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As in the title. It may be very simple, but I'm having difficulty finding the proper substitution.

Cameron Buie
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2 Answers2

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You don't even need to solve it, just note that if $f_1(x)$ is some solution, then so is $f_1(x) + c$ for any $c \in \Bbb R$. Thus we have proven $$ \text{There are solutions }\implies\text{ There are solutions with }f(0) \neq 0,1 $$ which is all we really need.

Arthur
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With $y=x=z/2$ you obtain $f(z)=z^2$

Redundant Aunt
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    Don't you get $f(z)-f(0) = z^2$? The question is actually about what $f(0)$ is. In fact $f(x) = x^2 +c$ is a solution for any $c\in \mathbb R$. – Jorik Dec 23 '15 at 13:12
  • But what's the value of f(0) in our case? –  Dec 23 '15 at 13:20
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    @hetajr There is not enough information to tell. If we are only given that $f(x+y)-f(x-y) = 4xy$ then $f(0)$ can be any number. See Arthur's answer as well. – Jorik Dec 23 '15 at 13:26
  • This answer seems to implicitly assume that $f(0)=0$. The reasoning seems circular. – Element118 Dec 23 '15 at 14:37