Let the assertion $P(x,y)$ If $f(a) = f(b)$, $P(a,0)$ and $P(b,0)$ implies $|a| = |b|$ So $P(0,0)$ implies $f(f(0)^2) = f(0)$ which means $f(0)^2 = 0$ and $f(0) = 0$ Also, no other number can have $0$ as its function value $P(0,x)$ implies $f(x) = f(|x|)$ which also means that $f$ is even Also, $P(x,0)$ implies that $f(f(x)^2) = x^2$ so for all positive real $t$, $x$ such that $f(x) = t$ exists. Lets show that $f(x) \ge 0$ for $x \ge 0$ If $x_0 \ge 0$ such that $f(x_0) < 0$ exists, Let $y_0$ as a real such that $f(x_0) \ -y_0 ^2$ $P(y_0,x_0) : f(f(y_0)^2 + x_0) = y+0^2 + f(x_0) = 0$ so $x_0 +f(y_0)^2 = 0$ Hence, $x_0 < 0$ and this is contradiction Hence, $f(x) >0$ for all $x > 0$ For $x,y>0$, $t$ such that $f(t)^2 = x$ exists $P(t,y) : f(x + y) = f(x) + f(y)$ So $f(x+y) > f(x)$ and $f$ is strictly increasing over $\mathbb{R}^+$ Hence, $f(x) = f(1)x$ for all $x > 0$ and $f(1) = 1$ by substituting this Therefore, only possible solution is $f(x) = |x|$ for all $x$ which indeed fits the equation.

Seungjun_Lee wrote: Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that $$f(f(x)^2 + |y|) = x^2 + f(y)$$

Seungjun_Lee wrote: Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that $$f(f(x)^2 + |y|) = x^2 + f(y)$$ Let $P(x,y)$ be the given assertion. if $f(a) = f(b)$ then from $P(a,0)$ and $P(b,0)$ we see that $|a|=|b|$. So $P(0,0)$ gives us that $f(0) = 0$. $P(0,y)\implies f(y) = f(|y|)$ and $f$ is even. Let $g:\mathbb R_{\ge0}\to \mathbb R_{\ge0}$ be such that $g(x) = f(x)$. $P(x,0)\implies g(g(x)^2) = x^2$, thus $g$ is bijective. Let $Q(x,y) : g(g(x)^2 + y) = g(g(x)^2) + f(y)$. Let $x,y\ge 0$ let $g(t) = \sqrt{x}$. $Q(t,y)\implies g(x+y) = g(x) + g(y)$ Thus $g$ is additive and bounded so it is linear. $g(x) = g(1)x$. Substituting into $P$ we see that $g(x) = x$. So $\boxed{f(x) = |x|}$ is the only solution.