Existence of solutions to a nonlinear algebraic equation

Question

How can we prove that equation (1) has solutions for every $p$. I mean, is there an analytic method that can be used to show that there exist solutions for every $p$ for this nonlinear equation:

\begin{equation} \left\{ \begin{array}{ccc} x_1^2 x_2 \cdots x_{p-1} x_p+x_1 &=& 1 \, ,\\ &&\\ x_1 x_2^2 x_3 \cdots x_{p-1} x_p+x_1 x_2 &=& 1 \, ,\\ &&\\ x_1 x_2 x_3^2 x_4 \cdots x_{p-1} x_p+x_1 x_2 x_3&=& 1\, , \\ &&\\ \vdots & \vdots & \vdots \\ &&\\ x_1 x_2 x_3 x_4 \cdots x_{p-1} x_p^2+x_1 x_2 \cdots x_p&=& 1\, . \end{array} \right. \end{equation}

For example, let ($p=3$). Then we have:

\begin{equation} \left\{ \begin{array}{ccc} x_1^2 x_2 x_3+x_1 &=& 1 \, ,\\ &&\\ x_1 x_2^2 x_3 +x_1 x_2 &=& 1 \, ,\\ &&\\ x_1 x_2 x_3^2 +x_1 x_2 x_3&=& 1\, . \end{array} \right. \end{equation}

So is there an analytic method for existence of solution for equation (2). For this example I used MAPLE software and got these solutions:

\begin{equation*} \left\{ \begin{array}{ccc} a &=& 0.658418845314095780 \, ,\\ &&\\ b &=& 0.849466898144101812 \, ,\\ &&\\ c&=& 0.927561975482924960 \, . \end{array} \right. \end{equation*}

Answer 1

Note that no $x_k$ can vanish. If you put $ u :=x_1x_2\dots x_p $ the system gives inductively, for $k=1,\dots,p$:

$$\frac{1}{x_1 x_2\dots x_k}=1+u+\dots+u^k . $$

In particular $u$ solves $$\frac{1}{u}=1+u+\dots+u^p . $$ Incidentally, $u\neq 1$, so we can express the solutions to the systems simply as $$x_k=\frac{u^{k}-1}{u^{k+1}-1},$$ in terms of the solutions $u\neq 1$ to the equation

$$u^{p+2}-2u+1=0. $$

Also note that, by elementary arguments, if $p$ is even the latter equation has a unique real solution (besides $u=1$), which is positive, while if $p$ is odd it has a positive and a negative solution (besides $u=1$). So the system has one or two real solutions according whether $p$ is even or odd, and in any case a unique solution if $x_k$ are assumed to be real and positive.

To complete the analytic solution, we may solve $u^{p+2}-2u+1=0$ by series using the Lagrange inversion formula to invert $u-u^{p+2}/2$ at $1/2$: for the solution $0<u_p<1$ one gets

$$u_p=\frac{1}{2}\sum_{k=0}^\infty {pk+2k\choose k} \frac{2^{-pk-2k}}{pk + k+1}.$$

For instance, $$u_3=\frac{1}{2}\sum_{k=0}^\infty {5k\choose k} \frac{32^{-k}}{4k+1}=0.5187900635...$$ that corresponds to the solutions $x_1=a, x_2=b, x_3=c$ you got by MAPLE.