why is it constant?

Question

$f$ is a function from $\Bbb R$ to $\Bbb R$: $\frac{f(x+y)}{(x+y)} - (x+y)^2 = \frac{f(x-y)}{(x-y)} - (x-y)^2$ for all $x$ and $y$

the solution book just says "Thus $\frac{f(x)}{x} - x^2$ is a constant"

how did they conclude that ?

Answer 1

If $x=\frac {w+z}2$ and $y=\frac {w-z}2$ we have $x+y=w$ and $x-y=z$ and the functional equation becomes

$$\frac {f(w)}w-w^2=\frac {f(z)}z-z^2$$ But $w$ and $z$ were any (non-zero) numbers we chose, so the expression must be constant (aside from the case of zero).

Answer 2

Hint: For any $u,v$ you can write $u=x+y$ and $v=x-y$. Why?