Finding all polynomials $P(x)$ satisfying $(P(x)+P(\frac{1}{x}))^2 =P(x^2)P(\frac{1}{x^2})$ 
Find all polynomials $P(x)$ satisfying
$$\left(P(x)+P\left(\frac{1}{x}\right)\right)^2 =P(x^2)P\left(\frac{1}{x^2}\right)$$

If $P'(x)\neq 0$, then $P(x) = a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$
$\Rightarrow$  ($a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0+a_n\frac{1}{x^n}+a_{n-1}\frac{1}{x^{n-1}}+...+a_1\frac{1}{x}+a_0)^2 = (a_nx^{2n}+a_{n-1}x^{2n-2}+...+a_1x^2+a_0)( a_n\frac{1}{x^{2n}}+a_{n-1}\frac{1}{x^{2(n-1)}}+...+a_1\frac{1}{x^2}+a_0)$
Considering the coefficient of $x^{2n-1}$ we see  : $2a_na_{n-1} = 0$ $\Rightarrow  a_{n-1}= 0$
Considering the coefficient of $x^{2n-3}$ we see  : $2a_na_{n-3}+2a_{n-1}a_{n-2}= 0$
$\Rightarrow  a_{n-3}= 0$
Similarly, we can prove $a_{n-5}=0;a_{n-7}=0;a_{n-9}=0;...$
But at this point I have no further ideas, it is certain that all the coefficients of $P(x)$ cannot simultaneously be zero except $a_n$ (because it is easy to see that $P(x) = Cx^n$ does not satisfy the requirement)
Looking forward to getting help from everyone. Thank you very much! I sincerely apologize for my mistake. I have corrected my post, please forgive my ignorance.
 A: We claim that the only solutions of the equation is $P(x) = 0$.
Let $P(x) = \sum_{k} a_k x^k$ satisfy the given equation. Write $d = \deg P $.

Lemma. $P(x)$ is an even polynomial.
Proof. The conclusion is obvious if $P(x)$ is constant, so we only consider the case where $P(x)$ is non-constant.
First, we show that $P(x)$ is either even or odd. Otherwise, there exists a smallest positive odd integer $m$ such that $a_{d-m} \neq 0$. Then by comparing the coefficients of $x^{2d-m}$ in both sides of the equation, we get $ 2a_d a_{d-m} = 0, $, and so, $a_{d-m} = 0$, a contradiction.
Next, by comparing the coefficients of $x^{2d}$ in both sides of the equation, we get $a_d^2 = a_d a_0$, and so, $a_0 = a_d$. This cannot happen if $P(x)$ is odd, and therefore the lemma is proved. $\square$

Now suppose $P(x)$ solves the equation.

*

*Assume $P(x)$ is non-constant. By the above lemma, $P(x)$ is an even polynomial. In particular, $Q(x) = P(\sqrt{x})$ is still a polynomial in $x$ and satisfies
$$ (Q(x) + Q(x^{-1}))^2 = (P(x^{1/2} + P(x^{-1/2}))^2 = P(x)P(x^{-1}) = Q(x^2)Q(x^{-2}). $$
So $Q(x)$ also solves the equation, and in particular, $\deg Q = \frac{1}{2}\deg P$ is also an even integer. Repeating this argument, we find that $\frac{1}{2^k}\deg P$ is an even integer for all $k \geq 0$, which is impossible.


*The above contradiction tells that $P(x)$ must be constant. Then by plugging $P(x) = a$ to the equation, we find that $4a^2 = a^2$, and hence $a = 0$. Therefore $P(x) = 0$.
A: $P(x)=0$ is the only solution.
Proof:
Let us assume that a non-zero polynomial solution exists. Then it has some degree $n$. Since if $P$ is constant, it is clear that it must be $0$, we'll look at non-constant solutions. So $n\geq 1$.
Now, in the relation you've written above, check the constant term in both polynomials. Equating, we get:
$$2a_n^2+2a_{n-1}^2+...+4a_0^2=a_n^2+a_{n-1}^2+...+a_0^2$$
This means that $a_n=a_{n-1}=...a_0=0$, since sum of square terms is equal to $0$.
Thus, non-zero solution cannot exist.
