Find all polynomials P(x) with real coefficients such that

P(x2)=P(x)P(x+2).

Supposerris a root of the polynomial. Thenp(r2)=p(r)p(r+1)=0 andp((r−1)^2)=p(r−1)p(r)=0 so r^2and(r−1)^2 are roots as well.
In order to avoid generating an infinite sequence of distinct roots, each the square of the previous one, we need|r|=0 or r|=1. Similarly we need|r−1|=0 or11, and|r^2−1|=0 or11. It's not hard to show that the only possible roots are00and11.
For example
So letp(x)=ax^m(x−1)^np(x)=ax^m(x−1)^n. By considering the leading coefficient ofp(x^2)−p(x)p(x+1)p(x^2)−p(x)p(x+1), we find thata=1a=1. The zero ofp(x^2)p(x^2)at00has order2m, while the zero of p(x)p(x+1)there has orderm+n som=n.
And finally, we find thatp(x)=x^m(x−1)^m does satisfy the equation.

Supposerris a root of the polynomial. Thenp(r2)=p(r)p(r+1)=0 andp((r−1)^2)=p(r−1)p(r)=0 so r^2and(r−1)^2 are roots as well.
In order to avoid generating an infinite sequence of distinct roots, each the square of the previous one, we need|r|=0 or r|=1. Similarly we need|r−1|=0 or11, and|r^2−1|=0 or11. It's not hard to show that the only possible roots are00and11.
For example
So letp(x)=ax^m(x−1)^np(x)=ax^m(x−1)^n. By considering the leading coefficient ofp(x^2)−p(x)p(x+1)p(x^2)−p(x)p(x+1), we find thata=1a=1. The zero ofp(x^2)p(x^2)at00has order2m, while the zero of p(x)p(x+1)there has orderm+n som=n.
And finally, we find thatp(x)=x^m(x−1)^m does satisfy the equation.

The degree of(x2)(x2)and(x)(x+2)(x)(x+2)is same which is22.

This implies one of the polynomials would be linear. Let the polynomial be ax+b.

Now for the given condition,

ax^2+b=(ax+b)(ax+2a+b) ,ax^2+b=(ax+b)(ax+2a+b)

Comparing a & b from both sides we get,a=1,b=−1.a=1,b=−1.

Hence, one polynomial can be,x–1.

This process can complicate things at higher degree polynomial, but you can find polynomials till degree 3 with this method