socrates wrote: Determine all functions $f:\Bbb{R}\to\Bbb{R}$ such that \[ f(x^2 + f(y)) = (f(x) + y^2)^ 2 \] , for all $x,y\in \Bbb{R}.$ Let $P(x,y)$ be the assertion $f(x^2+f(y))=(f(x)+y^2)^2$ Let $a=f(0)$ If $a<0$, then $P(0,\sqrt{-a})$ $\implies$ $f(f(\sqrt{-a}))=0$ and $P(0,f(\sqrt{-a}))$ $\implies$ $a=(a+f(\sqrt{-a})^2)^2>0$ and so contradiction. So $a\ge 0$ If $f(u)=f(v)$ subtracting $P(0,v)$ from $P(0,u)$ gives $(u^2-v^2)(u^2+v^2+2a)=0$ and so $|u|=|v|$ (since $a\ge 0$) Comparing $P(x,0)$ and $P(-x,0)$, we get $f(-x)=\pm f(x)$ $\forall x$ if $f(-x)=-f(x)$ for some $x$, then comparing $P(1,x)$ and $P(1,-x)$, we get $f(1+f(x))=f(1-f(x))$ and so $|1+f(x)|=|1-f(x)|$ and so $f(x)=0$ So $f(-x)=f(x)$ $\forall x$ If $x\ge f(y)$ for some $x,y$, $P(\sqrt{x-f(y)},y)$ $\implies$ $f(x)\ge 0$ So, if $f(x)<0$ for some $x$, $x<f(y)$ $\forall y$ and so $x<f(x)<0$ which is impossible since $f(x)=f(-x)$ and both $x,-x$ cant be negative; So $f(x)\ge 0$ $\forall x$ $P(0,0)$ $\implies$ $f(a)=a^2$ $P(\sqrt{f(x)},a)$ $\implies$ $f(f(x)+a^2)=(f(\sqrt{f(x)})+a^2)^2$ $P(a,x)$ $\implies$ $f(a^2+f(x))=(a^2+x^2)^2$ and so $f(\sqrt{f(x)})=x^2$ As a consequence, $f(x)$ is a bijection from $[0,+\infty)\to[0,+\infty)$ and $\exists u\ge 0$ such that $f(f(u))=0$ $P(0,u)$ $\implies$ $0=(u^2+a)^2$ and so $u=a=0$ Let then $g(x)=\sqrt{f(x)}$. We got that $g(x)$ is a bijective function from $[0,+\infty)\to[0,+\infty)$ with $g(0)=0$ $P(x,y)$ may be written $Q(x,y)$ : $g(x^2+g(y)^2)=g(x)^2+y^2$ $Q(x,0)$ $\implies$ $g(x^2)=g(x)^2$ and $Q(x,y)$ may be written $R(x,y)$ : $g(x+g(y))=g(x)+y$ $\forall x,y\ge 0$ This implies $g(g(y))=y$ and $g(x+y)=g(x)+g(y)$ And so $g(x)$ is an increasing solution (since $\ge 0$) of additive Cauchy equation and we get $g(x)=g(1)x$ Plugging this in $g(x^2)=g(x)^2$, we get $g(1)=1$ and $g(x)=x$ Hence the result : $\boxed{f(x)=x^2}$ $\forall x$ which indeed is a solution.

1) $f(x)\geq0,\forall x\geq0$. If there exists $f(y)<0$, $P(x,y)$ implies when $x\geq 0>f(y)$, we have $f(x)\geq0$. 2) If $f(d)=d^2$ for some $d\geq0$, $f(x)\geq d,\forall x\geq d$. $P(x,d)$ for $x\geq0$ implies $f(x)\geq d^4,\forall x\geq d^2$. Now $P(x,0)$ for $x\geq d$ implies $f^2(x)=f(x^2+f(0))\geq d^4$ or $f(x)\geq d^2,\forall x\geq d$. 3) $f$ is increasing when $x\geq0$. Suppose $f(x)<f(y)$ for some $x>y\geq0$. We have $a=f(y)+x^2>d=f(x)+x^2>b=f(x)+y^2\geq0$. Notice $f(a)=b^2<d^2=f(d)$, contradiction with 2) 4) $f(x)=x^2,\forall x\geq0$ Fix $x>y\geq0$. Let $a=f(y)+x^2$ and $b=f(x)+y^2$. We have $f(a)=b^2$ and $f(b)=a^2$. 3) implies $a=b$. So $f(x)-x^2$ is constant when $x\geq0$. Looking at $P(x,y)$ for $x,y\geq0$ we see that constant has to be zero. 5) $f(x)=x^2$ For any $y$, we can pick large $x>0$ such that $x^2+f(y)>0$. Then 4) implies $f(y)=y^2$.