Find $f:\mathbb{R}\to\mathbb{R}$ such that $f(x)+f(x-2) = x^2-2x+1$ for all $x\in\mathbb{R}$, given that $f$ is is an even function.
I have tried putting $$x = x+2$$ and also $$x=-x$$but that gives the same thing, I don't know I am unable to make any progress. I also tried differentiating the function, because we were taught sometimes it could be solved like that, and I also know differentiation of an even function is an odd function, but it's also not helping.
Suppose $p$ is a real polynomial that satisfies $p(x)+p(x-2)=x^2-2x+1$. Note that the leading terms of $p(x)$ and $p(x-2)$ do not cancel out when they sum together. Thus, $p$ must be a quadratic polynomial. By letting $p(x)=ax^2+bx+c$, we can find that $p(x)=\frac{x^2-1}{2}$ as in the comments.
Now, suppose $f$ is an even function that satisfies $$ f(x)+f(x-2)=x^2-2x+1. \tag{1} $$ We can write $f=p+g$, where $g$ is another function. Substituting back into $(1)$, we have $$ g(x)=-g(x-2) \tag{2} $$ for all $x\in\mathbb{R}$. Since $f$ and $p$ are even functions, $g$ is also even, so $$ g(x)=g(-x) \tag{3} $$ for all $x\in\mathbb{R}$. Note that equations $(2)$ and $(3)$ are also sufficient conditions for $p+g$ being a solution.
Equations $(2)$ and $(3)$ show that $g$ is completely determined by its image on $[0,1)$, and $g(x)=0$ for all $x\in 1+2\mathbb{Z}$. This is not hard to show. First, by equation $(2)$, $g$ is completely determined by its image on $[0,2)$. Now, for all $x\in (1,2)$, we have $g(x)=-g(x-2)=-g(2-x)$. Since $2-x\in (0,1)$ for all $x\in (1,2)$, the image of $g$ on $(1,2)$ is determined by the image of $g$ on $(0,1)$. Also, $g(1)=-g(2-1)=-g(1)$, so $g(1)=0$ and hence $g(x)=0$ for all $x\in 1+2\mathbb{Z}$ by equation $(2)$.
Since equations $(2)$ and $(3)$ are sufficient conditions for $p+g$ being a solution, any arbitrary function $h:[0,1)\to \mathbb{R}$ induces a solution as shown above. In other words,