Find all injective functions from reals to reals that satisfy $f(x + f(y)) = f(x) - f(y)$

Question

Find all injective functions from reals to reals that satisfy $f(x + f(y)) = f(x) - f(y)$, from reals to reals.

I tried subbing in values to cancel some variables out, like $y = x$ and $y=-f(x)$, but they did not give me a solution where I can find a function as $x+constant$, which I presume is the answer.

Any help is appreciated.

Answer 1

$$x = y \implies f(x+f(x)) = 0$$

let $a$ be the one real number such that $f(a) = 0$ (it must exist since $f(x+f(x)) = 0$ and it must be unique because $f$ is injective) :

then

$$x + f(x) = a \implies f(x) = a - x$$ for all $x \in \mathbb R$.

So:

$$a - (x + (a-y)) = a-x - a+ y \iff a -x -a + y = a-x - a +y$$

which is true for any $a \in \mathbb R$, so the only solution is $f(x) =a-x$ for any given $a$.