I'm looking to find a root $(x\_1,\dots,x\_n)$ of a polynomial $p \in {\mathbb R}[x\_1,\dots,x\_n]$ such that $0 \leq x\_i < 1$ for all $i$. Further, I know in advance that setting $x\_1 = \cdots = x\_n$ is a root of $p$, but wish to avoid this root. How can I find at least one of these roots, preferably using a computer?


In more detail, $p$ is formed by taking two homogeneous polynomials $q\_1, q\_2$ (of degrees $d\_1$ and $d\_2$) where every coefficient is either zero or one, letting $D$ be the least common multiple of $d\_1$ and $d\_2$, and setting $p = q\_1^{D/d\_1} - q\_2^{D/d\_2}$. This is encoding the simultaneous solution to the equations $$q\_1(x) = y^{d\_1}, \quad q\_2(x) = y^{d\_2}$$ over $(x\_1,\dots,x\_n, y) \in [0,1)^{n+1}$.



Non-Uniform Root of Polynomial in Open Cube
If you can find points $u, v \in [0,1]^n$ where $p(u) < 0$ and $p(v) > 0$, there will be
a solution of the form $tu + (1-t) v$ for some $t \in [0,1]$, and that can be found by solving a polynomial in one variable. If $n > 2$, generically the line $tu+(1-t)v$ will not intersect the diagonal. With a bit of luck, you might find $u$ and $v$ by random search and/or local optimization. Of course, this won't work if $p$ is always $\ge 0$ or always $\le 0$.


