Proving $f_1(x)-f_1(z)=f_2(x)-f_2(z)$ for all $x,z$ implies $f_1=f_2+c$?

Question

I am looking for advice on how to prove this?

Specifically, it is clearly satisfied when $f_1=f_2+c$, and not satisfied in some other cases like $f_1=\alpha f_2$, but to show it is true I would need to show that, out of all possible transformations of $f_1$, only $f_1=f_2+c$ has the property that $f_1(x)-f_1(z)=f_2(x)-f_2(z)$. I'm not sure how to rule out all other possible transformations

(I feel like this might just follow by some definition or something though?)

Note: You can take the domain to be $\mathbb{R}$

Answer 1

It's enough that the equality holds for one $\,z\,$, say $\,z=0\,$, then $\;f_1(x) = f_2(x) + \big(f_1(0)-f_2(0)\big)\,$.