Find all polynomials $f$ such that $f(x+m)=f(x)+m$

Question

I apologize if this is a very basic question but as someone with very limited experience solving functional equations, I'm more interested in the methods employed to tackle these kinds of problems, since I have a more advanced problem I'm trying to solve. I've tried to guess and check but haven't gotten beyond obvious solutions like $x+b$ and can't prove they're the only form a solution can take. squaring both sides and trying to find a way to group the RHS into terms of $f(x)+m$, but can't figure out where to go from there or if this is even a useful approach. To clarify, $f$ must be a polynomial and $m$ is some nonzero constant. $$(f(x)+m)^2=f(x)^2+2mf(x)+m^2=f(x)(f(x)+m)+m(f(x)+m)$$ Would appreciate some hints or problem-solving strategies: thanks!

Answer 1

Consider $g(x)=f(x)-x$; then $g$ is also a polynomial, and $g(x+m)= f(x+m) - x - m = f(x)-x=g(x)$ for all $m$, so $g$ is a constant function (this is the only periodic polynomial): that is, $g(x)=c$, so $f(x)=x+c$. You ask for general problem solving strategies for functional equations: one is "guess the answer and consider (unknown function - your guess) or (unknown function /your guess) depending on the context".

Answer 2

By a simple recurrence, you can show that $f(km) = f(0)+km$ for all integer $k$. Now a polynomial of degree $d$ grows like $Cx^d$. There is a $C$ and an $R$ such that for all $x > R$, $$|f(x)| \geq C |x|^d$$ This shows that the degree is equal to 0 or 1. A degree zero polynomial is constant, so $f$ is of degree one as you thought.