types of solutions of the functional equation $f\left(x^2+y^2\right)=f(x)^2+f(y)^2$ 
Are functions $0$, $\frac{1}{2}$, $x$ and $|x|$ the only continuous solutions of $$f\left(x^2+y^2\right)=f(x)^2+f(y)^2\text?$$


Does the equation have discontinuous solutions?

[Edit: This one is solved by  @Gae.S. in a comment below.]
(own problem, partially inspired by Is there a continuous function $f$ satisfying $f^{2}(x) = f(x^{2})$, $f(0)=1$ and $ f(1)=0$?)
 A: It can be shown that the only functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \left( x ^ 2 + y ^ 2 \right) = f ( x ) ^ 2 + f ( y ) ^ 2 \tag 0 \label 0 $$
are those of the following form:

*

*$ f ( x ) = 0 $;

*$ f ( x ) = \frac 1 2 $;

*$ f ( x ) = \begin {cases} x & x \notin A \\ - x & x \in A \end{cases} $ for some $ A \subseteq \mathbb R ^ - $.

Among these, the continuous ones are exactly those you've mentioned. It's straightforward to check that these are all solutions. We prove that they are the only ones.
Substituting $ - x $ for $ x $ in \eqref{0} and comparing to \eqref{0} itself, we have
$$ f ( - x ) ^ 2 = f ( x ) ^ 2 \text . \tag 1 \label 1 $$
Defining $ a = f ( 0 ) $ and setting $ y = 0 $ in \eqref{0} we have
$$ f \left( x ^ 2 \right) = f ( x ) ^ 2 + a ^ 2 \text . \tag 2 \label 2 $$
\eqref{2} can be used to rewrite the right-hand side of \eqref{0} and get
$$ f \left( x ^ 2 + y ^ 2 \right) = f \left( x ^ 2 \right) + f \left( y ^ 2 \right) - 2 a ^ 2 \text , $$
which by defining $ g : \mathbb R ^ { 0 + } \to \mathbb R $ with $ g ( x ) = f ( x ) - 2 a ^ 2 $, can be transformed to
$$ g ( x + y ) = g ( x ) + g ( y ) \tag 3 \label 3 $$
for all $ x , y \in \mathbb R ^ { 0 + } $. \eqref{2} shows that $ g $ is bounded below; in particular $ g ( x ) \ge - a ^ 2 $ for all $ x \in \mathbb R ^ { 0 + } $. Using this fact together with \eqref{3}, shows that $ g $ must be of the form
$$ g ( x ) = b x \tag 4 \label 4 $$
for some constant $ b \in \mathbb R ^ { 0 + } $ and all $ x \in \mathbb R ^ { 0 + } $ (see here). Letting $ x = 0 $ in \eqref{2}, we see that $ a \in \{ 0 , \frac 1 2 \} $.

Case $ f ( 0 ) = 0 $:
Putting $ x = 1 $ in \eqref{2} and using \eqref{4}, we see that $ b \in \{ 0 , 1 \} $. If $ b = 0 $, then using \eqref{1} we can conclude that $ f $ is constantly zero on all $ \mathbb R $, which gives one of the mentioned solutions. If $ b = 1 $, we have $ f ( x ) = x $ for all $ x \in \mathbb R ^ { 0 + } $, and using \eqref{1} we can conclude that for any $ x \in \mathbb R ^ - $ we have $ f ( x ) \in \{ - x , x \} $. As we checked before, this can be chosen arbitrarily for negative $ x $, and leads to another class of solutions mentioned above.

Case $ f ( 0 ) = \frac 1 2 $:
Putting $ x = 1 $ in \eqref{2} and using \eqref{4}, we see that $ b = 0 $, which gives the constant solution $ f ( x ) = \frac 1 2 $.
