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Adam
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Let $R$ be a commutative, associative ring with $1$ and let $\tau: R[x^{\pm 1}]\to R[x^{\pm 1}]$ be the $R$-algebra involution $\tau(x)=x^{-1}.$ Then it is natural to ask which elements of the invariant subalgebra $R[x^{\pm 1}]^\tau$ are of the form $p\cdot \tau(p)$ for some $p\in R[x^{\pm 1}]$.

This problem reduces to a question whether each $a_n(x^n+x^{-n})+ ... + a_1(x+x^{-1})+a_0\in R[x^{\pm 1}]^\tau$ in the form $$a_n(x^n+x^{-n})+ ... + a_1(x+x^{-1})+a_0\in R[x^{\pm 1}]^\tau$$ equals $p(x)p(x^{-1}),$ for some $p(x)=\sum_{i=0}^n c_ix^i$, which leads to the system of equations: $$\begin{cases} c_0^2+...+c_n^2=a_0\\ c_0c_1+...+c_{n-1}c_n=a_1\\ ...\\ c_0c_n=a_n. \end{cases} $$

I suspect that this problem has been studied already and I am hoping to be pointed in the right direction. I expect that this system have a solution for algebraically closed fields $R$ (at least when characteristic $\ne 2$), but don't know the proof. However, the most interesting case for me is $R=\mathbb Z$. I wonder if there is a manageable description of $a_0,...,a_n\in \mathbb Z$ for which the above system has a solution in integers.

Let $R$ be a commutative, associative ring with $1$ and let $\tau: R[x^{\pm 1}]\to R[x^{\pm 1}]$ be the $R$-algebra involution $\tau(x)=x^{-1}.$ Then it is natural to ask which elements of the invariant subalgebra $R[x^{\pm 1}]^\tau$ are of the form $p\cdot \tau(p)$ for some $p\in R[x^{\pm 1}]$.

This problem reduces to a question whether each $a_n(x^n+x^{-n})+ ... + a_1(x+x^{-1})+a_0\in R[x^{\pm 1}]^\tau$ in the form $p(x)p(x^{-1}),$ for some $p(x)=\sum_{i=0}^n c_ix^i$, which leads to the system of equations: $$\begin{cases} c_0^2+...+c_n^2=a_0\\ c_0c_1+...+c_{n-1}c_n=a_1\\ ...\\ c_0c_n=a_n. \end{cases} $$

I suspect that this problem has been studied already and I am hoping to be pointed in the right direction. I expect that this system have a solution for algebraically closed fields $R$ (at least when characteristic $\ne 2$), but don't know the proof. However, the most interesting case for me is $R=\mathbb Z$.

Let $R$ be a commutative, associative ring with $1$ and let $\tau: R[x^{\pm 1}]\to R[x^{\pm 1}]$ be the $R$-algebra involution $\tau(x)=x^{-1}.$ Then it is natural to ask which elements of the invariant subalgebra $R[x^{\pm 1}]^\tau$ are of the form $p\cdot \tau(p)$ for some $p\in R[x^{\pm 1}]$.

This problem reduces to a question whether each $$a_n(x^n+x^{-n})+ ... + a_1(x+x^{-1})+a_0\in R[x^{\pm 1}]^\tau$$ equals $p(x)p(x^{-1}),$ for some $p(x)=\sum_{i=0}^n c_ix^i$, which leads to the system of equations: $$\begin{cases} c_0^2+...+c_n^2=a_0\\ c_0c_1+...+c_{n-1}c_n=a_1\\ ...\\ c_0c_n=a_n. \end{cases} $$

I suspect that this problem has been studied already and I am hoping to be pointed in the right direction. I expect that this system have a solution for algebraically closed fields $R$ (at least when characteristic $\ne 2$), but don't know the proof. However, the most interesting case for me is $R=\mathbb Z$. I wonder if there is a manageable description of $a_0,...,a_n\in \mathbb Z$ for which the above system has a solution in integers.

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Adam
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  • 13

Laurent polynomials of the form $p(x)\cdot p(x^{-1})$

Let $R$ be a commutative, associative ring with $1$ and let $\tau: R[x^{\pm 1}]\to R[x^{\pm 1}]$ be the $R$-algebra involution $\tau(x)=x^{-1}.$ Then it is natural to ask which elements of the invariant subalgebra $R[x^{\pm 1}]^\tau$ are of the form $p\cdot \tau(p)$ for some $p\in R[x^{\pm 1}]$.

This problem reduces to a question whether each $a_n(x^n+x^{-n})+ ... + a_1(x+x^{-1})+a_0\in R[x^{\pm 1}]^\tau$ in the form $p(x)p(x^{-1}),$ for some $p(x)=\sum_{i=0}^n c_ix^i$, which leads to the system of equations: $$\begin{cases} c_0^2+...+c_n^2=a_0\\ c_0c_1+...+c_{n-1}c_n=a_1\\ ...\\ c_0c_n=a_n. \end{cases} $$

I suspect that this problem has been studied already and I am hoping to be pointed in the right direction. I expect that this system have a solution for algebraically closed fields $R$ (at least when characteristic $\ne 2$), but don't know the proof. However, the most interesting case for me is $R=\mathbb Z$.