Polynomial satisfying $ P \big(P (x)\big)=P (x)+ P\big(x^2\big)$ If $P(x)$ is a polynomial with integer coefficients such that for all integer $x$, $$P (P (x)) = P (x)+P (x^2).$$ 
I've tried solving it putting it as a function instead. Not much though. How do you find such polynomials?
 A: Let the polynomial be of degree n $P(P(x))$ has a maximum degree $(x^n)^n$ i.e., $n^2$ and on the right hand side similarly it has a max degree of $2n$. So now compare this max degrees $n^2=2n$ which implies n =2. Now try solving it by considering the polynomial as quadratic.
A: Hints. First note that either $P\equiv 0$ or it is a second degree polynomial, with leading coefficient 1. 
To do this, observe first that if $P$ is constant, then $P\equiv 0$. 
Next, if the degree $d$ of $P$ is at least 1, then the degree of the left-hand side of 
$$
P(x)+P\big(x^2\big)=P\big(P(x)\big),
$$
is $2d$, while the order of the right hand side is $d^2$, and thus $2d=d^2$, i.e., $d=0$ or $d=2$.
Next, if $a_2x^2$ is the leading term of $P$, then the leading term of the left-hand side is $a_2x^2$, while leading term of the right-hand side is $a_2^2x^2$. 
Hence $P(x)=x^2+ax+b$. Then plug this in the equation to determine the values of $a$ and $b$. Now, the third order term in the left hand side is zero, while in the right-hand side is $2ax^3$. Thus $a=0$. It is not hard to get that $b=0$, as well.
Answer. It turns out that $P\equiv 0$.
