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I took the function $f(x)=x^2$ and noted that $f$ satisfies the following functional equation (from well-known equality $(x+1)^2=x^2+2x+1$):$$f(x+1)=f(x)+2x+1.$$

Here's the question: is there another solution $f: \mathbb{R} \to \mathbb{R}$ for this equation? I tried to prove it myself, but didn't succeed. Could you help, please?

Also I wonder: there is another functional equation for $f(x)=x^2$: $$f(x+y)=f(x)+2xy+f(y).$$ Is it related to the previous one? Can we deduce the second from the first?

UPD1: the fist question was answered in the comments. What about the second one?

UPD2: the second eqution has others solutions too. For instance, take $f(x)=x^2+ax$, where $a$ is an arbitrary real number. See Functional equation $f(x+y)=f(x)+2xy+f(y)$ .

lnv619
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  • For the first question. Hint: Lots of solutions exists. Show that for an arbitrarily defined function $g(x): [0,1 ) \rightarrow R$, we can extend that to a function $f(x) : R \rightarrow R$ such that $g(x) = f(x)$ on $x\in [0, 1)$. $\quad$ Please refrain from asking multiple questions in the same post. You can split them up as needed. – Calvin Lin Nov 19 '23 at 17:58
  • Thank you very much for your answer! It won't happen again. :) – lnv619 Nov 19 '23 at 18:10
  • Is $f(x)$ continuous? – NN2 Nov 19 '23 at 18:39
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    The first equation notably has $x^2+c$, for any $c \in \mathbb R$, as solution. When $c \ne 0$, this is not solution of the second equation. The solutions $x^2+ax$ you found for the second equation are not solution of the first equation, when $a \ne 0$. So the two equations are not related. – Jean-Armand Moroni Nov 19 '23 at 18:43
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    If $f(1)=1$, then the second equation implies the first. Otherwise, neither equation implies the other. – Geoffrey Trang Nov 19 '23 at 18:44
  • This is equivalent to $f(x+1)-(x+1)^2=f(x)-x^2.$ So $g(x)=f(x)-x^2$ can be any function satisfying $g(x+1)=g(x).$ That is, any function of period $1.$ – Thomas Andrews Nov 19 '23 at 18:44
  • Given a $g(x)$ of period $1,$ you get $f(x)=x^2+g(x)$ is an example. In particular, $g$ any constant function works. – Thomas Andrews Nov 19 '23 at 18:47

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