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

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!

• Use Mean Value Theorem. Mar 9, 2023 at 20:43
• I think $m$ can be any non-zero number; the crucial requirement is that $f(x+m)=f(x) + m$ for all $x$.
– mcd
Mar 9, 2023 at 20:44
• Yes, I meant that m can be any non-zero integer: I'll edit the post to clarify that Mar 9, 2023 at 21:21

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".
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.