & 1 & \cdots & 1 \\ 1 & 1 & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1 \end{pmatrix} = \frac{1}{n} \begin{pmatrix} n-1 & -1 & \cdots & -1 \\ -1 & n-1 & \cdots & -1 \\ \vdots & \vdots & \ddots & \vdots \\ -1 & -1 & \cdots & n-1 \end{pmatrix} . $$ Equation (3) can then be reduced to $ (B\textbf{C}_p-2\theta_2 I_n)\textbf{x} =\frac{2a\theta_2}{n}\textbf{1}_n.$ All that is left to do is to solve the following system of $n+1$ equations with $n+1$ variables: \begin{equation} \tag{4} \begin{aligned} (B\textbf{C}_p-2\theta_2 I_n)\textbf{x} &=\frac{2a\theta_2}{n}\textbf{1}_n,\\ \textbf{x}^\top\textbf{x}&=b, \end{aligned} \end{equation} which, to the best of my knowledge, can only be solved generally by numerical methods such as the LSQI method mentioned in Golub and Van Loan's book Matrix computations. However, given the fact that $\textbf{C}_p$ is quite specially defined, I hope that a closed-form solution, or at least a lower bound on the solution, can be achieved.
