Find all real functions that satisfy: $f(x+y) = (f(x))^2 + (f(y))^2$, $x$, $y$ $\in$ $\mathbb{R}$

Question

I am trying to find all real functions that satisfy the property: $f(x+y) = (f(x))^2 + (f(y))^2$, $x$, $y$ real numbers. I tried to substitute $x$ and $y$ with $0$ but end up with nothing, then I tried substituting $y$ with $x$ and I get $f(2x) = 2(f(x))^2$, then I tried to prove that $f(0) = 0$ and I think I am missing an important step.

How can I proceed from here? If it isn't the wrong already.

Answer 1

Setting $x=y=0$ tells us that $f(0)=0$ or $f(0)=\frac 1 2$.

Now, $y=0$ gives us $$f(x)=f(x)^2+f(0)^2\tag1$$

Write $z=f(x)$. This presents us with $z^2-z+f(0)^2=0$. Therefore, the image of $x$ under $f$ is always a root of this equation. Can you take it from here, case by case?

Answer 2

$x = y = 0$ gives $f(0) = (f(0))^2 + (f(0))^2$, and since $f(x)$, which has the solutions $f(0) = 0, f(0) = \frac{1}{2}$. Case $1$ ($f(0) = 0$) gives that, if $y = -x$: $f(x + -x) = (f(x))^2 + (f(y))^2 \iff 0 = (f(x))^2 + (f(y))^2$, but since $f$ is a real valued function, both terms in the $RHS$ of the equation have to equal $0$, and since we made no assumptions about $x$, this means that $f(x) = 0 \ \forall x \in \mathbf{R}$. Case $2$ ($f(0) = \frac{1}{2}$) gives that, if $y = 0$, $f(x) = (f(x))^2 + \frac{1}{2}$, which has the only solution $f(x) = \frac{1}{2}$. This means that the valid solutions are $f(x) = 0, f(x) = \frac{1}{2}$.