Let $P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$ with $a_n \ne 0$ and all $a_i \in \mathbb{R}$ so that it is a polynomial of degree $n$. We start by considering the leading term of each expression: $P(2P(x)) = P(2(a_nx^n + ...)) = a_n(2a_nx^n + ...)^n + ... = a_n(2^na_n^n(x^n)^n + ...) + ... = 2^na_n^{n+1}x^{n^2} + ...$ $2P(P(x)) = 2P(a_nx^n + ...) = 2[a_n(a_nx^n + ...)^n + ...] = 2a_n(a_n^n(x^n)^n + ...) + ... = 2a_n^{n+1}x^{n^2} + ...$ $2(P(x))^2 = 2(a_nx^n + ...)^2 = 2(a_n^2(x^n)^2 + ...) = 2a_n^2x^{2n} + ...$ For the two sides to be always equal, the coefficients of each term must be equal. In particular, the leading coefficients must be equal. On the left side (expanded), we have a polynomial of degree $n^2$. On the right side, we have the sum of a polynomial of degree $n^2$ and a polynomial of degree $2n$, so the combined degree will be $\max(n^2, 2n)$. If $n>2$, then $n^2 > 2n$, so the expression $2(P(x))^2$ does not contribute to the leading coefficient of the expanded polynomial on the right side. Comparing leading coefficients, we have $2^na_n^{n+1} = 2a_n^{n+1}$. We assumed in our original definition that $a_n \ne 0$, so we can divide both sides by $a_n^{n+1}$ to get $2^n = 2$. The only solution to this is $n = 1$, but we assumed that $n>2$, hence we have a contradiction, so this case is not possible. Next we consider $n < 2$. If $n = 1$, then $n^2 < 2n$, so the left side will have a degree (1) less than that of the right side (2), so this case is not possible. If $n = 0$, then $P(x) = y$, for some constant $y$. Substituting this into the original polynomial equation yields $y = 2y + 2y^2$. Rearranging gives $2y^2 + y = 0$, so $y = 0$ or $y = -\frac{1}{2}$. We can easily check to see that both work, so $\boxed{P(x) = 0}$ and $\boxed{P(x) = -\frac{1}{2}}$ are possible solutions. Finally, we take the case $n = 2$. Then comparing leading coefficients gives us $4a_2^3 = 2a_2^3 + 2a_2^2$. Rearranging and dividing out a factor of $2$ gives $a_2^3 - a_2^2 = 0$, or $a_2^2(a_2 - 1) = 0$. Since we must have $a_2 \ne 0$, the only solution is $a_2 = 1$. Hence our polynomial takes the form $P(x) = x^2 + bx + c$. We now substitute this into our original polynomial equation the following: \[P(2(x^2 + bx + c)) = 2P(x^2 + bx + c) + 2(x^2 + bx + c)^2\] \[(2(x^2 + bx + c))^2 + b[2(x^2 + bx + c)] + c = 2[(x^2 + bx +c)^2 + b(x^2 + bx + c) + c] + 2(x^2 + bx + c)^2\] \[4(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + c = 2(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + c + 2(x^2 + bx + c)^2\] \[4(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + c = 4(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + c\] Both sides are identically equal, so all polynomials of the form $\boxed{P(x) = x^2 + bx + c}$ for any $b, c \in \mathbb{R}$ also work. SidenoteIs there a notation that I can use for leading coefficients so that I don't have to keep writing $...$?

Let $P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$ with $a_n \ne 0$ and all $a_i \in \mathbb{R}$ so that it is a polynomial of degree $n$. We start by considering the leading term of each expression: $P(2P(x)) = P(2(a_nx^n + ...)) = a_n(2a_nx^n + ...)^n + ... = a_n(2^na_n^n(x^n)^n + ...) + ... = 2^na_n^{n+1}x^{n^2} + ...$ $2P(P(x)) = 2P(a_nx^n + ...) = 2[a_n(a_nx^n + ...)^n + ...] = 2a_n(a_n^n(x^n)^n + ...) + ... = 2a_n^{n+1}x^{n^2} + ...$ $2(P(x))^2 = 2(a_nx^n + ...)^2 = 2(a_n^2(x^n)^2 + ...) = 2a_n^2x^{2n} + ...$ For the two sides to be always equal, the coefficients of each term must be equal. In particular, the leading coefficients must be equal. On the left side (expanded), we have a polynomial of degree $n^2$. On the right side, we have the sum of a polynomial of degree $n^2$ and a polynomial of degree $2n$, so the combined degree will be $\max(n^2, 2n)$. If $n>2$, then $n^2 > 2n$, so the expression $2(P(x))^2$ does not contribute to the leading coefficient of the expanded polynomial on the right side. Comparing leading coefficients, we have $2^na_n^{n+1} = 2a_n^{n+1}$. We assumed in our original definition that $a_n \ne 0$, so we can divide both sides by $a_n^{n+1}$ to get $2^n = 2$. The only solution to this is $n = 1$, but we assumed that $n>2$, hence we have a contradiction, so this case is not possible. Next we consider $n < 2$. If $n = 1$, then $n^2 < 2n$, so the left side will have a degree (1) less than that of the right side (2), so this case is not possible. If $n = 0$, then $P(x) = y$, for some constant $y$. Substituting this into the original polynomial equation yields $y = 2y + 2y^2$. Rearranging gives $2y^2 + y = 0$, so $y = 0$ or $y = -\frac{1}{2}$. We can easily check to see that both work, so $\boxed{P(x) = 0}$ and $\boxed{P(x) = -\frac{1}{2}}$ are possible solutions. Finally, we take the case $n = 2$. Then comparing leading coefficients gives us $4a_2^3 = 2a_2^3 + 2a_2^2$. Rearranging and dividing out a factor of $2$ gives $a_2^3 - a_2^2 = 0$, or $a_2^2(a_2 - 1) = 0$. Since we must have $a_2 \ne 0$, the only solution is $a_2 = 1$. Hence our polynomial takes the form $P(x) = x^2 + bx + c$. We now substitute this into our original polynomial equation the following: \[P(2(x^2 + bx + c)) = 2P(x^2 + bx + c) + 2(x^2 + bx + c)^2\]\[(2(x^2 + bx + c))^2 + b[2(x^2 + bx + c)] + c = 2[(x^2 + bx +c)^2 + b(x^2 + bx + c) + c] + 2(x^2 + bx + c)^2\]\[4(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + c = 2(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + c + 2(x^2 + bx + c)^2\]\[4(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + c = 4(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + c\] Both sides are identically equal, so all polynomials of the form $\boxed{P(x) = x^2 + bx + c}$ for any $b, c \in \mathbb{R}$ also work. SidenoteIs there a notation that I can use for leading coefficients so that I don't have to keep writing $...$?