I am interested in solving the following system of $n$ equations:


$$x\_j^2 = \sum\_{i=1}^n B\_{ij} x\_i $$


for all $j\in\{1,\dots,n\}$, where $n$ is a positive integer and all the $0\leq B\_{ij}\leq 1$ are known constants. The system has a trivial solution at $x=0$. Taking the square root of the equation, we can use ~~the contraction mapping theorem~~ Theorem 3 in this [note](http://www.ssc.wisc.edu/~jkennan/research/FixedPoint21.pdf) to find that there is a unique real non-trivial positive solution. I'm hoping that this solution can be expressed in a simple way but I haven't been able to do so.



Does this system have a closed-form solution? $x_j^2 = \sum_{i=1}^n B_{ij} x_i$
Sorry, there cannot be a simple solution. For example,
taking $n=3$ and $B\_{ij} = (i+j-1)/6$, we compute numerically
(by iterating the contraction mapping as you suggest)
$$
(x\_1,x\_2,x\_3) = (1.26922421\ldots, 1.54095434\ldots, 1.77148256\ldots)
$$
and then (using **algdep** in **gp**) that these satisfy irreducible
sextic equations, respectively
$$
3888 x^6 - 2592 x^5 - 3888 x^4 + 936 x^3 + 210 x^2 + 90 x + 7 = 0,
$$
$$
3888 x^6 - 7776 x^5 + 1080 x^4 + 2592 x^3 - 129 x^2 + 162 x - 10 = 0,
$$
$$
3888 x^6 - 12960 x^5 + 12960 x^4 - 4176 x^3 + 678 x^2 - 342 x + 1 = 0
$$
with maximal Galois group $S\_6$, and thus not solvable in radicals because
the group $S\_6$ is not solvable.


*Added later*: some further notes . . .


1) The calculation reported above is not quite a complete proof:
to finish it, we can choose one of the variables, say $x\_1$,
which we guessed is a zero of the polynomial $P$,
then write the others as polynomials in that variable
(whose coefficients we can surmise in **gp** using **lindep**),
and verify that the equations $x\_j^2 = \sum\_j B\_{ij} x\_i$ all hold
modulo $P$. Here we find
$$
x\_2 = \frac1{9013}
 (81648 X^5 - 19764 X^4 - 101628 X^3 - 4182 X^2 + 7570 X + 1170),
$$
$$
x\_3 = \frac1{9013}
 (-54432 X^5 + 13176 X^4 + 67752 X^3 + 20814 X^2 - 8051 X - 780).
$$


2) The formulas for our example become somewhat simpler when written as
polynomials satisfied by $6x\_i$ (to counteract the denominators of $6$
in $B\_{ij} = (i+j-1)/6$); for instance $6x\_1$ is a root of 
$x^6 - 4 x^5 - 36 x^4 + 52 x^3 + 70 x^2 + 180 x + 84 = 0$.


3) Usually one expects that an $n$-variable system of quadratic
equations should yield solutions satisfying equations of degree
$2^n-1$ (subtracting $1$ for the trivial solution at the origin),
so $7$ for $n=3$; we got $6$ because our $B\_{ij}$ form a
degeenrate matrix, making the origin a solution of multiplicity $2$.


4) The contraction mapping argument is not actually correct:
extracting square roots yields an *expansion* mapping for small $x\_i$.
The existence of a positive solution can nevertheless be proved
using the fixed-point theorem on the $(n-1)$-simplex
$\Delta = \{ x\_i \geq 0, \sum\_i x\_i = 1 \}$:
the composite of the linear transformation $B$,
the coordinatewise square root, and the scaling
$(x\_i) \mapsto (x\_i) / \sum\_i x\_i$ gives a continuous map $\Delta \to \Delta$;
any fixed point has all coordinates $x\_i$ positive, and can be scaled to
a solution $c(x\_i)$ of the original system. Uniqueness might still hold
(it does for $n=1$ (trivially) and $n=2$, and I've not been able to find
a counterexample for $n=3$ either), but would have to be proved some other way.


