Find all functions $ f : \mathbb R \to \mathbb R $ that satisfy $$ f \left( x ^ 2 + 2 y f ( x ) \right) + f \left( y ^ 2 \right) = f ( x + y ) ^ 2 $$ for all $ x , y \in \mathbb R $.
I tried a couple of standard approaches (for example, making $ f ( x ) = 0 $, or $ f ( x ) = 1 $), but non of them yielded a result. Could someone help me?
On
You can show that the only functions $ f : \mathbb R \to \mathbb R $ satisfying $$ f \left( x ^ 2 + 2 y f ( x ) \right) + f \left( y ^ 2 \right) = f ( x + y ) ^ 2 \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $, are the identity function $ f ( x ) = x $, and the constant functions $ f ( x ) = 0 $ and $ f ( x ) = 2 $. It's straightforward to verify that these are indeed solution. We try to prove the converse.
Setting $ y = 0 $ in \eqref{0} gives $$ f \left( x ^ 2 \right) + f ( 0 ) = f ( x ) ^ 2 \text , \tag 1 \label 1 $$ while plugging $ x = 0 $ in it shows $$ f \big( 2 y f ( 0 ) \big) + f \left( y ^ 2 \right) = f ( y ) ^ 2 \text . \tag 2 \label 2 $$ We can see that $ f ( 0 ) \in \{ 0 , 2 \} $ by putting $ x = 0 $ in \eqref{1}. If $ f ( 0 ) = 2 $, comparing \eqref{1} and \eqref{2}, we get that $ f $ is the constant function with the value $ 2 $, which is one of the mentioned solutions. So, from now on, we consider the case $ f ( 0 ) = 0 $, which in turn makes both \eqref{1} and \eqref{2} equivalent to $$ f \left( x ^ 2 \right) = f ( x ) ^ 2 \tag 3 \label 3 $$ for all $ x \in \mathbb R $. Assume that there is some $ a \in \mathbb R \setminus \{ 0 \} $ with $ f ( a ) = 0 $. Without loss of generality, we can assume that $ a > 0 $; since by \eqref{3} we have $ f ( - a ) ^ 2 = f \left( ( - a ) ^ 2 \right) = f \left( a ^ 2 \right) = f ( a ) ^ 2 = 0 $, and we could take $ - a $ instead of $ a $. Letting $ x = a $ in \eqref{0} and using \eqref{3} we have $ f ( y + a ) ^ 2 = f ( y ) ^ 2 $ for all $ y \in \mathbb R $. Using this and induction, we get $ | f ( x + n a ) | = | f ( x ) | $ for all $ x \in \mathbb R $ and all $ n \in \mathbb Z $, and in particular $ f ( n a ) = 0 $ for all $ n \in \mathbb Z $. Given this, we can substitute $ x + n a $ for $ x $ and $ - x $ for $ y $ in \eqref{0} and use \eqref{3} to get $ f \left( ( x + n a ) ^ 2 - 2 x f ( x + n a ) \right) + f ( x ) ^ 2 = 0 $. Note that \eqref{3} shows that $ f $ takes nonnegative values at nonnegative points. Thus, for any $ x \in \mathbb R $, if we choose $ n $ large enough, namely $ n \ge \frac { \left| \sqrt { 2 | x f ( x ) | } - x \right| } a $, then both the terms on the left-hand side of the last equation will be nonnegative, and thus both of them must be equal to $ 0 $. This shows that in this case, $ f $ must be the constant function with the value $ 0 $, which is another one of the mentioned solutions. Lastly, consider the other case in which $ 0 $ is the only point at which $ f $ takes the value $ 0 $. Letting $ y = - \frac x 2 $ in \eqref{0} we have $ f \left( x ^ 2 - x f ( x ) \right) = 0 $, and hence $ x ^ 2 - x f ( x ) = 0 $, which gives $ f ( x ) = x $ for all $ x \in \mathbb R \setminus \{ 0 \} $. As we also had $ f ( 0 ) = 0 $, $ f $ must be the identity function, which is the remaining mentioned solution, and we're done.
Setting $y = 0$ yields $$f(x^2) + f(0) = f(x)^2$$
Setting $x = 0$ yields $$f(2f(0)\cdot y) + f(y^2) = f(y)^2$$
Set $x = y = t$ then subtract the two equations to get that
$$f(2f(0)\cdot t) = f(0)$$ for all $t$. This means that $f$ must be a constant function or $f(0) = 0$. Assuming it's a constant function, and plugging in $f(x) = c$ into the original functional equation yields $$2c = c^2$$ for which the only solutions are $c = 0$ and $c = 2$. So the solutions from these are $f(x) = 0$ and $f(x) = 2$
If $f(0) = 0$, then the condition from setting $y = 0$ is $$f(x^2) = f(x)^2$$ for which the only solutions are functions of the form $f(x) = x^c$. Set $y = -x$ and plug in $f(x) = x^c$ to get
$$(x^2-2x^{c+1})^c+x^{2c}=0$$
The only value of $c$ that this holds true for all $x$ is $c = 1$. So the only solution with $f(x)$ not constant is $f(x) = x$.
Thus the set of solutions is $$f(x) = \{ 0, 2, x \}$$