TITLE: Systems of simultaneous real quadratic equations
QUESTION [5 upvotes]: Starting from a problem in spectral graph theory, I got dragged into a problem in combinatorial matrix theory about constructing $n\times n$ real orthogonal matrices with a specified pattern of zero/non-zero entries.  (I can expand on this if required.)
Anyway, I have a construction which appears to work (which I have tested in Maple for some small values of $n$), but to prove it in full generality would require showing that a certain system of simultaneous real quadratic equations has a real solution.
I know very little about algebraic geometry and, well, nothing at all about real algebraic geometry, but I was wondering if there were any existing results in the literature which may help with this---as it stands I'm not even sure where to start looking!
[edited after Alexandre's answer]
The aim (in the first instance) is to construct an orthogonal $n\times n$ matrix with zero diagonal entries and non-zero off-diagonal entries. The idea is to use the following matrix: $$ A=\left[ \begin{array}{c|ccc} 0 & 1 & \cdots & 1 \\ \hline 1 & & & \\ \vdots && B & \\ 1 &&& \end{array} \right] $$
where the rows of $B$ are formed of the $n-1$ cyclic permutations of $v=(0,x_1,\ldots,x_{n-2})$.  For this to be orthogonal (up to scaling by $\sqrt{n-1}$), the inner product of two distinct rows must be $0$, and the norm of each row must be $\sqrt{n-1}$.  This gives the equations $\sum_{i=1}^{n-2} x_i = 0$, $\sum_{i=1}^{n-2} x_i^2 = n-2$, and $\lceil (n-2)/2\rceil$ other equations arising from inner products of rows containing cyclic permutations of $v$ (from adjacent rows, rows 2 apart, 3 apart, etc.).  Also, we have the constraint that any solution for $x_1,\ldots.x_n$ may not contain a zero.
For example, when $n=5$, Maple finds the solution $x_2=1$, $x_1,x_3 = \frac{-1\pm \sqrt{3}}{2}$.

REPLY [3 votes]: This question is linked to the discrete Fourier transform. If $u=[1,\cdots,1]^T$, then $A=\begin{pmatrix}0&u^T\\u&B\end{pmatrix}$ and the equation $AA^T=(n-1)I_n$ is equivalent to $Bu=0,BB^T=(n-1)I_{n-1}-uu^T$. The spectrum of $BB^T$ is $0$ and $n-1$ ($n-2$ times). Let $F$ be the $(n-2)\times(n-2)$ matrix defined by $f_{j,k}=\exp(\dfrac{-2i\pi jk}{n-1})$ (the partial Fourier matrix associated to the non-zero eigenvalues of $B$). $B,B^T$ commute and they are simultaneously diagonalizable (using the complete Fourier matrix) and similar to $B'=diag(F[x_1,\cdots,x_{n-2}]^T)$ and to ${B'}^T=diag(F[x_{n-2},\cdots,x_1]^T)$. The eigenvalues of $BB^T$ are the $B'_{i,i}{B'}^T_{i,i}$. Finally our vector $X$ is solution of the quadratic system: for every $i\leq n-2$, $(\sum_{j=1}^{n-2}f_{i,j}x_j)(\sum_{j=1}^{n-2}f_{i,j}x_{n-1-j})=n-1$. 
EDIT: Here $X$ is real; then the equations can be rewritten: for every $i\leq n-2$, $|\sum_{j=1}^{n-2}f_{i,j}x_j|=\sqrt{n-1}$.