Find all polynomials $P(x)$ with $P(x)P(1/x)=P(x)+P(1/x)$ 
Find all polynomials $P(x)$ with
  $$P(x)P({1\over x})=P(x)+P({1\over x})$$

First I choose $x=1$, so $P(1)=0$ or $P(1)=2$. So I choose $x=-1$ too, but it's the same. I'm very stuck on this because when I substitute any $x$ it's got reciprocal always.
 A: Rewrite the equation like this:
$$ P(x)P({1\over x})-P(x)-P({1\over x})+1=1 $$
so $$\Big(P(x)-1\Big)\Big(P({1\over x})-1\Big)=1$$
Put $Q(x)= P(x)-1$ and we get $$Q(x)Q({1\over x})=1$$
If $Q(x)= a_nx^n+...a_1x+a_0$ and $a_n\ne 0$ we have $$ Q(x)x^nQ({1\over x})=x^n$$ so
$$(a_nx^n+...a_1x+a_0)(a_0x^n+...a_{n-1}x+a_n) = x^n$$ 
If any $a_k\ne 0$ for $k<n$ then we have on a left the degree $>n$ which is impossibile. So $$Q(x)=ax^n$$
for some number $a$. Since for $x=1$ we get $a^2 =Q(1)^2=1$ and thus $P(x) = \pm x^n+1$.
A: Consider P(x) to be a n-degree polynomial. P(x) = $a_{n}x^{n} + a_{n-1}x^{n-1} +.....+a_{0}$. now, P(x)*P($\frac{1}{x}$) = $a_{0}*a_{n}x^{n} + (a_{1}*a_{n} + a_{0}*a_{n-1})x^{n-1} +.....+a_{0}a_{n}x^{-n}$. Compare the coefficients on LHS and RHS, you'll get the desired result. P(x) = 1 + $x^{n}$ or P(x) = 1 - $x^{n}$. Note that P(x) $\equiv$ 0 is also a solution.
A: 
Find all polynomials $P\in \mathbb{R}[x]$ such that $P(x)P\left(\frac 1x\right) = P(x)+P\left(\frac 1x\right)$.

The null polynomial satisfies the relation. Let $P(x)=\sum_{i=0}^n a_ix^i\in \mathbb{R}[x]$ with $n\geq 0$ and $a_n\neq 0$. Thus
$$
P\left(\frac 1x\right) = \frac{1}{x^n} \sum_{i=0}^n a_{n-i}x^i.
$$
Note that
$$
P(x)+P\left(\frac 1x\right) = \frac{1}{x^n} \left(\sum_{i=0}^{n-1} a_{n-1}x^i + 2a_0x^n + \sum_{i=1}^n a_ix^{n+i}\right),
$$
and that
$$
P(x)P\left(\frac 1x\right) = \sum_{i=0}^n \left(\sum_{j=0}^i a_ja_{n+j-i}\right)x^i + \sum_{i=n+1}^{2n}c_ix^i,
$$
for some $c_i\in \mathbb{R}$. Since two polynomials are equal if and only if they have the same coefficients for the same powers of $x$, we have:


*

*$a_0a_n=a_n$ that implies $a_0=1$ since $a_n\neq 0$;

*$a_i=0$ for all $i=1,\,\ldots,\,n-1$, that can be proved inductively;

*$a_0^2+a_1^2+\cdots +a_n^2=2a_0$, that is $a_n^2=1$, that is $a_n=1$ or $a_n=-1$.
Thus a polynomial $P$ satisfies the relation if and only if it is the null polynomial or if it is of the form $P(x)=1\pm x^n$.
