Edit. I missed the condition that $x_i$ can be negative. Nevertheless, I still keep the solution below in case someone can salvage it. All ones. In what follows, $x_{n+1}\triangleq x_1$. Note that by Cauchy-Schwarz \[ 2\left(x_i^2+\frac{1}{x_{i+1}^2}\right)\ge \left(x_i+\frac{1}{x_{i+1}}\right)^2. \]Consequently, multiplying both sides by $2^n$, we deduce \[ 2^n \prod_{1\le i\le n}\left(x_i+\frac{1}{x_{i+1}}\right)\ge \prod_{1\le i\le n}\left(x_i+\frac{1}{x_{i+1}}\right)^2. \]We hence obtain \[ 2^n \ge \prod_{1\le i\le n}\left(x_i+\frac{1}{x_{i+1}}\right). \]Now, notice, on the other hand, that \[ x_i+\frac{1}{x_{i+1}}\ge 2\sqrt{\frac{x_i}{x_{i+1}}} \implies \prod_{1\le i\le n} \left(x_i+\frac{1}{x_{i+1}}\right) \ge 2^n \prod_{1\le i\le n}\sqrt{\frac{x_i}{x_{i+1}}}=2^n. \]Thus we must retain equalities everywhere. Now, $x_i = \frac{1}{x_{i+1}}$. If we set $x_1=t$, we obtain the numbers are alternating $t,\frac1t,t,\frac1t$ and so on. For $n$ odd, this is possible only when $t=1$. Suppose $n$ is even. The right evaluates $\frac{2^n}{t^n}$, and the left evaluates $\frac{2^n}{t^{2n}}$. Hence, $t^{2n}=t^{n}$. From here we deduce $t=1$. Thus $(x_1,\dots,x_n)=(1,1,\dots,1)$ is the only tuple enjoying the desired property.

Above: $x_i$ may be negative.

Let $b_i=x_ix_{i+1}$ with $i=1,2,...,n$ and $x_{n+1}=x_1$ Multipliying two side by $x_1^2x_2^2...x_n^2$ we get: $x_1x_2...x_n(b_1+1)(b_2+1)...(b_n+1)=(b_1^2+1)(b_2^2+1)...(b_n^2+1)$ but $b_1b_2...b_n=(x_1x_2...x_n)^2$ Square both sides and we get $b_1b_2...b_n(b_1+1)^2(b_2+1)^2...(b_n+1)^2=(b_1^2+1)^2(b_2^2+1)^2...(b_n^2+1)^2$ Using the following inequality we obtain: $x(x+1)^2\leq(x^2+1)^2\Leftrightarrow (x-1)^2(x^2+x+1)\geq 0$ therefore $(b_1^2+1)^2(b_2^2+1)^2...(b_n^2+1)^2\geq b_1b_2...b_n(b_1+1)^2(b_2+1)^2...(b_n+1)^2$ equality only when $b_1=b_2=....b_n=1$ If $n$ is odd $x_1x_2=x_nx_1\Rightarrow x_2=x_n, x_3=x_{n-1},...,x_{\frac{n+1}{2}}=x_{\frac{n+1}{2}+1}$ therefore $x_{\frac{n+1}{2}}=x_{\frac{n+1}{2}+1}=\pm1$ if $x_{\frac{n+1}{2}}=1$ then $x_1=x_2=...=x_n=1$ and this works Otherwise $x_{\frac{n+1}{2}}=-1$ then $x_1=x_2=...=x_n=-1$ and this clearly dont work Therefore if $n$ is odd $(x_1,x_2,...,x_n)=(1,1,....,1)$ If $n$ is even $x_{i}=k$ for a constant $k$ and $i$ odd Also $x_i=j$ for a constant $j$ and $i$ even therefore if $n=2m$ we have: $k^{2m}j^{2m}=(kj)^{2m}=1$ then if $m\neq 0$ we have $k=\frac{1}{j}$ or and this clearly works Therefore if $n$ is even $(x_1,x_2,...,x_n)=(j,\frac{1}{j},...,j,\frac{1}{j})$ works

@above $(x_1,x_2,...,x_n)=(x,1/x,..,x, 1/x)$ is also a solution for all even $n$

plagueis wrote: @above $(x_1,x_2,...,x_n)=(x,1/x,..,x, 1/x)$ is also a solution for all even $n$ Yes, I thought the conclusion was that they were all the same, but I didn't check yesterday. What do you think now?

Looks good now

plagueis wrote: Let $n\ge 2$ be a positive integer. Let $x_1, x_2, \dots, x_n$ be non-zero real numbers satisfying the equation \[\left(x_1+\frac{1}{x_2}\right)\left(x_2+\frac{1}{x_3}\right)\dots\left(x_n+\frac{1}{x_1}\right)=\left(x_1^2+\frac{1}{x_2^2}\right)\left(x_2^2+\frac{1}{x_3^2}\right)\dots\left(x_n^2+\frac{1}{x_1^2}\right).\]Find all possible values of $x_1, x_2, \dots, x_n$. Proposed by Victor Domínguez Personal opinion. I think this has far less detail then P5, but I really enjoyed solving this. To simplify things, we can rewrite this as \[ \left( \prod_{i = 1}^n x_i \right) \left( \prod_{j = 1}^{n} (x_j x_{j + 1} + 1) \right) = \prod_{k = 1}^n (x_{k}^2 x_{k + 1}^2 + 1) \]where indices are considered modulo $n$. Now, we intend to solve this using inequality. (For more info, see motivation below). We know that by simple CS, \[ x_k^2 x_{k + 1}^2 + 1 \ge \frac{1}{2} (|x_k x_{k + 1} | + 1)^2 \]Now, let us consider the following chain of inequalities: \begin{align*} \prod_{k = 1}^n (x_k^2 x_{k + 1}^2 + 1) &\ge \prod_{k = 1}^n \frac{1}{2} (|x_k x_{k + 1}| + 1)^2 \\ &= \frac{1}{2^n} \left( \prod_{k = 1}^n ( |x_kx_{k + 1}| + 1 ) \right)^2 \\ &= \frac{1}{2^n} \left( \prod_{k = 1}^n (|x_k x_{k + 1}| + 1 ) \right) \cdot 2^n \left( \prod_{\ell = 1}^n \sqrt{|x_{\ell}x_{\ell + 1}|} \right) \\ &= \left( \prod_{k = 1}^n ( |x_k x_{k + 1} | + 1) \right) \left| \prod_{\ell = 1}^n x_{\ell} \right| \\ &\ge \left( \prod_{k = 1}^n |x_k x_{k + 1} + 1 | \right) \left| \prod_{\ell = 1}^n x_{\ell} \right| \\ &= \left| \left( \prod_{k = 1}^n (x_k x_{k + 1} + 1) \right) \cdot \prod_{\ell = 1}^n x_{\ell} \right| \\ &\ge \left( \prod_{k = 1}^n (x_k x_{k + 1} + 1) \right) \cdot \left( \prod_{\ell = 1}^n x_{\ell} \right) \end{align*}Since the statement of the problem is the equality of the above equality, then we just need to consider the equality cases. Equality occurs when: 1) $|x_kx_{k + 1} | = 1$ for all $k$ (CS argument). 2) $x_k x_{k + 1}$ and $1$ has the same sign for all $k$ (Triangle Inequality argument). 3) $\prod_{\ell = 1}^n x_{\ell} \left( \prod_{k = 1}^n (x_k x_{k + 1} + 1) \right) > 0$. ($|x| = x$ iff $x > 0$) By 1) and 2), we have \[ x_k x_{k + 1} = 1 \ \text{for all } k = 1, 2, \dots, n \]For odd $n$, this simplifies to $x_1 = x_2 = \dots = x_n = 1$ or $x_1 = x_2 = \dots = x_n = -1$, where the latter is false by 3). For even $n$, this simplifies to the solution $x_1 = x_3 = \dots = x_{n - 1} = a$ and $x_2 = x_4 = \dots = x_{n} = \frac{1}{a}$. Checking back to the equation, this indeed satisfy. Therefore, the answer is \[ (x_1, x_2, \dots, x_n) = \begin{cases} \underbrace{(1, 1, 1, \dots, 1)}_{n \ \text{times}} \ &\text{if } n \ \text{is odd.} \\ \underbrace{\left(a, \frac{1}{a}, a, \frac{1}{a}, \dots, a, \frac{1}{a} \right)}_{n \ \text{times}} \ &\text{if } n \ \text{is even.}